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college_math.Matrix_Theory_and_Linear_Algebra | exercise.6.2.6 | Recall that the desired matrix has $i^{\text {th }}$ column equal to $\operatorname{proj}_{\mathbf{u}}\left(\mathbf{e}_{i}\right)=\frac{\mathbf{u} \cdot \mathbf{e}_{i}}{\|\mathbf{u}\|^{2}} \mathbf{u}$. Therefore, the matrix is
$$
\frac{1}{14}\left[\begin{array}{rrr}
1 & -2 & 3 \\
-2 & 4 & -6 \\
3 & -6 & 9
\end{array}\right]
$$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the matrix for $T(\mathbf{w})=\operatorname{proj}_{\mathbf{v}}(\mathbf{w})$, where $\mathbf{v}=[1,-2,3]^{T}$. | $\boxed{Recall that the desired matrix has $i^{\text {th }}$ column equal to $\operatorname{proj}_{\mathbf{u}}\left(\mathbf{e}_{i}\right)=\frac{\mathbf{u} \cdot \mathbf{e}_{i}}{\|\mathbf{u}\|^{2}} \mathbf{u}$. Therefore, the matrix is
$$
\frac{1}{14}\left[\begin{array}{rrr}
1 & -2 & 3 \\
-2 & 4 & -6 \\
3 & -6 & 9
\end{array}\right]
}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.8.5.5 | $A=\left[\begin{array}{rrr}0 & 0 & 2 \\ -1 & 1 & 2 \\ -1 & 0 & 3\end{array}\right]$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $A=\left[\begin{array}{lll}-2 & 0 & 6 \\ -3 & 1 & 6 \\ -3 & 0 & 7\end{array}\right]$. Find a square root of $A$. | $\boxed{A=\left[\begin{array}{rrr}0 & 0 & 2 \\ -1 & 1 & 2 \\ -1 & 0 & 3\end{array}\right]}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.5.3.3 | Yes, this is a subspace. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $\mathbf{w}, \mathbf{v}$ be given vectors in $\mathbb{R}^{4}$ and define
$$
M=\left\{\mathbf{u}=\left[\begin{array}{l}
u_{1} \\
u_{2} \\
u_{3} \\
u_{4}
\end{array}\right] \in \mathbb{R}^{4} \mid \mathbf{w} \cdot \mathbf{u}=0 \text { and } \mathbf{v} \cdot \mathbf{u}=0\right\}
$$
Is $M$ a subspace of $\mathbb{R}^{4}$ ? Explain. | $\boxed{Yes, this is a subspace.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.7 | $h=4$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find $h$ such that
$$
\left[\begin{array}{ll|l}
2 & h & 4 \\
3 & 6 & 7
\end{array}\right]
$$
is the augmented matrix of an inconsistent system. | $\boxed{h=4$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.6.5 | $\cos \theta=\frac{-10}{\sqrt{1+4+1} \sqrt{1+4+49}}$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find $\cos \theta$ where $\theta$ is the angle between the vectors
$$
\mathbf{u}=\left[\begin{array}{r}
1 \\
-2 \\
1
\end{array}\right], \mathbf{v}=\left[\begin{array}{r}
1 \\
2 \\
-7
\end{array}\right]
$$ | $\boxed{\cos \theta=\frac{-10}{\sqrt{1+4+1} \sqrt{1+4+49}}}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.4.4.15 | $A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right], B=\left[\begin{array}{lll}1 & 2 & 0 \\ 3 & 4 & 0 \\ 0 & 0 & 0\end{array}\right]$. Then $A B=\left[\begin{array}{lll}3 & 4 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 0\end{array}\right]$ and $B A=\left[\begin{array}{lll}2 & 1 & 0 \\ 4 & 3 & 0 \\ 0 & 0 & 0\end{array}\right]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find $3 \times 3$-matrices $A$ and $B$ such that $A B \neq B A$. | $\boxed{A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right], B=\left[\begin{array}{lll}1 & 2 & 0 \\ 3 & 4 & 0 \\ 0 & 0 & 0\end{array}\right]$. Then $A B=\left[\begin{array}{lll}3 & 4 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 0\end{array}\right]$ and $B A=\left[\begin{array}{lll}2 & 1 & 0 \\ 4 & 3 & 0 \\ 0 & 0 & 0\end{array}\right]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.7.15 | It means that if you place them so that they all have their tails at the same point, the three will lie in the same plane. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | What does it mean geometrically if the box product of three vectors equals zero? | $\boxed{It means that if you place them so that they all have their tails at the same point, the three will lie in the same plane.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.24 | Yes. It has a unique solution. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Suppose the coefficient matrix of a system of $n$ equations with $n$ variables has the property that every column is a pivot column. Does it follow that the system of equations must have a solution? If so, must the solution be unique? Explain. | $\boxed{Yes. It has a unique solution.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.1.3 | We need $5 x-3 y=2 x-2 y$ and $4=2 y$. The unique solution is $x=\frac{2}{3}$ and $y=2$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find $x$ and $y$ so that $\mathbf{u}=[5 x-3 y, 4]^{T}$ and $\mathbf{v}=[2 x-2 y, 2 y]^{T}$ are equal in $\mathbb{R}^{2}$. | $\boxed{We need $5 x-3 y=2 x-2 y$ and $4=2 y$. The unique solution is $x=\frac{2}{3}$ and $y=2$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.9.4.6 | Yes, because the set of 5 linearly independent vectors can be extended to a basis $B$ of $V$. But since $V$ is 5-dimensional, $B$ has only 5 elements, which must be the original 5 vectors. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $V$ be a 5-dimensional vector space. If you have 5 linearly independent vectors in $V$, can you conclude that the vectors span $V$ ? | $\boxed{Yes, because the set of 5 linearly independent vectors can be extended to a basis $B$ of $V$. But since $V$ is 5-dimensional, $B$ has only 5 elements, which must be the original 5 vectors.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.4.1 | The best approximation is $\mathbf{v}^{\prime}=3 \mathbf{u}_{1}-\mathbf{u}_{2}=\left[\begin{array}{r}-2 \\ 5 \\ 2\end{array}\right]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider $\mathbb{R}^{3}$ with the usual dot product. Let
$$
\mathbf{u}_{1}=\left[\begin{array}{r}
-1 \\
1 \\
1
\end{array}\right], \quad \mathbf{u}_{2}=\left[\begin{array}{r}
-1 \\
-2 \\
1
\end{array}\right], \quad \text { and } \quad \mathbf{v}=\left[\begin{array}{r}
-1 \\
5 \\
3
\end{array}\right]
$$
Note that $\mathbf{u}_{1}$ and $\mathbf{u}_{2}$ are orthogonal. Find the best approximation of $\mathbf{v}$ in $\operatorname{span}\left\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\}$. | $\boxed{The best approximation is $\mathbf{v}^{\prime}=3 \mathbf{u}_{1}-\mathbf{u}_{2}=\left[\begin{array}{r}-2 \\ 5 \\ 2\end{array}\right]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.3.5 | $(x, y, z)=(4,1,1)$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Solve the following system of equations by back substitution.
$$
\begin{array}{r}
x+3 y-2 z=5 \\
y+3 z=4 \\
z=1 .
\end{array}
$$ | $\boxed{(x, y, z)=(4,1,1)$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.9 | Any $h$ will work. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find $h$ such that
$$
\left[\begin{array}{ll|r}
1 & 1 & 4 \\
3 & h & 12
\end{array}\right]
$$
is the augmented matrix of a consistent system. | $\boxed{Any $h$ will work.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.5.6 | Solution is: $[x=4, y=-4, z=-2]$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Use Gauss-Jordan elimination to solve the system of equations $-19 x+8 y=-108,-71 x+$ $30 y=-404,-2 x+y=-12,4 x+z=14$. | $\boxed{Solution is: $[x=4, y=-4, z=-2]}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.5.3 | $y=2+2 x$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the points $\left(x_{1}, y_{1}\right)=(-1,0),\left(x_{2}, y_{2}\right)=(0,3),\left(x_{3}, y_{3}\right)=(1,3),\left(x_{4}, y_{4}\right)=(2,5)$, $\left(x_{5}, y_{5}\right)=(3,9)$. Find the least squares line for these points. | $\boxed{y=2+2 x$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.3.2 | $\mathbf{u}_{1}=\left[\begin{array}{l}0 \\ 1 \\ 1 \\ 0\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{c}3 \\ -1 \\ 1 \\ -1\end{array}\right]$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | In $\mathbb{R}^{4}$ with the usual dot product, find an orthogonal basis for
$$
\operatorname{span}\left\{\left[\begin{array}{l}
0 \\
1 \\
1 \\
0
\end{array}\right],\left[\begin{array}{c}
3 \\
0 \\
2 \\
-1
\end{array}\right]\right\}
$$ | $\boxed{\mathbf{u}_{1}=\left[\begin{array}{l}0 \\ 1 \\ 1 \\ 0\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{c}3 \\ -1 \\ 1 \\ -1\end{array}\right]}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.13 | The solution is: $x=\frac{1}{3}-\frac{1}{3} t, y=\frac{2}{3}+\frac{2}{3} t, z=\frac{1}{3}, w=t$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Determine if the system is consistent. If so, is the solution unique?
$$
\begin{gathered}
x+2 y+z-w=2 \\
x-y+z+w=0 \\
2 x+y-z=1 \\
4 x+2 y+z=3
\end{gathered}
$$ | $\boxed{The solution is: $x=\frac{1}{3}-\frac{1}{3} t, y=\frac{2}{3}+\frac{2}{3} t, z=\frac{1}{3}, w=t$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.5.2.14 | From $a(\mathbf{u}+\mathbf{v})+b(\mathbf{u}+\mathbf{w})+c(\mathbf{w}+\mathbf{v})=\mathbf{0}$ we get $(a+b) \mathbf{u}+(a+c) \mathbf{v}+(b+c) \mathbf{w}=\mathbf{0}$. Since $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are linearly independent, this last system has only the trivial solution, so $a+b=0, a+c=0$, and $b+c=0$. Solving, we find the unique solution $(a, b, c)=(0,0,0)$. So the vectors $\mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{w}$, and $\mathbf{w}+\mathbf{v}$ are linearly independent. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $\mathbf{u}, \mathbf{v}, \mathbf{w}$ be linearly independent vectors in $\mathbb{R}^{n}$. Are the vectors $\mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{w}$, and $\mathbf{w}+\mathbf{v}$ linearly independent? | $\boxed{From $a(\mathbf{u}+\mathbf{v})+b(\mathbf{u}+\mathbf{w})+c(\mathbf{w}+\mathbf{v})=\mathbf{0}$ we get $(a+b) \mathbf{u}+(a+c) \mathbf{v}+(b+c) \mathbf{w}=\mathbf{0}$. Since $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are linearly independent, this last system has only the trivial solution, so $a+b=0, a+c=0$, and $b+c=0$. Solving, we find the unique solution $(a, b, c)=(0,0,0)$. So the vectors $\mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{w}$, and $\mathbf{w}+\mathbf{v}$ are linearly independent.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.21 | Solution is: $[x=1, y=2, z=-5]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Solve the system of equations $8 x+2 y+3 z=-3,8 x+3 y+3 z=-1$, and $4 x+y+3 z=-9$. | $\boxed{Solution is: $[x=1, y=2, z=-5]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.7.7 | Let $P=(1,2,3), Q=(2,3,4)$, and $R=(3,4,5) . \overrightarrow{P Q} \times \overrightarrow{P R}=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] \times\left[\begin{array}{l}2 \\ 2 \\ 2\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$. The area of the triangle is 0. It means the three points are on a line. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the area of the triangle determined by the three points, $(1,2,3),(2,3,4)$ and $(3,4,5)$. Did something interesting happen here? What does it mean geometrically? | $\boxed{Let $P=(1,2,3), Q=(2,3,4)$, and $R=(3,4,5) . \overrightarrow{P Q} \times \overrightarrow{P R}=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] \times\left[\begin{array}{l}2 \\ 2 \\ 2\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$. The area of the triangle is 0. It means the three points are on a line.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.6.1 | $\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 4\end{array}\right] \cdot\left[\begin{array}{l}2 \\ 0 \\ 1 \\ 3\end{array}\right]=17$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find $\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 4\end{array}\right] \cdot\left[\begin{array}{l}2 \\ 0 \\ 1 \\ 3\end{array}\right]$. | $\boxed{\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 4\end{array}\right] \cdot\left[\begin{array}{l}2 \\ 0 \\ 1 \\ 3\end{array}\right]=17}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.4 | There might be a solution. If so, there are infinitely many. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the following augmented matrix in which $*$ denotes an arbitrary number and $\mathbf{I}$ denotes a non-zero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?
$$
\left[\begin{array}{ccccc|c}
\mathbf{\square} & * & * & * & * & * \\
0 & \mathbf{\square} & * & * & 0 & * \\
0 & 0 & 0 & 0 & \mathbf{\square} & 0 \\
0 & 0 & 0 & 0 & * & \mathbf{\square}
\end{array}\right]
$$ | $\boxed{There might be a solution. If so, there are infinitely many.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.2.2 | $f_{1} \perp f_{2}, f_{1} \perp f_{4}, f_{2} \perp f_{3}$, and $f_{3} \perp f_{4}$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | On $C[-1,1]$, which of the following functions are orthogonal to each other?
$$
f_{1}(x)=x, \quad f_{2}(x)=x^{2}, \quad f_{3}(x)=x^{3}-x, \quad f_{4}(x)=1-x^{4} .
$$ | $\boxed{f_{1} \perp f_{2}, f_{1} \perp f_{4}, f_{2} \perp f_{3}$, and $f_{3} \perp f_{4}$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.5.1 | $(x, y, z)=(-1,-1,2)$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the least squares approximation for the system of equations
$$
\begin{array}{rr}
x+2 y+2 z= & 5, \\
x+y-z= & 11, \\
x+2 y-z= & -18 \\
2 x-y+2 z= & 0 .
\end{array}
$$ | $\boxed{(x, y, z)=(-1,-1,2)$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.5.2 | $(x, y, z)=(2,2,-1)$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the least squares approximation for the system of equations
$$
\left[\begin{array}{rrr}
-1 & 2 & 1 \\
-1 & 0 & -1 \\
2 & 0 & 2 \\
0 & 0 & 2 \\
1 & 2 & 2
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
1 \\
1 \\
3 \\
-2 \\
4
\end{array}\right]
$$ | $\boxed{(x, y, z)=(2,2,-1)$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.18 | Solution is: $[x=2-4 t, y=-8 t, z=t]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Solve the system of equations $3 x-y+4 z=6, y+8 z=0$, and $-2 x+y=-4$. | $\boxed{Solution is: $[x=2-4 t, y=-8 t, z=t]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.9.8 | The principal axes are $\frac{1}{\sqrt{3}}\left[\begin{array}{r}1 \\ -1 \\ -1\end{array}\right], \frac{1}{\sqrt{2}}\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right], \frac{1}{\sqrt{6}}\left[\begin{array}{r}1 \\ -1 \\ 2\end{array}\right]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the principal axes of the ellipsoid $2 x^{2}+2 y^{2}+3 z^{2}+2 x z-2 y z$. | $\boxed{The principal axes are $\frac{1}{\sqrt{3}}\left[\begin{array}{r}1 \\ -1 \\ -1\end{array}\right], \frac{1}{\sqrt{2}}\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right], \frac{1}{\sqrt{6}}\left[\begin{array}{r}1 \\ -1 \\ 2\end{array}\right]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.7.6.6 | No. It has non-zero determinant $\operatorname{det}(A)=\cos ^{2} t+\sin ^{2} t=1$ for all $t$, so it is invertible for all $t$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the matrix
$$
A=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos t & -\sin t \\
0 & \sin t & \cos t
\end{array}\right]
$$
Does there exist a value of t for which this matrix fails to be invertible? Explain. | $\boxed{No. It has non-zero determinant $\operatorname{det}(A)=\cos ^{2} t+\sin ^{2} t=1$ for all $t$, so it is invertible for all $t$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.8.4.4 | The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]\right\} \text { for eigenvalue }-3, \quad\left\{\left[\begin{array}{c}
-2 \\
1 \\
1
\end{array}\right]\right\} \text { for eigenvalue } 3, \quad\left\{\left[\begin{array}{c}
-1 \\
0 \\
1
\end{array}\right]\right\} \text { for eigenvalue }-2
$$
The matrix $P$ needed to diagonalize the above matrix is
$$
\left[\begin{array}{rrr}
0 & -2 & -1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{array}\right]
$$
and the diagonal matrix $D$ is
$$
\left[\begin{array}{rrr}
-3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & -2
\end{array}\right]
$$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the eigenvalues and eigenvectors of the matrix
$$
\left[\begin{array}{rrr}
8 & 0 & 10 \\
-6 & -3 & -6 \\
-5 & 0 & -7
\end{array}\right] \text {. }
$$
One eigenvalue is -3. Diagonalize if possible. | $\boxed{The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]\right\} \text { for eigenvalue }-3, \quad\left\{\left[\begin{array}{c}
-2 \\
1 \\
1
\end{array}\right]\right\} \text { for eigenvalue } 3, \quad\left\{\left[\begin{array}{c}
-1 \\
0 \\
1
\end{array}\right]\right\} \text { for eigenvalue }-2
$$
The matrix $P$ needed to diagonalize the above matrix is
$$
\left[\begin{array}{rrr}
0 & -2 & -1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{array}\right]
$$
and the diagonal matrix $D$ is
$$
\left[\begin{array}{rrr}
-3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & -2
\end{array}\right]
}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.4.4.14 | $A=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right], B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find $2 \times 2$-matrices $A$ and $B$ such that $A \neq 0$ and $B \neq 0$, but $A B=0$. | $\boxed{A=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right], B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.6.2.7 | $$
\frac{1}{35}\left[\begin{array}{rrr}
1 & 5 & 3 \\
5 & 25 & 15 \\
3 & 15 & 9
\end{array}\right]
$$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the matrix for $T(\mathbf{w})=\operatorname{proj}_{\mathbf{v}}(\mathbf{w})$, where $\mathbf{v}=[1,5,3]^{T}$. | $\boxed{
\frac{1}{35}\left[\begin{array}{rrr}
1 & 5 & 3 \\
5 & 25 & 15 \\
3 & 15 & 9
\end{array}\right]
}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.25 | The last column must not be a pivot column. The remaining columns must each be pivot columns. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Suppose there is a unique solution to a system of linear equations. What must be true of the pivot columns in the augmented matrix? | $\boxed{The last column must not be a pivot column. The remaining columns must each be pivot columns.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.4.5.20 | The matrix $A$ is right invertible. Two possible right inverses are
$$
\left[\begin{array}{cc}
1 & -2 \\
0 & 1 \\
0 & 0
\end{array}\right] \text { and }\left[\begin{array}{cc}
-2 & 1 \\
0 & 1 \\
1 & -1
\end{array}\right]
$$
The matrices $B$ and $C$ are not right invertible. The matrix $D$ is right invertible with inverse
$$
\left[\begin{array}{cc}
-3 & 2 \\
2 & -1
\end{array}\right]
$$
Since $D$ is square, its right inverse is actually an inverse, and therefore unique. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Which of the following matrices is right invertible? Find a right inverse if one exists. If possible, find two different right inverses.
$$
A=\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0
\end{array}\right] \quad B=\left[\begin{array}{lll}
1 & 1 & 2 \\
2 & 2 & 4
\end{array}\right] \quad C=\left[\begin{array}{ll}
1 & 2 \\
1 & 0 \\
0 & 1
\end{array}\right] \quad D=\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]
$$ | $\boxed{The matrix $A$ is right invertible. Two possible right inverses are
$$
\left[\begin{array}{cc}
1 & -2 \\
0 & 1 \\
0 & 0
\end{array}\right] \text { and }\left[\begin{array}{cc}
-2 & 1 \\
0 & 1 \\
1 & -1
\end{array}\right]
$$
The matrices $B$ and $C$ are not right invertible. The matrix $D$ is right invertible with inverse
$$
\left[\begin{array}{cc}
-3 & 2 \\
2 & -1
\end{array}\right]
$$
Since $D$ is square, its right inverse is actually an inverse, and therefore unique.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.4.2 | The best approximation is $\mathbf{v}^{\prime}=\mathbf{u}_{1}+2 \mathbf{u}_{2}-\mathbf{u}_{3}=\left[\begin{array}{r}1 \\ 3 \\ -2 \\ 4\end{array}\right]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider $\mathbb{R}^{4}$ with the usual dot product. Let
$$
\mathbf{u}_{1}=\left[\begin{array}{l}
0 \\
0 \\
1 \\
3
\end{array}\right], \quad \mathbf{u}_{2}=\left[\begin{array}{l}
1 \\
1 \\
0 \\
0
\end{array}\right], \quad \mathbf{u}_{3}=\left[\begin{array}{r}
1 \\
-1 \\
3 \\
-1
\end{array}\right], \quad \text { and } \quad \mathbf{v}=\left[\begin{array}{r}
6 \\
-2 \\
-5 \\
5
\end{array}\right]
$$
Note that $\mathbf{u}_{1}, \mathbf{u}_{2}$, and $\mathbf{u}_{3}$ are orthogonal. Find the best approximation of $\mathbf{v}$ in $\operatorname{span}\left\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\}$. | $\boxed{The best approximation is $\mathbf{v}^{\prime}=\mathbf{u}_{1}+2 \mathbf{u}_{2}-\mathbf{u}_{3}=\left[\begin{array}{r}1 \\ 3 \\ -2 \\ 4\end{array}\right]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.7.6.9 | $$
\operatorname{det}\left[\begin{array}{ccc}
e^{t} & e^{-t} \cos t & e^{-t} \sin t \\
e^{t} & -e^{-t} \cos t-e^{-t} \sin t & -e^{-t} \sin t+e^{-t} \cos t \\
e^{t} & 2 e^{-t} \sin t & -2 e^{-t} \cos t
\end{array}\right]=5 e^{-t} \neq 0
$$
and so this matrix is always invertible. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the matrix
$$
A=\left[\begin{array}{ccc}
e^{t} & e^{-t} \cos t & e^{-t} \sin t \\
e^{t} & -e^{-t} \cos t-e^{-t} \sin t & -e^{-t} \sin t+e^{-t} \cos t \\
e^{t} & 2 e^{-t} \sin t & -2 e^{-t} \cos t
\end{array}\right]
$$
Does there exist a value of t for which this matrix fails to be invertible? Explain. | $\boxed{
\operatorname{det}\left[\begin{array}{ccc}
e^{t} & e^{-t} \cos t & e^{-t} \sin t \\
e^{t} & -e^{-t} \cos t-e^{-t} \sin t & -e^{-t} \sin t+e^{-t} \cos t \\
e^{t} & 2 e^{-t} \sin t & -2 e^{-t} \cos t
\end{array}\right]=5 e^{-t} \neq 0
$$
and so this matrix is always invertible.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.30 | These are not legitimate row operations. They do not preserve the solution set of the system. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the system $-5 x+2 y-z=0$ and $-5 x-2 y-z=0$. Both equations equal zero and so $-5 x+2 y-z=-5 x-2 y-z$ which is equivalent to $y=0$. Does it follow that $x$ and $z$ can equal anything? Notice that when $x=1, z=-4$, and $y=0$ are plugged in to the equations, the equations do not equal 0 . Why? | $\boxed{These are not legitimate row operations. They do not preserve the solution set of the system.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.3.1 | $(x, y)=(1,0)$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Use elementary operations to solve the system of equations
$$
\begin{aligned}
& 3 x+y=3 \\
& x+2 y=1
\end{aligned}
$$ | $\boxed{(x, y)=(1,0)$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.9.3.2 | No. It is not closed under scalar multiplication. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the set of all vectors $\left[\begin{array}{l}x \\ y\end{array}\right] \in \mathbb{R}^{2}$ such that $x+y \geq 0$. Is this a subspace of $\mathbb{R}^{2}$ ? | $\boxed{No. It is not closed under scalar multiplication.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.7.6.11 | $$
\left[\begin{array}{ccc}
e^{t} & \cos t & \sin t \\
e^{t} & -\sin t & \cos t \\
e^{t} & -\cos t & -\sin t
\end{array}\right]^{-1}=\left[\begin{array}{ccc}
\frac{1}{2} e^{-t} & 0 & \frac{1}{2} e^{-t} \\
\frac{1}{2} \cos t+\frac{1}{2} \sin t & -\sin t & \frac{1}{2} \sin t-\frac{1}{2} \cos t \\
\frac{1}{2} \sin t-\frac{1}{2} \cos t & \cos t & -\frac{1}{2} \cos t-\frac{1}{2} \sin t
\end{array}\right]
$$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the inverse, if it exists, of the matrix
$$
A=\left[\begin{array}{ccc}
e^{t} & \cos t & \sin t \\
e^{t} & -\sin t & \cos t \\
e^{t} & -\cos t & -\sin t
\end{array}\right]
$$ | $\boxed{
\left[\begin{array}{ccc}
e^{t} & \cos t & \sin t \\
e^{t} & -\sin t & \cos t \\
e^{t} & -\cos t & -\sin t
\end{array}\right]^{-1}=\left[\begin{array}{ccc}
\frac{1}{2} e^{-t} & 0 & \frac{1}{2} e^{-t} \\
\frac{1}{2} \cos t+\frac{1}{2} \sin t & -\sin t & \frac{1}{2} \sin t-\frac{1}{2} \cos t \\
\frac{1}{2} \sin t-\frac{1}{2} \cos t & \cos t & -\frac{1}{2} \cos t-\frac{1}{2} \sin t
\end{array}\right]
}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.5.9 | $$
\frac{1}{\|\mathbf{u}\|} \mathbf{u}=\frac{1}{\sqrt{5}}\left[\begin{array}{l}
1 \\
2
\end{array}\right], \quad \frac{1}{\|\mathbf{v}\|} \mathbf{v}=\frac{1}{\sqrt{17}}\left[\begin{array}{r}
-2 \\
3 \\
2
\end{array}\right], \quad \frac{1}{\|\mathbf{w}\|} \mathbf{w}=\frac{1}{6}\left[\begin{array}{r}
5 \\
-3 \\
1 \\
-1
\end{array}\right]
$$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Normalize the following vectors.
$$
\mathbf{u}=\left[\begin{array}{l}
1 \\
2
\end{array}\right], \quad \mathbf{v}=\left[\begin{array}{r}
-2 \\
3 \\
2
\end{array}\right], \quad \mathbf{w}=\left[\begin{array}{r}
5 \\
-3 \\
1 \\
-1
\end{array}\right]
$$ | $\boxed{
\frac{1}{\|\mathbf{u}\|} \mathbf{u}=\frac{1}{\sqrt{5}}\left[\begin{array}{l}
1 \\
2
\end{array}\right], \quad \frac{1}{\|\mathbf{v}\|} \mathbf{v}=\frac{1}{\sqrt{17}}\left[\begin{array}{r}
-2 \\
3 \\
2
\end{array}\right], \quad \frac{1}{\|\mathbf{w}\|} \mathbf{w}=\frac{1}{6}\left[\begin{array}{r}
5 \\
-3 \\
1 \\
-1
\end{array}\right]
}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.9.1.13 | Let $f(i)$ be the $i^{\text {th }}$ component of a vector $\mathbf{x} \in \mathbb{R}^{n}$. Thus a typical element in $\mathbb{R}^{n}$ is $(f(1), \ldots, f(n))$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $X=\{1,2, \ldots, n\}$, and consider the space Func $_{X, \mathbb{R}}$ of real-valued functions defined on $X$. Explain how Func $_{X, \mathbb{R}}$ can be considered as $\mathbb{R}^{n}$. | $\boxed{Let $f(i)$ be the $i^{\text {th }}$ component of a vector $\mathbf{x} \in \mathbb{R}^{n}$. Thus a typical element in $\mathbb{R}^{n}$ is $(f(1), \ldots, f(n))$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.4.3 | The system
$$
\left[\begin{array}{l}
4 \\
4 \\
4
\end{array}\right]=a_{1}\left[\begin{array}{r}
3 \\
1 \\
-1
\end{array}\right]+a_{2}\left[\begin{array}{r}
8 \\
0 \\
-1
\end{array}\right]+a_{3}\left[\begin{array}{r}
2 \\
-2 \\
1
\end{array}\right]
$$
has no solution. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Decide whether
$$
\mathbf{v}=\left[\begin{array}{l}
4 \\
4 \\
4
\end{array}\right]
$$
is a linear combination of the vectors
$$
\mathbf{u}_{1}=\left[\begin{array}{r}
3 \\
1 \\
-1
\end{array}\right], \quad \mathbf{u}_{2}=\left[\begin{array}{r}
8 \\
0 \\
-1
\end{array}\right] \quad \text { and } \quad \mathbf{u}_{3}=\left[\begin{array}{r}
2 \\
-2 \\
1
\end{array}\right]
$$
If yes, find the coefficients. | $\boxed{The system
$$
\left[\begin{array}{l}
4 \\
4 \\
4
\end{array}\right]=a_{1}\left[\begin{array}{r}
3 \\
1 \\
-1
\end{array}\right]+a_{2}\left[\begin{array}{r}
8 \\
0 \\
-1
\end{array}\right]+a_{3}\left[\begin{array}{r}
2 \\
-2 \\
1
\end{array}\right]
$$
has no solution.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.8.2.8 | Yes. $\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]$ works. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Is it possible for a non-zero matrix to have only 0 as an eigenvalue? | $\boxed{Yes. $\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]$ works.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.4.7.1 | $X^{T} Y=\left[\begin{array}{rrr}0 & -1 & -2 \\ 0 & -1 & -2 \\ 0 & 1 & 2\end{array}\right], X Y^{T}=1$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $X=\left[\begin{array}{lll}-1 & -1 & 1\end{array}\right]$ and $Y=\left[\begin{array}{lll}0 & 1 & 2\end{array}\right]$. Find $X^{T} Y$ and $X Y^{T}$ if possible. | $\boxed{X^{T} Y=\left[\begin{array}{rrr}0 & -1 & -2 \\ 0 & -1 & -2 \\ 0 & 1 & 2\end{array}\right], X Y^{T}=1}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.4.4.13 | $A=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right], B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right], C=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find $2 \times 2$-matrices $A, B$, and $C$ such that $A \neq 0, C \neq B$, but $A C=A B$. | $\boxed{A=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right], B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right], C=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.4.9.1 | Ciphertext: "ZRUJPZVAEJTWOXGJZV". | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Encrypt the message "Rendezvous at dawn" using the Hill cipher with block size 3 and encryption matrix
$$
A=\left[\begin{array}{lll}
2 & 1 & 1 \\
1 & 3 & 1 \\
1 & 1 & 4
\end{array}\right]
$$ | $\boxed{Ciphertext: "ZRUJPZVAEJTWOXGJZV".}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.5.7 | The rank of the coefficient matrix is 3 , so both systems have a unique solution. The solution of the first system is $(x, y, z)=(1,0,1)$, and the solution of the second system is $(x, y, z)=(1,1,2)$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Solve the following two systems of equations simultaneously, by using a single augmented matrix with two constant vectors.
$$
\begin{array}{rrr}
x+2 y-z=0 & x+2 y-z=1 \\
2 x+3 y+z=3 & 2 x+3 y+z=7 \\
x-y+2 z=3 & x-y+2 z=4
\end{array}
$$ | $\boxed{The rank of the coefficient matrix is 3 , so both systems have a unique solution. The solution of the first system is $(x, y, z)=(1,0,1)$, and the solution of the second system is $(x, y, z)=(1,1,2)$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.7.14 | Yes. It will involve the sum of a product of integers and so it will be an integer. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Suppose $\mathbf{u}, \mathbf{v}$, and $\mathbf{w}$ are three vectors whose components are all integers. Can you conclude the volume of the parallelepiped determined from these three vectors will always be an integer? | $\boxed{Yes. It will involve the sum of a product of integers and so it will be an integer.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.7.6.7 | $\operatorname{det}(A)=\left|\begin{array}{ccc}1 & t & t^{2} \\ 0 & 1 & 2 t \\ t & 0 & 2\end{array}\right|=t^{3}+2$, and so $A$ has no inverse when $t=-\sqrt[3]{2}$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the matrix
$$
A=\left[\begin{array}{rrr}
1 & t & t^{2} \\
0 & 1 & 2 t \\
t & 0 & 2
\end{array}\right]
$$
Does there exist a value of t for which this matrix fails to be invertible? Explain. | $\boxed{\operatorname{det}(A)=\left|\begin{array}{ccc}1 & t & t^{2} \\ 0 & 1 & 2 t \\ t & 0 & 2\end{array}\right|=t^{3}+2$, and so $A$ has no inverse when $t=-\sqrt[3]{2}$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.4.1.2 | $x=2, y=-1, z=2$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find scalars $x, y, z$ such that the following two matrices are equal.
$$
\left[\begin{array}{rr}
x & -1 \\
2 & 4
\end{array}\right] \text { and }\left[\begin{array}{ll}
2 & y \\
z & 4
\end{array}\right]
$$ | $\boxed{x=2, y=-1, z=2$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.7.3 | $\left[\begin{array}{l}1 \\ 1 \\ 3\end{array}\right] \times\left[\begin{array}{r}-7 \\ -2 \\ 2\end{array}\right]=\left[\begin{array}{r}8 \\ -23 \\ 5\end{array}\right]$. The area of the parallelogram is $8 \sqrt{3}$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the area of the parallelogram determined by the vectors $\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right],\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right]$. | $\boxed{\left[\begin{array}{l}1 \\ 1 \\ 3\end{array}\right] \times\left[\begin{array}{r}-7 \\ -2 \\ 2\end{array}\right]=\left[\begin{array}{r}8 \\ -23 \\ 5\end{array}\right]$. The area of the parallelogram is $8 \sqrt{3}$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.10.4.8 | We have $T\left(\mathbf{v}_{1}\right)=\mathbf{v}_{1}, T\left(\mathbf{v}_{2}\right)=\mathbf{0}$, and $T\left(\mathbf{v}_{3}\right)=\mathbf{0}$. Therefore
$$
[T]_{B, B}=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] .
$$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $\mathbf{v}=\left[\begin{array}{r}1 \\ -2 \\ 3\end{array}\right]$ and consider the linear function $T(\mathbf{w})=\operatorname{proj}_{\mathbf{v}}(\mathbf{w})$. Find the matrix of $T$ with respect to the basis
$$
B=\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\}=\left\{\left[\begin{array}{r}
1 \\
-2 \\
3
\end{array}\right],\left[\begin{array}{l}
2 \\
1 \\
0
\end{array}\right],\left[\begin{array}{l}
3 \\
0 \\
1
\end{array}\right]\right\}
$$ | $\boxed{We have $T\left(\mathbf{v}_{1}\right)=\mathbf{v}_{1}, T\left(\mathbf{v}_{2}\right)=\mathbf{0}$, and $T\left(\mathbf{v}_{3}\right)=\mathbf{0}$. Therefore
$$
[T]_{B, B}=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] .
}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.10.4.2 | $[\mathbf{x}]_{B}=\left[\begin{array}{r}2 \\ 1 \\ -1\end{array}\right]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $B=\left\{\left[\begin{array}{r}1 \\ -1 \\ 2\end{array}\right],\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right],\left[\begin{array}{r}-1 \\ 0 \\ 2\end{array}\right]\right\}$ be a basis of $\mathbb{R}^{3}$ and let $\mathbf{x}=\left[\begin{array}{r}5 \\ -1 \\ 4\end{array}\right]$ be a vector in $\mathbb{R}^{2}$. Find $[\mathbf{x}]_{B}$. | $\boxed{[\mathbf{x}]_{B}=\left[\begin{array}{r}2 \\ 1 \\ -1\end{array}\right]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.7.7.5 | By Cramer's rule, we have
$$
y=\frac{\left|\begin{array}{ccc}
1 & t & 1 \\
1 & s & t^{2} \\
1 & 1 & s^{2}
\end{array}\right|}{\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & t & t^{2} \\
1 & s & s^{2}
\end{array}\right|}=\frac{s^{3}+t^{3}+1-2-t s^{2}-t^{2}}{t s^{2}+t^{2}+s-t-s t^{2}-s^{2}} .
$$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the value of $y$ in the following system of equations:
$$
\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & t & t^{2} \\
1 & s & s^{2}
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
t \\
s \\
1
\end{array}\right] .
$$ | $\boxed{By Cramer's rule, we have
$$
y=\frac{\left|\begin{array}{ccc}
1 & t & 1 \\
1 & s & t^{2} \\
1 & 1 & s^{2}
\end{array}\right|}{\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & t & t^{2} \\
1 & s & s^{2}
\end{array}\right|}=\frac{s^{3}+t^{3}+1-2-t s^{2}-t^{2}}{t s^{2}+t^{2}+s-t-s t^{2}-s^{2}} .
}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.9.3.8 | This is not a subspace. $\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ is in it, but $5\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ is not. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $U=\left\{[x, y, z]^{T} \in \mathbb{R}^{3}|| x \mid \leq 4\right\}$. Is $U$ a subspace of $\mathbb{R}^{3}$ ? | $\boxed{This is not a subspace. $\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ is in it, but $5\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ is not.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.2 | A solution exists and is unique. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the following augmented matrix in which $*$ denotes an arbitrary number and denotes a non-zero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?
$$
\left[\begin{array}{ccc|c}
\mathbf{\square} & * & * & * \\
0 & \mathbf{0} & * & * \\
0 & 0 & \mathbf{\square} & *
\end{array}\right]
$$ | $\boxed{A solution exists and is unique.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.5.2 | $\sqrt{27}$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Find the distance between the points $P=(1,3,-1,0)$ and $Q=(2,2,3,3)$ in $\mathbb{R}^{4}$. | $\boxed{\sqrt{27}$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.3.2.2 | We have
$$
\left[\begin{array}{l}
x \\
y \\
z \\
w
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
0 \\
0
\end{array}\right]+\left(1-r_{1}\right)\left[\begin{array}{l}
1 \\
0 \\
0 \\
1
\end{array}\right]+\left(r_{1}+r_{2}\right)\left[\begin{array}{r}
-1 \\
-1 \\
1 \\
0
\end{array}\right]=\left[\begin{array}{l}
2 \\
2 \\
0 \\
1
\end{array}\right]+r_{1}\left[\begin{array}{r}
-2 \\
-1 \\
1 \\
-1
\end{array}\right]+r_{2}\left[\begin{array}{r}
-1 \\
-1 \\
1 \\
0
\end{array}\right] .
$$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the following vector equation for a plane in $\mathbb{R}^{4}$ :
$$
\left[\begin{array}{l}
x \\
y \\
z \\
w
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
0 \\
0
\end{array}\right]+t\left[\begin{array}{l}
1 \\
0 \\
0 \\
1
\end{array}\right]+s\left[\begin{array}{r}
-1 \\
-1 \\
1 \\
0
\end{array}\right] .
$$
Find a new vector equation for the same plane by doing the change of parameters $t=1-r_{1}, s=r_{1}+r_{2}$. | $\boxed{We have
$$
\left[\begin{array}{l}
x \\
y \\
z \\
w
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
0 \\
0
\end{array}\right]+\left(1-r_{1}\right)\left[\begin{array}{l}
1 \\
0 \\
0 \\
1
\end{array}\right]+\left(r_{1}+r_{2}\right)\left[\begin{array}{r}
-1 \\
-1 \\
1 \\
0
\end{array}\right]=\left[\begin{array}{l}
2 \\
2 \\
0 \\
1
\end{array}\right]+r_{1}\left[\begin{array}{r}
-2 \\
-1 \\
1 \\
-1
\end{array}\right]+r_{2}\left[\begin{array}{r}
-1 \\
-1 \\
1 \\
0
\end{array}\right] .
}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.17 | Solution is: $[x=1-2 t, z=1, y=t]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Solve the system of equations $7 x+14 y+15 z=22,2 x+4 y+3 z=5$, and $3 x+6 y+10 z=$ 13. | $\boxed{Solution is: $[x=1-2 t, z=1, y=t]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.2.7.8 | $(\mathbf{i} \times \mathbf{j}) \times \mathbf{j}=\mathbf{k} \times \mathbf{j}=-\mathbf{i}$. However, $\mathbf{i} \times(\mathbf{j} \times \mathbf{j})=\mathbf{0}$ and so the cross product is not associative. The expression $\mathbf{u} \times \mathbf{v} \times \mathbf{w}$ has no meaning. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Is $\mathbf{u} \times(\mathbf{v} \times \mathbf{w})=(\mathbf{u} \times \mathbf{v}) \times \mathbf{w}$ ? What is the meaning of $\mathbf{u} \times \mathbf{v} \times \mathbf{w}$ ? Explain. Hint: Try $(\mathbf{i} \times \mathbf{j}) \times \mathbf{j}$. | $\boxed{(\mathbf{i} \times \mathbf{j}) \times \mathbf{j}=\mathbf{k} \times \mathbf{j}=-\mathbf{i}$. However, $\mathbf{i} \times(\mathbf{j} \times \mathbf{j})=\mathbf{0}$ and so the cross product is not associative. The expression $\mathbf{u} \times \mathbf{v} \times \mathbf{w}$ has no meaning.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.8.5.4 | $B=\left[\begin{array}{rr}-1 & -2 \\ 3 & 4\end{array}\right]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $A=\left[\begin{array}{rr}-5 & -6 \\ 9 & 10\end{array}\right]$. Find a square root of $A$, i.e., find a matrix $B$ such that $B^{2}=A$. | $\boxed{B=\left[\begin{array}{rr}-1 & -2 \\ 3 & 4\end{array}\right]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.4.5.2 | Yes $B=C$. Multiply $A B=A C$ on the left by $A^{-1}$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Suppose $A B=A C$ and $A$ is an invertible $n \times n$-matrix. Does it follow that $B=C$ ? Explain why or why not. | $\boxed{Yes $B=C$. Multiply $A B=A C$ on the left by $A^{-1}$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.19 | Solution is: $[x=-1, y=2, z=-1]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Solve the system of equations $9 x-2 y+4 z=-17,13 x-3 y+6 z=-25$, and $-2 x-z=3$. | $\boxed{Solution is: $[x=-1, y=2, z=-1]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.9.3.3 | No. It is not closed under addition. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the set of all vectors $\left[\begin{array}{l}x \\ y\end{array}\right] \in \mathbb{R}^{2}$ such that $x y=0$. Is this a subspace of $\mathbb{R}^{2}$ ? | $\boxed{No. It is not closed under addition.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.23 | No. This would lead to $0=1$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Suppose a system of linear equations has an augmented matrix with 2 rows and 4 columns and the last column is a pivot column. Could the system of linear equations be consistent? Explain. | $\boxed{No. This would lead to $0=1$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.10.4.7 | Recall that $\operatorname{proj}_{\mathbf{v}}(\mathbf{w})=\frac{\mathbf{v}^{\bullet} \mathbf{w}}{\|\mathbf{v}\|^{2}} \mathbf{v}$. The desired matrix has $i^{\text {th }}$ column equal to $\operatorname{proj}_{\mathbf{v}}\left(\mathbf{e}_{i}\right)$. Therefore, the desired matrix is
$$
\frac{1}{14}\left[\begin{array}{rrr}
1 & -2 & 3 \\
-2 & 4 & -6 \\
3 & -6 & 9
\end{array}\right] .
$$ | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $\mathbf{v}=\left[\begin{array}{r}1 \\ -2 \\ 3\end{array}\right]$ and consider the linear function $T(\mathbf{w})=\operatorname{proj}_{\mathbf{v}}(\mathbf{w})$. Find the matrix of $T$ with respect to the standard basis of $\mathbb{R}^{3}$. | $\boxed{Recall that $\operatorname{proj}_{\mathbf{v}}(\mathbf{w})=\frac{\mathbf{v}^{\bullet} \mathbf{w}}{\|\mathbf{v}\|^{2}} \mathbf{v}$. The desired matrix has $i^{\text {th }}$ column equal to $\operatorname{proj}_{\mathbf{v}}\left(\mathbf{e}_{i}\right)$. Therefore, the desired matrix is
$$
\frac{1}{14}\left[\begin{array}{rrr}
1 & -2 & 3 \\
-2 & 4 & -6 \\
3 & -6 & 9
\end{array}\right] .
}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.10.9 | Orthogonal: $[0, i, 2]^{T},[1,2, i]^{T}$. Orthonormal: $\frac{1}{\sqrt{5}}[0, i, 2]^{T}, \frac{1}{\sqrt{6}}[1,2, i]^{T}$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider $\mathbb{C}^{3}$ with the complex dot product. Let $\mathbf{v}_{1}=\left[\begin{array}{c}0 \\ i \\ 2\end{array}\right]$ and $\mathbf{v}_{2}=\left[\begin{array}{c}1 \\ 1+i \\ 3 i+2\end{array}\right]$. Use the Gram-Schmidt procedure to find an orthogonal basis for $\operatorname{span}\left\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\}$. Then find an orthonormal basis. | $\boxed{Orthogonal: $[0, i, 2]^{T},[1,2, i]^{T}$. Orthonormal: $\frac{1}{\sqrt{5}}[0, i, 2]^{T}, \frac{1}{\sqrt{6}}[1,2, i]^{T}$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.11.2.6 | We have $\mathbf{v}=a_{1} \mathbf{u}_{1}+a_{2} \mathbf{u}_{2}+a_{3} \mathbf{u}_{3}$ where $a_{1}=\frac{\left\langle\mathbf{u}_{1}, \mathbf{v}\right\rangle}{\left\langle\mathbf{u}_{1}, \mathbf{u}_{1}\right\rangle}=\frac{1}{2}$. So the first coordinate is $\frac{1}{2}$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Suppose $B=\left\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\}$ is an orthogonal basis of $\mathbb{R}^{3}$. We have been told that
$$
\mathbf{u}_{1}=\left[\begin{array}{l}
1 \\
1 \\
0
\end{array}\right],
$$
but it is not known what $\mathbf{u}_{2}$ and $\mathbf{u}_{3}$ are. Find the first coordinate of the vector
$$
\mathbf{v}=\left[\begin{array}{l}
1 \\
0 \\
2
\end{array}\right]
$$
with respect to the basis $B$. | $\boxed{We have $\mathbf{v}=a_{1} \mathbf{u}_{1}+a_{2} \mathbf{u}_{2}+a_{3} \mathbf{u}_{3}$ where $a_{1}=\frac{\left\langle\mathbf{u}_{1}, \mathbf{v}\right\rangle}{\left\langle\mathbf{u}_{1}, \mathbf{u}_{1}\right\rangle}=\frac{1}{2}$. So the first coordinate is $\frac{1}{2}$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.5.1.2 | It is the plane $2 x+3 y-z=0$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Describe the span of the vectors $\mathbf{u}=\left[\begin{array}{l}1 \\ 0 \\ 2\end{array}\right]$ and $\mathbf{v}=\left[\begin{array}{r}2 \\ -1 \\ 1\end{array}\right]$ in $\mathbb{R}^{3}$. | $\boxed{It is the plane $2 x+3 y-z=0$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.1 | The solution exists but is not unique. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Consider the following augmented matrix in which $*$ denotes an arbitrary number and denotes a non-zero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?
$$
\left[\begin{array}{lllll|l}
\mathbf{\square} & * & * & * & * & * \\
0 & \mathbf{\square} & * & * & 0 & * \\
0 & 0 & \mathbf{\square} & * & * & * \\
0 & 0 & 0 & 0 & \mathbf{\square} & *
\end{array}\right]
$$ | $\boxed{The solution exists but is not unique.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.1.4.20 | Solution is: $[x=2, y=4, z=5]$. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Solve the system of equations $65 x+84 y+16 z=546,81 x+105 y+20 z=682$, and $84 x+110 y+21 z=713$. | $\boxed{Solution is: $[x=2, y=4, z=5]$.}$ |
college_math.Matrix_Theory_and_Linear_Algebra | exercise.4.4.10 | There is no possible choice of $k$ which will make these matrices commute. | Creative Commons Attribution License (CC BY) | college_math.linear_algebra | Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 2 \\ 1 & k\end{array}\right]$. Is it possible to choose $k$ such that $A B=B A$? If so, what should $k$ equal? | $\boxed{There is no possible choice of $k$ which will make these matrices commute.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.1.2.44 | The last column must not be a pivot column. The remaining columns must each be pivot columns. | Creative Commons License (CC BY) | college_math.linear_algebra | Suppose there is a unique solution to a system of linear equations. What must be true of the pivot columns in the augmented matrix? | $\boxed{The last column must not be a pivot column. The remaining columns must each be pivot columns.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.4.12.23 | $$
\begin{aligned}
\vec{F}_{1} \bullet\left[\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
0
\end{array}\right] 10+\vec{F}_{2} \bullet\left[\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
0
\end{array}\right] 10 & =\left(\vec{F}_{1}+\vec{F}_{2}\right) \bullet\left[\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
0
\end{array}\right] 10 \\
& =\left[\begin{array}{r}
6 \\
4 \\
-4
\end{array}\right] \bullet\left[\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
0
\end{array}\right] 10 \\
& =50 \sqrt{2}
\end{aligned}
$$ | Creative Commons License (CC BY) | college_math.linear_algebra | An object moves 10 meters in the direction of $\vec{j}+\vec{i}$. There are two forces acting on this object, $\vec{F}_{1}=\vec{i}+2 \vec{j}+2 \vec{k}$, and $\vec{F}_{2}=5 \vec{i}+2 \vec{j}-6 \vec{k}$. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why? | $\boxed{
\begin{aligned}
\vec{F}_{1} \bullet\left[\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
0
\end{array}\right] 10+\vec{F}_{2} \bullet\left[\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
0
\end{array}\right] 10 & =\left(\vec{F}_{1}+\vec{F}_{2}\right) \bullet\left[\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
0
\end{array}\right] 10 \\
& =\left[\begin{array}{r}
6 \\
4 \\
-4
\end{array}\right] \bullet\left[\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \\
0
\end{array}\right] 10 \\
& =50 \sqrt{2}
\end{aligned}
}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.9.9.13 | $$
\frac{1}{10}\left[\begin{array}{lll}
1 & 0 & 3 \\
0 & 0 & 0 \\
3 & 0 & 9
\end{array}\right]
$$ | Creative Commons License (CC BY) | college_math.linear_algebra | Find the matrix for $T(\vec{w})=\operatorname{proj}_{\vec{v}}(\vec{w})$ where $\vec{v}=\left[\begin{array}{lll}1 & 0 & 3\end{array}\right]^{T}$. | $\boxed{
\frac{1}{10}\left[\begin{array}{lll}
1 & 0 & 3 \\
0 & 0 & 0 \\
3 & 0 & 9
\end{array}\right]
}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.1.2.18 | Solution is: $\left[w=\frac{3}{2} y-1, x=\frac{2}{3}-\frac{1}{2} y, z=\frac{1}{3}\right]$ | Creative Commons License (CC BY) | college_math.linear_algebra | Determine if the system is consistent. If so, is the solution unique?
$$
\begin{gathered}
x+2 y+z-w=2 \\
x-y+z+w=0 \\
2 x+y-z=1 \\
4 x+2 y+z=3
\end{gathered}
$$ | $\boxed{Solution is: $\left[w=\frac{3}{2} y-1, x=\frac{2}{3}-\frac{1}{2} y, z=\frac{1}{3}\right]}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.9.4.1 | This is not a subspace. $\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 1\end{array}\right]$ is in it, but $20\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 1\end{array}\right]$ is not. | Creative Commons License (CC BY) | college_math.linear_algebra | Let $M=\left\{\vec{u}=\left(u_{1}, u_{2}, u_{3}, u_{4}\right) \in \mathbb{R}^{4}:\left|u_{1}\right| \leq 4\right\}$. Is $M$ a subspace of $\mathbb{R}^{4}$ ? | $\boxed{This is not a subspace. $\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 1\end{array}\right]$ is in it, but $20\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 1\end{array}\right]$ is not.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.1.2.37 | Solution is: $[x=-1, y=-5, z=4]$ | Creative Commons License (CC BY) | college_math.linear_algebra | Find the solution to the system of equations, $-8 x+2 y+5 z=18,-8 x+3 y+5 z=13$, and $-4 x+y+5 z=19$. | $\boxed{Solution is: $[x=-1, y=-5, z=4]}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.4.9.2 | $\left[\begin{array}{r}3 \\ 0 \\ -3\end{array}\right] \times\left[\begin{array}{r}-4 \\ 0 \\ -2\end{array}\right]=\left[\begin{array}{r}0 \\ 18 \\ 0\end{array}\right]$. So the area is 9. | Creative Commons License (CC BY) | college_math.linear_algebra | Find the area of the triangle determined by the three points, $(1,2,3),(4,2,0)$ and $(-3,2,1)$. | $\boxed{\left[\begin{array}{r}3 \\ 0 \\ -3\end{array}\right] \times\left[\begin{array}{r}-4 \\ 0 \\ -2\end{array}\right]=\left[\begin{array}{r}0 \\ 18 \\ 0\end{array}\right]$. So the area is 9.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.5.9.13 | Solution is:
$$
\left[\begin{array}{c}
-\frac{1}{2} s-\frac{1}{2} t \\
\frac{1}{2} s-\frac{1}{2} t \\
s \\
t
\end{array}\right]
$$
for $s, t \in \mathbb{R}$. A basis is
$$
\left\{\left[\begin{array}{r}
-1 \\
1 \\
2 \\
0
\end{array}\right],\left[\begin{array}{c}
-1 \\
1 \\
0 \\
1
\end{array}\right]\right\}
$$ | Creative Commons License (CC BY) | college_math.linear_algebra | Write the solution set of the following system as a linear combination of vectors
$$
\left[\begin{array}{rrrr}
1 & 1 & 0 & 1 \\
1 & -1 & 1 & 0 \\
3 & 1 & 1 & 2 \\
3 & 3 & 0 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z \\
w
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0
\end{array}\right]
$$ | $\boxed{Solution is:
$$
\left[\begin{array}{c}
-\frac{1}{2} s-\frac{1}{2} t \\
\frac{1}{2} s-\frac{1}{2} t \\
s \\
t
\end{array}\right]
$$
for $s, t \in \mathbb{R}$. A basis is
$$
\left\{\left[\begin{array}{r}
-1 \\
1 \\
2 \\
0
\end{array}\right],\left[\begin{array}{c}
-1 \\
1 \\
0 \\
1
\end{array}\right]\right\}
}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.5.2.11 | Recall that $\operatorname{proj}_{\vec{u}}(\vec{v})=\frac{\vec{v} \vec{u}}{\|\vec{u}\|^{2}} \vec{u}$ and so the desired matrix has $i^{t h}$ column equal to $\operatorname{proj}_{\vec{u}}\left(\vec{e}_{i}\right)$. Therefore, the matrix desired is
$$
\frac{1}{14}\left[\begin{array}{rrr}
1 & -2 & 3 \\
-2 & 4 & -6 \\
3 & -6 & 9
\end{array}\right]
$$ | Creative Commons License (CC BY) | college_math.linear_algebra | Find the matrix for $T(\vec{w})=\operatorname{proj}_{\vec{v}}(\vec{w})$ where $\vec{v}=\left[\begin{array}{lll}1 & -2 & 3\end{array}\right]^{T}$. | $\boxed{Recall that $\operatorname{proj}_{\vec{u}}(\vec{v})=\frac{\vec{v} \vec{u}}{\|\vec{u}\|^{2}} \vec{u}$ and so the desired matrix has $i^{t h}$ column equal to $\operatorname{proj}_{\vec{u}}\left(\vec{e}_{i}\right)$. Therefore, the matrix desired is
$$
\frac{1}{14}\left[\begin{array}{rrr}
1 & -2 & 3 \\
-2 & 4 & -6 \\
3 & -6 & 9
\end{array}\right]
}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.7.1.14 | Yes. $\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]$ works. | Creative Commons License (CC BY) | college_math.linear_algebra | Is it possible for a nonzero matrix to have only 0 as an eigenvalue? | $\boxed{Yes. $\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]$ works.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.1.2.5 | Solution is : $\{g=60, I=90, b=200, s=50\}$ | Creative Commons License (CC BY) | college_math.linear_algebra | Four times the weight of Gaston is 150 pounds more than the weight of Ichabod. Four times the weight of Ichabod is 660 pounds less than seventeen times the weight of Gaston. Four times the weight of Gaston plus the weight of Siegfried equals 290 pounds. Brunhilde would balance all three of the others. Find the weights of the four people. | $\boxed{Solution is : $\{g=60, I=90, b=200, s=50\}}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.9.9.11 | Recall that $\operatorname{proj}_{\vec{u}}(\vec{v})=\frac{\vec{v} \bullet \vec{u}}{\|\vec{u}\|^{2}} \vec{u}$ and so the desired matrix has $i^{t h}$ column equal to $\operatorname{proj}_{\vec{u}}\left(\vec{e}_{i}\right)$. Therefore, the matrix desired is
$$
\frac{1}{14}\left[\begin{array}{rrr}
1 & -2 & 3 \\
-2 & 4 & -6 \\
3 & -6 & 9
\end{array}\right]
$$ | Creative Commons License (CC BY) | college_math.linear_algebra | Find the matrix for $T(\vec{w})=\operatorname{proj}_{\vec{v}}(\vec{w})$ where $\vec{v}=\left[\begin{array}{lll}1 & -2 & 3\end{array}\right]^{T}$. | $\boxed{Recall that $\operatorname{proj}_{\vec{u}}(\vec{v})=\frac{\vec{v} \bullet \vec{u}}{\|\vec{u}\|^{2}} \vec{u}$ and so the desired matrix has $i^{t h}$ column equal to $\operatorname{proj}_{\vec{u}}\left(\vec{e}_{i}\right)$. Therefore, the matrix desired is
$$
\frac{1}{14}\left[\begin{array}{rrr}
1 & -2 & 3 \\
-2 & 4 & -6 \\
3 & -6 & 9
\end{array}\right]
}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.5.9.15 | Solution is: $\left[\begin{array}{r}-\hat{t} \\ \hat{t} \\ \hat{t} \\ 0\end{array}\right]$, a basis is $\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 0\end{array}\right]$. | Creative Commons License (CC BY) | college_math.linear_algebra | Write the solution set of the following system as a linear combination of vectors
$$
\left[\begin{array}{rrrr}
1 & 1 & 0 & 1 \\
2 & 1 & 1 & 2 \\
1 & 0 & 1 & 1 \\
0 & -1 & 1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z \\
w
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0
\end{array}\right]
$$ | $\boxed{Solution is: $\left[\begin{array}{r}-\hat{t} \\ \hat{t} \\ \hat{t} \\ 0\end{array}\right]$, a basis is $\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 0\end{array}\right]$.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.6.3.2 | Solution is: $i \sqrt{3}+1,1-i \sqrt{3},-2$ | Creative Commons License (CC BY) | college_math.linear_algebra | Find the complex cube roots of 8. | $\boxed{Solution is: $i \sqrt{3}+1,1-i \sqrt{3},-2}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.4.2.3 | $$
\left[\begin{array}{r}
4 \\
4 \\
-3
\end{array}\right]=2\left[\begin{array}{r}
3 \\
1 \\
-1
\end{array}\right]-\left[\begin{array}{r}
2 \\
-2 \\
1
\end{array}\right]
$$ | Creative Commons License (CC BY) | college_math.linear_algebra | Decide whether $\vec{v}=\left[\begin{array}{r}4 \\ 4 \\ -3\end{array}\right]$ is a linear combination of the vectors $\vec{u}_{1}=\left[\begin{array}{r}3 \\ 1 \\ -1\end{array}\right]$ and $\vec{u}_{2}=\left[\begin{array}{r}2 \\ -2 \\ 1\end{array}\right]$. | $\boxed{
\left[\begin{array}{r}
4 \\
4 \\
-3
\end{array}\right]=2\left[\begin{array}{r}
3 \\
1 \\
-1
\end{array}\right]-\left[\begin{array}{r}
2 \\
-2 \\
1
\end{array}\right]
}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.5.9.5 | Solution is: $\left[\begin{array}{r}-4 \hat{t} \\ -2 \hat{t} \\ \hat{t}\end{array}\right]$. A basis is $\left[\begin{array}{r}-4 \\ -2 \\ 1\end{array}\right]$ | Creative Commons License (CC BY) | college_math.linear_algebra | Write the solution set of the following system as a linear combination of vectors.
$$
\left[\begin{array}{lll}
1 & -1 & 2 \\
1 & -2 & 0 \\
3 & -4 & 4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
$$ | $\boxed{Solution is: $\left[\begin{array}{r}-4 \hat{t} \\ -2 \hat{t} \\ \hat{t}\end{array}\right]$. A basis is $\left[\begin{array}{r}-4 \\ -2 \\ 1\end{array}\right]}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.7.3.1 | First we write $A=P D P^{-1}$.
$$
\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]=\left[\begin{array}{rr}
-1 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{rr}
-1 & 0 \\
0 & 3
\end{array}\right]\left[\begin{array}{rr}
-\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2}
\end{array}\right]
$$
Therefore $A^{10}=P D^{10} P^{-1}$.
$$
\begin{aligned}
{\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]^{10} } & =\left[\begin{array}{rr}
-1 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{rr}
-1 & 0 \\
0 & 3^{10}
\end{array}\right]\left[\begin{array}{rr}
-\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2}
\end{array}\right] \\
& =\left[\begin{array}{ll}
29525 & 29524 \\
29524 & 29525
\end{array}\right]
\end{aligned}
$$ | Creative Commons License (CC BY) | college_math.linear_algebra | Let $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$. Diagonalize A to find $A^{10}$. | $\boxed{First we write $A=P D P^{-1}$.
$$
\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]=\left[\begin{array}{rr}
-1 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{rr}
-1 & 0 \\
0 & 3
\end{array}\right]\left[\begin{array}{rr}
-\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2}
\end{array}\right]
$$
Therefore $A^{10}=P D^{10} P^{-1}$.
$$
\begin{aligned}
{\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]^{10} } & =\left[\begin{array}{rr}
-1 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{rr}
-1 & 0 \\
0 & 3^{10}
\end{array}\right]\left[\begin{array}{rr}
-\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2}
\end{array}\right] \\
& =\left[\begin{array}{ll}
29525 & 29524 \\
29524 & 29525
\end{array}\right]
\end{aligned}
}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.2.1.36 | $\left[\begin{array}{ll}0 & 1 \\ 5 & 3\end{array}\right]^{-1}=\left[\begin{array}{cc}-\frac{3}{5} & \frac{1}{5} \\ 1 & 0\end{array}\right]$ | Creative Commons License (CC BY) | college_math.linear_algebra | Let
$$
A=\left[\begin{array}{ll}
0 & 1 \\
5 & 3
\end{array}\right]
$$
Find $A^{-1}$ if possible. If $A^{-1}$ does not exist, explain why. | $\boxed{\left[\begin{array}{ll}0 & 1 \\ 5 & 3\end{array}\right]^{-1}=\left[\begin{array}{cc}-\frac{3}{5} & \frac{1}{5} \\ 1 & 0\end{array}\right]}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.1.2.39 | Solution is: $[x=1, y=5, z=3]$ | Creative Commons License (CC BY) | college_math.linear_algebra | Find the solution to the system of equations, $-9 x+15 y=66,-11 x+18 y=79,-x+y=4$, and $z=3$. | $\boxed{Solution is: $[x=1, y=5, z=3]}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.3.2.9 | $$
\operatorname{det}\left[\begin{array}{ccc}
e^{t} & \cosh t & \sinh t \\
e^{t} & \sinh t & \cosh t \\
e^{t} & \cosh t & \sinh t
\end{array}\right]=0
$$
and so this matrix fails to have a nonzero determinant at any value of $t$. | Creative Commons License (CC BY) | college_math.linear_algebra | Consider the matrix
$$
A=\left[\begin{array}{ccc}
e^{t} & \cosh t & \sinh t \\
e^{t} & \sinh t & \cosh t \\
e^{t} & \cosh t & \sinh t
\end{array}\right]
$$
Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain. | $\boxed{
\operatorname{det}\left[\begin{array}{ccc}
e^{t} & \cosh t & \sinh t \\
e^{t} & \sinh t & \cosh t \\
e^{t} & \cosh t & \sinh t
\end{array}\right]=0
$$
and so this matrix fails to have a nonzero determinant at any value of $t$.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.1.2.42 | No. This would lead to $0=1$. | Creative Commons License (CC BY) | college_math.linear_algebra | Suppose a system of linear equations has a $2 \times 4$ augmented matrix and the last column is a pivot column. Could the system of linear equations be consistent? Explain. | $\boxed{No. This would lead to $0=1$.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.7.2.8 | The eigenvalues are distinct because they are the $n^{\text {th }}$ roots of 1 . Hence if $X$ is a given vector with
$$
X=\sum_{j=1}^{n} a_{j} V_{j}
$$
then
$$
A^{n m} X=A^{n m} \sum_{j=1}^{n} a_{j} V_{j}=\sum_{j=1}^{n} a_{j} A^{n m} V_{j}=\sum_{j=1}^{n} a_{j} V_{j}=X
$$
so $A^{n m}=I$. | Creative Commons License (CC BY) | college_math.linear_algebra | Suppose the characteristic polynomial of an $n \times n$ matrix $A$ is $1-X^{n}$. Find $A^{m n}$ where $m$ is an integer. | $\boxed{The eigenvalues are distinct because they are the $n^{\text {th }}$ roots of 1 . Hence if $X$ is a given vector with
$$
X=\sum_{j=1}^{n} a_{j} V_{j}
$$
then
$$
A^{n m} X=A^{n m} \sum_{j=1}^{n} a_{j} V_{j}=\sum_{j=1}^{n} a_{j} A^{n m} V_{j}=\sum_{j=1}^{n} a_{j} V_{j}=X
$$
so $A^{n m}=I$.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.1.2.14 | Any $h$ will work. | Creative Commons License (CC BY) | college_math.linear_algebra | Find $h$ such that
$$
\left[\begin{array}{rr|r}
1 & 1 & 4 \\
3 & h & 12
\end{array}\right]
$$
is the augmented matrix of a consistent system. | $\boxed{Any $h$ will work.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.3.1.4 | $$
\left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 1 & 3 \\
2 & 1 & 1
\end{array}\right|=6
$$ | Creative Commons License (CC BY) | college_math.linear_algebra | Find the following determinant by expanding along the first row and second column.
$$
\left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 1 & 3 \\
2 & 1 & 1
\end{array}\right|
$$ | $\boxed{
\left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 1 & 3 \\
2 & 1 & 1
\end{array}\right|=6
}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.7.2.3 | The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{l}
-6 \\
-1 \\
-2
\end{array}\right]\right\} \leftrightarrow 6,\left\{\left[\begin{array}{c}
-5 \\
-2 \\
2
\end{array}\right]\right\} \leftrightarrow-3,\left\{\left[\begin{array}{c}
-8 \\
-2 \\
3
\end{array}\right]\right\} \leftrightarrow-2
$$
The matrix $P$ needed to diagonalize the above matrix is
$$
\left[\begin{array}{rrr}
-6 & -5 & -8 \\
-1 & -2 & -2 \\
2 & 2 & 3
\end{array}\right]
$$
and the diagonal matrix $D$ is
$$
\left[\begin{array}{rrr}
6 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & -2
\end{array}\right]
$$ | Creative Commons License (CC BY) | college_math.linear_algebra | Find the eigenvalues and eigenvectors of the matrix
$$
\left[\begin{array}{rrr}
89 & 38 & 268 \\
14 & 2 & 40 \\
-30 & -12 & -90
\end{array}\right]
$$
One eigenvalue is -3. Diagonalize if possible. | $\boxed{The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{l}
-6 \\
-1 \\
-2
\end{array}\right]\right\} \leftrightarrow 6,\left\{\left[\begin{array}{c}
-5 \\
-2 \\
2
\end{array}\right]\right\} \leftrightarrow-3,\left\{\left[\begin{array}{c}
-8 \\
-2 \\
3
\end{array}\right]\right\} \leftrightarrow-2
$$
The matrix $P$ needed to diagonalize the above matrix is
$$
\left[\begin{array}{rrr}
-6 & -5 & -8 \\
-1 & -2 & -2 \\
2 & 2 & 3
\end{array}\right]
$$
and the diagonal matrix $D$ is
$$
\left[\begin{array}{rrr}
6 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & -2
\end{array}\right]
}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.3.2.10 | $$
\operatorname{det}\left[\begin{array}{ccc}
e^{t} & e^{-t} \cos t & e^{-t} \sin t \\
e^{t} & -e^{-t} \cos t-e^{-t} \sin t & -e^{-t} \sin t+e^{-t} \cos t \\
e^{t} & 2 e^{-t} \sin t & -2 e^{-t} \cos t
\end{array}\right]=5 e^{-t} \neq 0
$$
and so this matrix is always invertible. | Creative Commons License (CC BY) | college_math.linear_algebra | Consider the matrix
$$
A=\left[\begin{array}{ccc}
e^{t} & e^{-t} \cos t & e^{-t} \sin t \\
e^{t} & -e^{-t} \cos t-e^{-t} \sin t & -e^{-t} \sin t+e^{-t} \cos t \\
e^{t} & 2 e^{-t} \sin t & -2 e^{-t} \cos t
\end{array}\right]
$$
Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain. | $\boxed{
\operatorname{det}\left[\begin{array}{ccc}
e^{t} & e^{-t} \cos t & e^{-t} \sin t \\
e^{t} & -e^{-t} \cos t-e^{-t} \sin t & -e^{-t} \sin t+e^{-t} \cos t \\
e^{t} & 2 e^{-t} \sin t & -2 e^{-t} \cos t
\end{array}\right]=5 e^{-t} \neq 0
$$
and so this matrix is always invertible.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.3.1.9 | It does not change the determinant. This was just taking the transpose. | Creative Commons License (CC BY) | college_math.linear_algebra | An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.
$$
\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \rightarrow \cdots \rightarrow\left[\begin{array}{ll}
a & c \\
b & d
\end{array}\right]
$$ | $\boxed{It does not change the determinant. This was just taking the transpose.}$ |
college_math.A_First_Course_in_Linear_Algebra | exercise.2.1.35 | $\left[\begin{array}{rr}2 & 1 \\ -1 & 3\end{array}\right]^{-1}=\left[\begin{array}{rr}\frac{3}{7} & -\frac{1}{7} \\ \frac{1}{7} & \frac{2}{7}\end{array}\right]$ | Creative Commons License (CC BY) | college_math.linear_algebra | Let
$$
A=\left[\begin{array}{rr}
2 & 1 \\
-1 & 3
\end{array}\right]
$$
Find $A^{-1}$ if possible. If $A^{-1}$ does not exist, explain why. | $\boxed{\left[\begin{array}{rr}2 & 1 \\ -1 & 3\end{array}\right]^{-1}=\left[\begin{array}{rr}\frac{3}{7} & -\frac{1}{7} \\ \frac{1}{7} & \frac{2}{7}\end{array}\right]}$ |