A.E: Linear Algebra (Exercises)
- Page ID
- 73042
A.1: Vectors, Mappings, and Matrices
On a piece of graph paper draw the vectors:
- \(\begin{bmatrix} 2 \\ 5 \end{bmatrix}\)
- \(\begin{bmatrix} -2 \\ -4 \end{bmatrix}\)
- \((3,-4)\)
On a piece of graph paper draw the vector \((1,2)\) starting at (based at) the given point:
- based at \((0,0)\)
- based at \((1,2)\)
- based at \((0,-1)\)
On a piece of graph paper draw the following operations. Draw and label the vectors involved in the operations as well as the result:
- \(\begin{bmatrix} 1 \\ -4 \end{bmatrix} + \begin{bmatrix} 2 \\ 3 \end{bmatrix}\)
- \(\begin{bmatrix} -3 \\ 2 \end{bmatrix} - \begin{bmatrix} 1 \\ 3 \end{bmatrix}\)
- \(3\begin{bmatrix} 2 \\ 1 \end{bmatrix}\)
Compute the magnitude of
- \(\begin{bmatrix} 7 \\ 2 \end{bmatrix}\)
- \(\begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}\)
- \((1,3,-4)\)
Compute
- \(\begin{bmatrix} 2 \\ 3 \end{bmatrix} + \begin{bmatrix} 7 \\ -8 \end{bmatrix}\)
- \(\begin{bmatrix} -2 \\ 3 \end{bmatrix} - \begin{bmatrix} 6 \\ -4 \end{bmatrix}\)
- \(-\begin{bmatrix} -3 \\ 2 \end{bmatrix}\)
- \(4\begin{bmatrix} -1 \\ 5 \end{bmatrix}\)
- \(5\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 9 \begin{bmatrix} 0 \\ 1 \end{bmatrix}\)
- \(3\begin{bmatrix} 1 \\ -8 \end{bmatrix} - 2 \begin{bmatrix} 3 \\ -1 \end{bmatrix}\)
Find the unit vector in the direction of the given vector
- \(\begin{bmatrix} 1 \\ -3 \end{bmatrix}\)
- \(\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}\)
- \((3,1,-2)\)
If \(\vec{x} = (1,2)\) and \(\vec{y}\) are added together, we find \(\vec{x}+\vec{y} = (0,2)\). What is \(\vec{y}\)?
Write \((1,2,3)\) as a linear combination of the standard basis vectors \(\vec{e}_1\), \(\vec{e}_2\), and \(\vec{e}_3\).
If the magnitude of \(\vec{x}\) is 4, what is the magnitude of
- \(0\vec{x}\)
- \(3\vec{x}\)
- \(-\vec{x}\)
- \(-4\vec{x}\)
- \(\vec{x}+\vec{x}\)
- \(\vec{x}-\vec{x}\)
Suppose a linear mapping \(F \colon {\mathbb R}^2 \to {\mathbb R}^2\) takes \((1,0)\) to \((2,-1)\) and it takes \((0,1)\) to \((3,3)\). Where does it take
- \((1,1)\)
- \((2,0)\)
- \((2,-1)\)
Suppose a linear mapping \(F \colon {\mathbb R}^3 \to {\mathbb R}^2\) takes \((1,0,0)\) to \((2,1)\), it takes \((0,1,0)\) to \((3,4)\), and it takes \((0,0,1)\) to \((5,6)\). Write down the matrix representing the mapping \(F\).
Suppose that a mapping \(F \colon {\mathbb R}^2 \to \mathbb{R}^2\) takes \((1,0)\) to \((1,2)\), \((0,1)\) to \((3,4)\), and \((1,1)\) to \((0,-1)\). Explain why \(F\) is not linear.
Let \({\mathbb R}^3\) represent the space of quadratic polynomials in \(t\): a point \((a_0,a_1,a_2)\) in \({\mathbb R}^3\) represents the polynomial \(a_0 + a_1 t + a_2 t^2\). Consider the derivative \(\frac{d}{dt}\) as a mapping of \({\mathbb R}^3\) to \({\mathbb R}^3\), and note that \(\frac{d}{dt}\) is linear. Write down \(\frac{d}{dt}\) as a \(3 \times 3\) matrix.
Compute the magnitude of
- \(\begin{bmatrix} 1 \\ 3 \end{bmatrix}\)
- \(\begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}\)
- \((-2,1,-2)\)
- Answer
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- \(\sqrt{10}\)
- \(\sqrt{14}\)
- \(3\)
Find the unit vector in the direction of the given vector
- \(\begin{bmatrix} -1 \\ 1 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}\)
- \((2,-5,2)\)
- Answer
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- \(\left[\begin{array}{c}{\frac{-1}{\sqrt{2}}}\\{\frac{1}{\sqrt{2}}}\end{array}\right]\)
- \(\left[\begin{array}{c}{\frac{1}{\sqrt{6}}}\\{\frac{-1}{\sqrt{6}}}\\{\frac{2}{\sqrt{6}}}\end{array}\right]\)
- \(\left(\frac{2}{\sqrt{33}},\frac{-5}{\sqrt{33}},\frac{2}{\sqrt{33}}\right)\)
Compute
- \(\begin{bmatrix} 3 \\ 1 \end{bmatrix} + \begin{bmatrix} 6 \\ -3 \end{bmatrix}\)
- \(\begin{bmatrix} -1 \\ 2 \end{bmatrix} - \begin{bmatrix} 2 \\ -1 \end{bmatrix}\)
- \(-\begin{bmatrix} -5 \\ 3 \end{bmatrix}\)
- \(2\begin{bmatrix} -2 \\ 4 \end{bmatrix}\)
- \(3\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 7 \begin{bmatrix} 0 \\ 1 \end{bmatrix}\)
- \(2\begin{bmatrix} 2 \\ -3 \end{bmatrix} - 6 \begin{bmatrix} 2 \\ -1 \end{bmatrix}\)
- Answer
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- \(\left[\begin{array}{c}{9}\\{-2}\end{array}\right]\)
- \(\left[\begin{array}{c}{-3}\\{3}\end{array}\right]\)
- \(\left[\begin{array}{c}{5}\\{-3}\end{array}\right]\)
- \(\left[\begin{array}{c}{-4}\\{8}\end{array}\right]\)
- \(\left[\begin{array}{c}{3}\\{7}\end{array}\right]\)
- \(\left[\begin{array}{c}{-8}\\{3}\end{array}\right]\)
If the magnitude of \(\vec{x}\) is 5, what is the magnitude of
- \(4\vec{x}\)
- \(-2\vec{x}\)
- \(-4\vec{x}\)
- Answer
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- \(20\)
- \(10\)
- \(20\)
Suppose a linear mapping \(F \colon {\mathbb R}^2 \to {\mathbb R}^2\) takes \((1,0)\) to \((1,-1)\) and it takes \((0,1)\) to \((2,0)\). Where does it take
- \((1,1)\)
- \((0,2)\)
- \((1,-1)\)
- Answer
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- \((3,-1)\)
- \((4,0)\)
- \((-1,-1)\)
A.2: Matrix Algebra
Add the following matrices
- \(\begin{bmatrix} -1 & 2 & 2 \\ 5 & 8 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 2 & 3 \\ 8 & 3 & 5 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 2 & 4 \\ 2 & 3 & 1 \\ 0 & 5 & 1 \end{bmatrix} + \begin{bmatrix} 2 & -8 & -3 \\ 3 & 1 & 0 \\ 6 & -4 & 1 \end{bmatrix}\)
Compute
- \(3\begin{bmatrix} 0 & 3 \\ -2 & 2 \end{bmatrix} + 6 \begin{bmatrix} 1 & 5 \\ -1 & 5 \end{bmatrix}\)
- \(2\begin{bmatrix} -3 & 1 \\ 2 & 2 \end{bmatrix} - 3 \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}\)
Multiply the following matrices
- \(\begin{bmatrix} -1 & 2 \\ 3 & 1 \\ 5 & 8 \end{bmatrix} \begin{bmatrix} 3 & -1 & 3 & 1 \\ 8 & 3 & 2 & -3 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 1 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 1 & 7 \\ 1 & 2 & 3 & -1 \\ 1 & -1 & 3 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 4 & 1 & 6 & 3 \\ 5 & 6 & 5 & 0 \\ 4 & 6 & 6 & 0 \end{bmatrix} \begin{bmatrix} 2 & 5 \\ 1 & 2 \\ 3 & 5 \\ 5 & 6 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 1 & 4 \\ 0 & 5 & 1 \end{bmatrix} \begin{bmatrix} 2 & 2 \\ 1 & 0 \\ 6 & 4 \end{bmatrix}\)
Compute the inverse of the given matrices
- \(\begin{bmatrix} -3 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 4 \\ 1 & 3 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 2 \\ 1 & 4 \end{bmatrix}\)
Compute the inverse of the given matrices
- \(\begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0.01 & 0 \\ 0 & 0 & 0 & -5 \end{bmatrix}\)
Add the following matrices
- \(\begin{bmatrix} 2 & 1 & 0 \\ 1 & 1 & -1 \end{bmatrix} + \begin{bmatrix} 5 & 3 & 4 \\ 1 & 2 & 5 \end{bmatrix}\)
- \(\begin{bmatrix} 6 & -2 & 3 \\ 7 & 3 & 3 \\ 8 & -1 & 2 \end{bmatrix} + \begin{bmatrix} -1 & -1 & -3 \\ 6 & 7 & 3 \\ -9 & 4 & -1 \end{bmatrix}\)
- Answer
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Compute
- \(2\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + 3 \begin{bmatrix} -1 & 3 \\ 1 & 2 \end{bmatrix}\)
- \(3\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} - 2 \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix}\)
- Answer
-
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Multiply the following matrices
- \(\begin{bmatrix} 2 & 1 & 4 \\ 3 & 4 & 4 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 6 & 3 \\ 3 & 5 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 3 & 3 \\ 2 & -2 & 1 \\ 3 & 5 & -2 \end{bmatrix} \begin{bmatrix} 6 & 6 & 2 \\ 4 & 6 & 0 \\ 2 & 0 & 4 \end{bmatrix}\)
- \(\begin{bmatrix} 3 & 4 & 1 \\ 2 & -1 & 0 \\ 4 & -1 & 5 \end{bmatrix} \begin{bmatrix} 0 & 2 & 5 & 0 \\ 2 & 0 & 5 & 2 \\ 3 & 6 & 1 & 6 \end{bmatrix}\)
- \(\begin{bmatrix} -2 & -2 \\ 5 & 3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 3 \\ 1 & 3 \end{bmatrix}\)
- Answer
-
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Compute the inverse of the given matrices
- \(\begin{bmatrix} 2 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}\)
- \(\begin{bmatrix} 4 & 2 \\ 4 & 4 \end{bmatrix}\)
- Answer
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Compute the inverse of the given matrices
- \(\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}\)
- \(\begin{bmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & -1 \end{bmatrix}\)
- \(\begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0.1 \end{bmatrix}\)
- Answer
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A.3: Elimination
Compute the reduced row echelon form for the following matrices:
- \(\begin{bmatrix} 1 & 3 & 1 \\ 0 & 1 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 3 & 3 \\ 6 & -3 \end{bmatrix}\)
- \(\begin{bmatrix} 3 & 6 \\ -2 & -3 \end{bmatrix}\)
- \(\begin{bmatrix} 6 & 6 & 7 & 7 \\ 1 & 1 & 0 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 9 & 3 & 0 & 2 \\ 8 & 6 & 3 & 6 \\ 7 & 9 & 7 & 9 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 1 & 3 & -3 \\ 6 & 0 & 0 & -1 \\ -2 & 4 & 4 & 3 \end{bmatrix}\)
- \(\begin{bmatrix} 6 & 6 & 5 \\ 0 & -2 & 2 \\ 6 & 5 & 6 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 2 & 0 & -1 \\ 6 & 6 & -3 & 3 \\ 6 & 2 & -3 & 5 \end{bmatrix}\)
Compute the inverse of the given matrices
- \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 1 \\ 0 & 2 & 1 \end{bmatrix}\)
Solve (find all solutions), or show no solution exists
- \(\begin{aligned} 4x_1+3x_2 & = -2 \\ -x_1+\phantom{3} x_2 & = 4 \end{aligned}\)
- \(\begin{aligned} x_1+5x_2+3x_3 & = 7 \\ 8x_1+7x_2+8x_3 & = 8 \\ 4x_1+8x_2+6x_3 & = 4 \end{aligned}\)
- \(\begin{aligned} 4x_1+8x_2+2x_3 & = 3 \\ -x_1-2x_2+3x_3 & = 1 \\ 4x_1+8x_2 \phantom{{}+3x_3} & = 2 \end{aligned}\)
- \(\begin{aligned} x+2y+3z & = 4 \\ 2 x-\phantom{2} y+3z & = 1 \\ 3 x+\phantom{2} y+6z & = 6 \end{aligned}\)
By computing the inverse, solve the following systems for \(\vec{x}\).
- \(\begin{bmatrix} 4 & 1 \\ -1 & 3 \end{bmatrix} \vec{x} = \begin{bmatrix} 13 \\ 26 \end{bmatrix}\)
- \(\begin{bmatrix} 3 & 3 \\ 3 & 4 \end{bmatrix} \vec{x} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}\)
Compute the rank of the given matrices
- \(\begin{bmatrix} 6 & 3 & 5 \\ 1 & 4 & 1 \\ 7 & 7 & 6 \end{bmatrix}\)
- \(\begin{bmatrix} 5 & -2 & -1 \\ 3 & 0 & 6 \\ 2 & 4 & 5 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 2 & 3 \\ -1 & -2 & -3 \\ 2 & 4 & 6 \end{bmatrix}\)
For the matrices in Exercise \(\PageIndex{A.3.5}\), find a linearly independent set of row vectors that span the row space (they don’t need to be rows of the matrix).
For the matrices in Exercise \(\PageIndex{A.3.5}\), find a linearly independent set of columns that span the column space. That is, find the pivot columns of the matrices.
Find a linearly independent subset of the following vectors that has the same span. \[\begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ -2 \\ -4 \end{bmatrix} , \quad \begin{bmatrix} -2 \\ 4 \\ 1 \end{bmatrix} , \quad \begin{bmatrix} -1 \\ 3 \\ -2 \end{bmatrix} \nonumber \]
Compute the reduced row echelon form for the following matrices:
- \(\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 1 \\ -2 & -2 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & -3 & 1 \\ 4 & 6 & -2 \\ -2 & 6 & -2 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 2 & 5 & 2 \\ 1 & -2 & 4 & -1 \\ 0 & 3 & 1 & -2 \end{bmatrix}\)
- \(\begin{bmatrix} -2 & 6 & 4 & 3 \\ 6 & 0 & -3 & 0 \\ 4 & 2 & -1 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 5 \end{bmatrix}\)
- Answer
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- \(\left[\begin{array}{ccc}{1}&{0}&{1}\\{0}&{1}&{0}\end{array}\right]\)
- \(\left[\begin{array}{cc}{1}&{0}\\{0}&{1}\end{array}\right]\)
- \(\left[\begin{array}{cc}{1}&{1}\\{0}&{0}\end{array}\right]\)
- \(\left[\begin{array}{ccc}{1}&{0}&{0}\\{0}&{1}&{-\frac{1}{3}}\\{0}&{0}&{0}\end{array}\right]\)
- \(\left[\begin{array}{cccc}{1}&{0}&{0}&{\frac{77}{15}}\\{0}&{1}&{0}&{-\frac{2}{15}}\\{0}&{0}&{1}&{-\frac{8}{5}}\end{array}\right]\)
- \(\left[\begin{array}{cccc}{1}&{0}&{-\frac{1}{2}}&{0}\\{0}&{1}&{\frac{1}{2}}&{\frac{1}{2}}\\{0}&{0}&{0}&{0}\end{array}\right]\)
- \(\left[\begin{array}{cccc}{0}&{0}&{0}&{0}\\{0}&{0}&{0}&{0}\end{array}\right]\)
- \(\left[\begin{array}{cccc}{1}&{2}&{3}&{0}\\{0}&{0}&{0}&{1}\end{array}\right]\)
Compute the inverse of the given matrices
- \(\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 4 & 0 \\ 2 & 2 & 3 \\ 2 & 4 & 1 \end{bmatrix}\)
- Answer
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- \(\left[\begin{array}{ccc}{0}&{-1}&{0}\\{1}&{0}&{0}\\{0}&{0}&{1}\end{array}\right]\)
- \(\left[\begin{array}{ccc}{0}&{0}&{1}\\{0}&{1}&{-1}\\{1}&{-1}&{0}\end{array}\right]\)
- \(\left[\begin{array}{ccc}{\frac{5}{2}}&{1}&{-3}\\{-1}&{-\frac{1}{2}}&{\frac{3}{2}}\\{-1}&{0}&{1}\end{array}\right]\)
Solve (find all solutions), or show no solution exists
- \(\begin{aligned} 4x_1+3x_2 & = -1 \\ 5x_1+6x_2 & = 4 \end{aligned}\)
- \(\begin{aligned} 5x+6y+5z & = 7 \\ 6x+8y+6z & = -1 \\ 5x+2y+5z & = 2 \end{aligned}\)
- \(\begin{aligned} a+\phantom{5}b+\phantom{6}c & = -1 \\ a+5b+6c & = -1 \\ -2a+5b+6c & = 8 \end{aligned}\)
- \(\begin{aligned} -2 x_1+2x_2+8x_3 & = 6 \\ x_2+\phantom{8}x_3 & = 2 \\ x_1+4x_2+\phantom{8}x_3 & = 7 \end{aligned}\)
- Answer
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- \(x_{1}=-2,\: x_{2}=\frac{7}{3}\)
- no solution
- \(a=-3,\: b=10,\: c=-8\)
- \(x_{3}\) is free, \(x_{1}=-1+3x_{3}\), \(x_{2}=2-x_{3}\)
By computing the inverse, solve the following systems for \(\vec{x}\).
- \(\begin{bmatrix} -1 & 1 \\ 3 & 3 \end{bmatrix} \vec{x} = \begin{bmatrix} 4 \\ 6 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 7 \\ 1 & 6 \end{bmatrix} \vec{x} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}\)
- Answer
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- \(\left[\begin{array}{c}{-1}\\{3}\end{array}\right]\)
- \(\left[\begin{array}{c}{-3}\\{1}\end{array}\right]\)
Compute the rank of the given matrices
- \(\begin{bmatrix} 7 & -1 & 6 \\ 7 & 7 & 7 \\ 7 & 6 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 3 & -1 \\ 6 & 3 & 1 \\ 4 & 7 & -1 \end{bmatrix}\)
- Answer
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- \(3\)
- \(1\)
- \(2\)
For the matrices in Exercise \(\PageIndex{A.3.13}\), find a linearly independent set of row vectors that span the row space (they don’t need to be rows of the matrix).
- Answer
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- \(\left[\begin{array}{ccc}{1}&{0}&{0}\end{array}\right]\), \(\left[\begin{array}{ccc}{0}&{1}&{0}\end{array}\right]\), \(\left[\begin{array}{ccc}{0}&{0}&{1}\end{array}\right]\)
- \(\left[\begin{array}{ccc}{1}&{1}&{1}\end{array}\right]\)
- \(\left[\begin{array}{ccc}{1}&{0}&{\frac{1}{3}}\end{array}\right]\), \(\left[\begin{array}{ccc}{0}&{1}&{-\frac{1}{3}}\end{array}\right]\)
For the matrices in Exercise \(\PageIndex{A.3.13}\), find a linearly independent set of columns that span the column space. That is, find the pivot columns of the matrices.
- Answer
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- \(\left[\begin{array}{c}{7}\\{7}\\{7}\end{array}\right]\), \(\left[\begin{array}{c}{-1}\\{7}\\{6}\end{array}\right]\), \(\left[\begin{array}{c}{7}\\{6}\\{2}\end{array}\right]\)
- \(\left[\begin{array}{c}{1}\\{1}\\{2}\end{array}\right]\)
- \(\left[\begin{array}{c}{0}\\{6}\\{4}\end{array}\right]\), \(\left[\begin{array}{c}{3}\\{3}\\{7}\end{array}\right]\)
Find a linearly independent subset of the following vectors that has the same span. \[\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} , \quad \begin{bmatrix} 3 \\ 1 \\ -5 \end{bmatrix} , \quad \begin{bmatrix} 0 \\ 3 \\ -1 \end{bmatrix} , \quad \begin{bmatrix} -3 \\ 2 \\ 4 \end{bmatrix} \nonumber \]
- Answer
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\(\left[\begin{array}{c}{3}\\{1}\\{-5}\end{array}\right]\), \(\left[\begin{array}{c}{0}\\{3}\\{-1}\end{array}\right]\)
A.4: Subspaces, Dimension, and The Kernel
For the following sets of vectors, find a basis for the subspace spanned by the vectors, and find the dimension of the subspace.
- \(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} , \quad \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ 0 \\ 5 \end{bmatrix} , \quad \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} , \quad \begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}\)
- \(\begin{bmatrix} -4 \\ -3 \\ 5 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ 3 \\ 3 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ 3 \\ 0 \end{bmatrix} , \quad \begin{bmatrix} 0 \\ 2 \\ 2 \end{bmatrix} , \quad \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ 3 \end{bmatrix} , \quad \begin{bmatrix} 0 \\ 2 \end{bmatrix} , \quad \begin{bmatrix} -1 \\ -1 \end{bmatrix}\)
- \(\begin{bmatrix} 3 \\ 1 \\ 3 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ 4 \\ -4 \end{bmatrix} , \quad \begin{bmatrix} -5 \\ -5 \\ -2 \end{bmatrix}\)
For the following matrices, find a basis for the kernel (nullspace).
- \(\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 5 \\ 1 & 1 & -4 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & -1 & -3 \\ 4 & 0 & -4 \\ -1 & 1 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} -4 & 4 & 4 \\ -1 & 1 & 1 \\ -5 & 5 & 5 \end{bmatrix}\)
- \(\begin{bmatrix} -2 & 1 & 1 & 1 \\ -4 & 2 & 2 & 2 \\ 1 & 0 & 4 & 3 \end{bmatrix}\)
Suppose a \(5 \times 5\) matrix \(A\) has rank 3. What is the nullity?
Suppose that \(X\) is the set of all the vectors of \({\mathbb{R}}^3\) whose third component is zero. Is \(X\) a subspace? And if so, find a basis and the dimension.
Consider a square matrix \(A\), and suppose that \(\vec{x}\) is a nonzero vector such that \(A \vec{x} = \vec{0}\). What does the Fredholm alternative say about invertibility of \(A\).
Consider \[M = \begin{bmatrix} 1 & 2 & 3 \\ 2 & ? & ? \\ -1 & ? & ? \end{bmatrix} . \nonumber \] If the nullity of this matrix is 2, fill in the question marks. Hint: What is the rank?
For the following sets of vectors, find a basis for the subspace spanned by the vectors, and find the dimension of the subspace.
- \(\begin{bmatrix} 1 \\ 2 \end{bmatrix} , \quad \begin{bmatrix} 1 \\ 1 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix} , \quad \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}\)
- \(\begin{bmatrix} 5 \\ 3 \\ 1 \end{bmatrix} , \quad \begin{bmatrix} 5 \\ -1 \\ 5 \end{bmatrix} , \quad \begin{bmatrix} -1 \\ 3 \\ -4 \end{bmatrix}\)
- \(\begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix} , \quad \begin{bmatrix} 4 \\ 4 \\ -3 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ 0 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ 0 \end{bmatrix} , \quad \begin{bmatrix} 3 \\ 0 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} , \quad \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\)
- Answer
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- \(\left[\begin{array}{c}{1}\\{2}\end{array}\right]\), \(\left[\begin{array}{c}{1}\\{1}\end{array}\right]\) dimension \(2\),
- \(\left[\begin{array}{c}{1}\\{1}\\{1}\end{array}\right]\), \(\left[\begin{array}{c}{1}\\{1}\\{2}\end{array}\right]\) dimension \(2\),
- \(\left[\begin{array}{c}{5}\\{3}\\{1}\end{array}\right]\), \(\left[\begin{array}{c}{5}\\{-1}\\{5}\end{array}\right]\), \(\left[\begin{array}{c}{-1}\\{3}\\{-4}\end{array}\right]\) dimension \(3\),
- \(\left[\begin{array}{c}{2}\\{2}\\{4}\end{array}\right]\), \(\left[\begin{array}{c}{2}\\{2}\\{3}\end{array}\right]\) dimension \(2\),
- \(\left[\begin{array}{c}{1}\\{1}\end{array}\right]\) dimension \(1\),
- \(\left[\begin{array}{c}{1}\\{0}\\{0}\end{array}\right]\), \(\left[\begin{array}{c}{0}\\{1}\\{2}\end{array}\right]\) dimension \(2\)
For the following matrices, find a basis for the kernel (nullspace).
- \(\begin{bmatrix} 2 & 6 & 1 & 9 \\ 1 & 3 & 2 & 9 \\ 3 & 9 & 0 & 9 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & -2 & -5 \\ -1 & 1 & 5 \\ -5 & 5 & -3 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & -5 & -4 \\ 2 & 3 & 5 \\ -3 & 5 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 4 & 4 \\ 0 & 1 & 1 \\ 0 & 5 & 5 \end{bmatrix}\)
- Answer
-
- \(\left[\begin{array}{c}{3}\\{-1}\\{0}\\{0}\end{array}\right]\), \(\left[\begin{array}{c}{3}\\{0}\\{3}\\{-1}\end{array}\right]\)
- \(\left[\begin{array}{c}{-1}\\{-1}\\{0}\end{array}\right]\)
- \(\left[\begin{array}{c}{1}\\{1}\\{-1}\end{array}\right]\)
- \(\left[\begin{array}{c}{-1}\\{0}\\{0}\end{array}\right]\), \(\left[\begin{array}{c}{0}\\{1}\\{-1}\end{array}\right]\)
Suppose the column space of a \(9 \times 5\) matrix \(A\) of dimension 3. Find
- Rank of \(A\).
- Nullity of \(A\).
- Dimension of the row space of \(A\).
- Dimension of the nullspace of \(A\).
- Size of the maximum subset of linearly independent rows of \(A\).
- Answer
-
- \(3\)
- \(2\)
- \(3\)
- \(2\)
- \(3\)
A.5: Inner Product and Projections
Find the \(s\) that makes the following vectors orthogonal: \((1,2,3)\), \((1,1,s)\).
Find the angle \(\theta\) between \((1,3,1)\), \((2,1,-1)\).
Given that \(\langle \vec{v} , \vec{w} \rangle = 3\) and \(\langle \vec{v} , \vec{u} \rangle = -1\) compute
- \(\langle \vec{u} , 2 \vec{v} \rangle\)
- \(\langle \vec{v} , 2 \vec{w} + 3 \vec{u} \rangle\)
- \(\langle \vec{w} + 3 \vec{u}, \vec{v} \rangle\)
Suppose \(\vec{v} = (1,1,-1)\). Find
- \(\operatorname{proj}_{\vec{v}}\bigl( (1,0,0) \bigr)\)
- \(\operatorname{proj}_{\vec{v}}\bigl( (1,2,3) \bigr)\)
- \(\operatorname{proj}_{\vec{v}}\bigl( (1,-1,0) \bigr)\)
Consider the vectors \((1,2,3)\), \((-3,0,1)\), \((1,-5,3)\).
- Check that the vectors are linearly independent and so form a basis.
- Check that the vectors are mutually orthogonal, and are therefore an orthogonal basis.
- Represent \((1,1,1)\) as a linear combination of this basis.
- Make the basis orthonormal.
Let \(S\) be the subspace spanned by \((1,3,-1)\), \((1,1,1)\). Find an orthogonal basis of \(S\) by the Gram-Schmidt process.
Starting with \((1,2,3)\), \((1,1,1)\), \((2,2,0)\), follow the Gram-Schmidt process to find an orthogonal basis of \({\mathbb{R}}^3\).
Find an orthogonal basis of \({\mathbb{R}}^3\) such that \((3,1,-2)\) is one of the vectors. Hint: First find two extra vectors to make a linearly independent set.
Using cosines and sines of \(\theta\), find a unit vector \(\vec{u}\) in \({\mathbb{R}}^2\) that makes angle \(\theta\) with \(\vec{\imath} = (1,0)\). What is \(\langle \vec{\imath} , \vec{u} \rangle\)?
Find the \(s\) that makes the following vectors orthogonal: \((1,1,1)\), \((1,s,1)\).
- Answer
-
\(s=-2\)
Find the angle \(\theta\) between \((1,2,3)\), \((1,1,1)\).
- Answer
-
\(\theta\approx 0.3876\)
Given that \(\langle \vec{v} , \vec{w} \rangle = 1\) and \(\langle \vec{v} , \vec{u} \rangle = -1\) and \(\lVert \vec{v} \rVert = 3\) and
- \(\langle 3 \vec{u} , 5 \vec{v} \rangle\)
- \(\langle \vec{v} , 2 \vec{w} + 3 \vec{u} \rangle\)
- \(\langle \vec{w} + 3 \vec{v}, \vec{v} \rangle\)
- Answer
-
- \(-15\)
- \(-1\)
- \(28\)
Suppose \(\vec{v} = (1,0,-1)\). Find
- \(\operatorname{proj}_{\vec{v}}\bigl( (0,2,1) \bigr)\)
- \(\operatorname{proj}_{\vec{v}}\bigl( (1,0,1) \bigr)\)
- \(\operatorname{proj}_{\vec{v}}\bigl( (4,-1,0) \bigr)\)
- Answer
-
- \(\left(-\frac{1}{2},0,\frac{1}{2}\right)\)
- \((0,0,0)\)
- \((2,0,-2)\)
The vectors \((1,1,-1)\), \((2,-1,1)\), \((1,-5,3)\) form an orthogonal basis. Represent the following vectors in terms of this basis:
- \((1,-8,4)\)
- \((5,-7,5)\)
- \((0,-6,2)\)
- Answer
-
- \((1,1,-1)-(2,-1,1)+2(1,-5,3)\)
- \(2(2,-1,1)+(1,-5,3)\)
- \(2(1,1,-1)-2(2,-1,1)+2(1,-5,3)\)
Let \(S\) be the subspace spanned by \((2,-1,1)\), \((2,2,2)\). Find an orthogonal basis of \(S\) by the Gram-Schmidt process.
- Answer
-
\((2,-1,1)\), \(\left(\frac{2}{3},\frac{8}{3},\frac{4}{3}\right)\)
Starting with \((1,1,-1)\), \((2,3,-1)\), \((1,-1,1)\), follow the Gram-Schmidt process to find an orthogonal basis of \({\mathbb{R}}^3\).
- Answer
-
\((1,1,-1)\), \((0,1,1)\), \(\left(\frac{4}{3},-\frac{2}{3},\frac{2}{3}\right)\)
A.6: Determinant
Compute the determinant of the following matrices:
- \(\begin{bmatrix} 3 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 1 & 0 \\ -2 & 7 & -3 \\ 0 & 2 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 1 & 3 \\ 8 & 6 & 3 \\ 7 & 9 & 7 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 2 & 5 & 7 \\ 0 & 0 & 2 & -3 \\ 3 & 4 & 5 & 7 \\ 0 & 0 & 2 & 4 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 1 & 2 & 0 \\ 1 & 1 & -1 & 2 \\ 1 & 1 & 2 & 1 \\ 2 & -1 & -2 & 3 \end{bmatrix}\)
For which \(x\) are the following matrices singular (not invertible).
- \(\begin{bmatrix} 2 & 3 \\ 2 & x \end{bmatrix}\)
- \(\begin{bmatrix} 2 & x \\ 1 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} x & 1 \\ 4 & x \end{bmatrix}\)
- \(\begin{bmatrix} x & 0 & 1 \\ 1 & 4 & 2 \\ 1 & 6 & 2 \end{bmatrix}\)
Compute \[\det \left( \begin{bmatrix} 2 & 1 & 2 & 3 \\ 0 & 8 & 6 & 5 \\ 0 & 0 & 3 & 9 \\ 0 & 0 & 0 & 1 \end{bmatrix}^{-1} \right) \nonumber \] without computing the inverse.
Suppose \[L = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ 7 & \pi & 1 & 0 \\ 2^8 & 5 & -99 & 1 \end{bmatrix} \qquad \text{and} \qquad U = \begin{bmatrix} 5 & 9 & 1 & -\sin(1) \\ 0 & 1 & 88 & -1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix} . \nonumber \] Let \(A = LU\). Compute \(\det(A)\) in a simple way, without computing what is \(A\). Hint: First read off \(\det(L)\) and \(\det(U)\).
Consider the linear mapping from \({\mathbb R}^2\) to \({\mathbb R}^2\) given by the matrix \(A = \left[ \begin{smallmatrix} 1 & x \\ 2 & 1 \end{smallmatrix} \right]\) for some number \(x\). You wish to make \(A\) such that it doubles the area of every geometric figure. What are the possibilities for \(x\) (there are two answers).
Suppose \(A\) and \(S\) are \(n \times n\) matrices, and \(S\) is invertible. Suppose that \(\det(A) = 3\). Compute \(\det(S^{-1}AS)\) and \(\det(SAS^{-1})\). Justify your answer using the theorems in this section.
Let \(A\) be an \(n \times n\) matrix such that \(\det(A)=1\). Compute \(\det(x A)\) given a number \(x\). Hint: First try computing \(\det(xI)\), then note that \(xA = (xI)A\).
Compute the determinant of the following matrices:
- \(\begin{bmatrix} -2 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & -2 \\ 1 & 3 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 9 & -11 \\ 0 & -1 & 5 \\ 0 & 0 & 3 \end{bmatrix}\)
- \(\begin{bmatrix} 2 & 1 & 0 \\ -2 & 7 & 3 \\ 1 & 1 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 5 & 1 & 3 \\ 4 & 1 & 1 \\ 4 & 5 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 3 & 2 & 5 & 7 \\ 0 & 0 & 2 & 0 \\ 0 & 4 & 5 & 0 \\ 2 & 1 & 2 & 4 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 2 & 1 & 0 \\ 1 & 2 & -3 & 4 \\ 5 & 6 & -7 & 8 \\ 1 & 2 & 3 & -2 \end{bmatrix}\)
- Answer
-
- \(-2\)
- \(8\)
- \(0\)
- \(-6\)
- \(-3\)
- \(28\)
- \(16\)
- \(-24\)
For which \(x\) are the following matrices singular (not invertible).
- \(\begin{bmatrix} 1 & 3 \\ 1 & x \end{bmatrix}\)
- \(\begin{bmatrix} 3 & x \\ 1 & 3 \end{bmatrix}\)
- \(\begin{bmatrix} x & 3 \\ 3 & x \end{bmatrix}\)
- \(\begin{bmatrix} x & 1 & 0 \\ 1 & 4 & 0 \\ 1 & 6 & 2 \end{bmatrix}\)
- Answer
-
- \(3\)
- \(9\)
- \(3\)
- \(\frac{1}{4}\)
Compute \[\det \left( \begin{bmatrix} 3 & 4 & 7 & 12 \\ 0 & -1 & 9 & -8 \\ 0 & 0 & -2 & 4 \\ 0 & 0 & 0 & 2 \end{bmatrix}^{-1} \right) \nonumber \] without computing the inverse.
- Answer
-
\(12\)
Find all the \(x\) that make the matrix inverse \[\begin{bmatrix} 1 & 2 \\ 1 & x \end{bmatrix}^{-1} \nonumber \] have only integer entries (no fractions). Note that there are two answers.
- Answer
-
\(1\) and \(3\)