2.3E: Matrix-Vector Multiplication Exercises
- Page ID
- 132802
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In each case find a system of equations that is equivalent to the given vector equation. (Do not solve the system.)
- \(x_{1} \left[ \begin{array}{r} 2 \\ -3 \\ 0 \end{array} \right] + x_{2} \left[ \begin{array}{r} 1 \\ 1 \\ 4 \end{array} \right] + x_{3} \left[ \begin{array}{r} 2 \\ 0 \\ -1 \end{array} \right] = \left[ \begin{array}{r} 5 \\ 6 \\ -3 \end{array} \right]\)
- \(x_{1} \left[ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \end{array} \right] + x_{2} \left[ \begin{array}{r} -3 \\ 8 \\ 2 \\ 1 \end{array} \right] + x_{3} \left[ \begin{array}{r} -3 \\ 0 \\ 2 \\ 2 \end{array} \right] + x_{4} \left[ \begin{array}{r} 3 \\ 2 \\ 0 \\ -2 \end{array} \right] = \left[ \begin{array}{r} 5 \\ 1 \\ 2 \\ 0 \end{array} \right]\)
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b. \( \begin{array}[t]{rrrrrrrrr} x_{1} & - & 3x_{2} & - & 3x_{3} & + & 3x_{4} & = & 5 \\ & & 8x_{2} & & & + & 2x_{4} & = & 1 \\ x_{1} & + & 2x_{2} & + & 2x_{3} & & & = & 2 \\ & & x_{2} & + & 2x_{3} & - & 5x_{4} & = & 0 \\ \end{array}\)
In each case find a vector equation that is equivalent to the given system of equations. (Do not solve the equation.)
- \( \begin{array}[t]{rrrrrrr} x_{1} & - & x_{2} & + & 3x_{3} & = & 5 \\ -3x_{1} & + & x_{2} & + & x_{3} & = & -6 \\ 5x_{1} & - & 8x_{2} & & & = & 9 \end{array}\)
- \( \begin{array}[t]{rrrrrrrrr} x_{1} & - & 2x_{2} & - & x_{3} & + & x_{4} & = & 5 \\ -x_{1} & & & + & x_{3} & - & 2x_{4} & = & -3 \\ 2x_{1} & - & 2x_{2} & + & 7x_{3} & & & = & 8 \\ 3x_{1} & - & 4x_{2} & + & 9x_{3} & - & 2x_{4} & = & 12 \end{array}\)
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b. \(x_{1} \left[ \begin{array}{r} 1 \\ -1 \\ 2 \\ 3 \end{array} \right] + x_{2} \left[ \begin{array}{r} -2 \\ 0 \\ -2 \\ -4 \end{array} \right] + x_{3} \left[ \begin{array}{r} -1 \\ 1 \\ 7 \\ 9 \end{array} \right] + x_{4} \left[ \begin{array}{r} 1 \\ -2 \\ 0 \\ -2 \end{array} \right] = \left[ \begin{array}{r} 5 \\ -3 \\ 8 \\ 12 \end{array} \right]\)
In each case compute \(A\mathbf{x}\) using: (i) Definition [def:002668]. (ii) Theorem [thm:002903].
- \(A = \left[ \begin{array}{rrr} 3 & -2 & 0 \\ 5 & -4 & 1 \end{array} \right]\) and \(\mathbf{x} = \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right]\).
- \(A = \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & -4 & 5 \end{array} \right]\) and \(\mathbf{x} = \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right]\).
- \(A = \left[ \begin{array}{rrrr} -2 & 0 & 5 & 4 \\ 1 & 2 & 0 & 3 \\ -5 & 6 & -7 & 8 \end{array} \right]\) and \(\mathbf{x} = \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]\).
- \(A = \left[ \begin{array}{rrrr} 3 & -4 & 1 & 6 \\ 0 & 2 & 1 & 5 \\ -8 & 7 & -3 & 0 \end{array} \right]\) and \(\mathbf{x} = \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]\).
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- \(A\mathbf{x} = \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & -4 & 5 \end{array} \right] \left[ \begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \end{array} \right] = x_{1} \left[ \begin{array}{r} 1 \\ 0 \end{array} \right] + x_{2} \left[ \begin{array}{r} 2 \\ -4 \end{array} \right] + x_{3} \left[ \begin{array}{r} 3 \\ 5 \end{array} \right] = \left[ \begin{array}{rrrrr} x_{1} & + & 2x_{2} & + & 3x_{3} \\ & - & 4x_{2} & + & 5x_{3} \end{array} \right]\)
- \(A\mathbf{x} = \left[ \begin{array}{rrrr} 3 & -4 & 1 & 6 \\ 0 & 2 & 1 & 5 \\ -8 & 7 & -3 & 0 \end{array} \right] \left[ \begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]\)
\({} = x_{1} \left[ \begin{array}{r} 3 \\ 0 \\ -8 \end{array} \right] + x_{2} \left[ \begin{array}{r} -4 \\ 2 \\ 7 \end{array} \right] + x_{3} \left[ \begin{array}{r} 1 \\ 1 \\ -3 \end{array} \right] + x_{4} \left[ \begin{array}{r} 6 \\ 5 \\ 0 \end{array} \right] = \left[ \begin{array}{rrrrrrr} 3x_{1} & - & 4x_{2} & + & x_{3} & + & 6x_{4} \\ & & 2x_{2} & + & x_{3} & + & 5x_{4} \\ -8x_{1} & + & 7x_{2} & - & 3x_{3} & & \\ \end{array} \right]\)
Let \(A = \left[ \begin{array}{cccc} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} & \mathbf{a}_{4} \end{array} \right]\) be the \(3 \times 4\) matrix given in terms of its columns \(\mathbf{a}_{1} = \left[ \begin{array}{r} 1 \\ 1 \\ -1 \end{array} \right]\), \(\mathbf{a}_{2} = \left[ \begin{array}{r} 3 \\ 0 \\ 2 \end{array} \right]\), \(\mathbf{a}_{3} = \left[ \begin{array}{r} 2 \\ -1 \\ 3 \end{array} \right]\), and \(\mathbf{a}_{4} = \left[ \begin{array}{r} 0 \\ -3 \\ 5 \end{array} \right]\). In each case either express \(\mathbf{b}\) as a linear combination of \(\mathbf{a}_{1}\), \(\mathbf{a}_{2}\), \(\mathbf{a}_{3}\), and \(\mathbf{a}_{4}\), or show that it is not such a linear combination. Explain what your answer means for the corresponding system \(A\mathbf{x} = \mathbf{b}\) of linear equations.
- \(\mathbf{b} = \left[ \begin{array}{r} 0 \\ 3 \\ 5 \end{array} \right]\)
- \(\mathbf{b} = \left[ \begin{array}{r} 4 \\ 1 \\ 1 \end{array} \right]\)
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b. To solve \(A\mathbf{x} = \mathbf{b}\) the reduction is \(\left[ \begin{array}{rrrr|r} 1 & 3 & 2 & 0 & 4 \\ 1 & 0 & -1 & -3 & 1 \\ -1 & 2 & 3 & 5 & 1 \end{array} \right] \rightarrow \left[ \begin{array}{rrrr|r} 1 & 0 & -1 & -3 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]\) so the general solution is \(\left[ \begin{array}{c} 1 + s + 3t \\ 1 - s -t \\ s \\ t \end{array} \right]\).
Hence \((1 + s + 3t)\mathbf{a}_{1} + (1 - s - t)\mathbf{a}_{2} + s\mathbf{a}_{3} + t\mathbf{a}_{4} = \mathbf{b}\) for any choice of \(s\) and \(t\). If \(s = t = 0\), we get \(\mathbf{a}_{1} + \mathbf{a}_{2} = \mathbf{b}\); if \(s = 1\) and \(t = 0\), we have \(2\mathbf{a}_{1} + \mathbf{a}_{3} = \mathbf{b}\).
In each case, express every solution of the system as a sum of a specific solution plus a solution of the associated homogeneous system.
- \( \begin{array}[t]{rrrrrrr} x & + & y & + & z & = & 2 \\ 2x & + & y & & & = & 3 \\ x & - & y & - & 3z & = & 0 \end{array}\)
- \( \begin{array}[t]{rrrrrrr} x & - & y & - & 4z & = & -4 \\ x & + & 2y & + & 5z & = & 2 \\ x & + & y & + & 2z & = & 0 \end{array}\)
- \( \begin{array}[t]{rrrrrrrrrrr} x_{1} & + & x_{2} & - & x_{3} & & & - & 5x_{5} & = & 2 \\ & & x_{2} & + & x_{3} & & & - & 4x_{5} & = & -1 \\ & & x_{2} & + & x_{3} & + & x_{4} & - & x_{5} & = & -1 \\ 2x_{1} & & & - & 4x_{3} & + & x_{4} & + & x_{5} & = & 6 \end{array}\)
- \( \begin{array}[t]{rrrrrrrrrrr} 2x_{1} & + & x_{2} & - & x_{3} & - & x_{4} & = & -1 \\ 3x_{1} & + & x_{2} & + & x_{3} & - & 2x_{4} & = & -2 \\ -x_{1} & - & x_{2} & + & 2x_{3} & + & x_{4} & = & 2 \\ -2x_{1} & - & x_{2} & & & + & 2x_{4} & = & 3 \end{array}\)
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- \(\left[ \begin{array}{r} -2 \\ 2 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} 1 \\ -3 \\ 1 \end{array} \right]\)
- \(\left[ \begin{array}{r} 3 \\ -9 \\ -2 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} -1 \\ 4 \\ 1 \\ 1 \end{array} \right]\)
If \(\mathbf{x}_{0}\) and \(\mathbf{x}_{1}\) are solutions to the homogeneous system of equations \(A\mathbf{x} = \mathbf{0}\), use Theorem [thm:002811] to show that \(s\mathbf{x}_{0} + t\mathbf{x}_{1}\) is also a solution for any scalars \(s\) and \(t\) (called a linear combination of \(\mathbf{x}_{0}\) and \(\mathbf{x}_{1}\)).
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We have \(A\mathbf{x}_{0} = \mathbf{0}\) and \(A\mathbf{x}_{1} = \mathbf{0}\) and so \(A(s\mathbf{x}_{0} + t\mathbf{x}_{1}) = s(A\mathbf{x}_{0}) + t(A\mathbf{x}_{1}) = s \cdot \mathbf{0} + t \cdot \mathbf{0} = \mathbf{0}\).
Assume that \(A \left[ \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right] = \mathbf{0} = A \left[ \begin{array}{r} 2 \\ 0 \\ 3 \end{array} \right]\). Show that \(\mathbf{x}_{0} = \left[ \begin{array}{r} 2 \\ -1 \\ 3 \end{array} \right]\) is a solution to \(A\mathbf{x} = \mathbf{b}\). Find a two-parameter family of solutions to \(A\mathbf{x} = \mathbf{b}\).
In each case write the system in the form \(A\mathbf{x} = \mathbf{b}\), use the gaussian algorithm to solve the system, and express the solution as a particular solution plus a linear combination of basic solutions to the associated homogeneous system \(A\mathbf{x} = \mathbf{0}\).
- =1pt \(\begin{array}[t]{rrrrrrrrrrr} x_{1} & - & 2x_{2} & + & x_{3} & + & 4x_{4} & - & x_{5} & = & 8 \\ -2x_{1} & + & 4x_{2} & + & x_{3} & - & 2x_{4} & - & 4x_{5} & = & -1 \\ 3x_{1} & - & 6x_{2} & + & 8x_{3} & + & 4x_{4} & - & 13x_{5} & = & 1 \\ 8x_{1} & - & 16x_{2} & + & 7x_{3} & + & 12x_{4} & - & 6x_{5} & = & 11 \end{array}\)
- =1pt \(\begin{array}[t]{rrrrrrrrrrr} x_{1} & - & 2x_{2} & + & x_{3} & + & 2x_{4} & + & 3x_{5} & = & -4 \\ -3x_{1} & + & 6x_{2} & - & 2x_{3} & - & 3x_{4} & - & 11x_{5} & = & 11 \\ -2x_{1} & + & 4x_{2} & - & x_{3} & + & x_{4} & - & 8x_{5} & = & 7 \\ -x_{1} & + & 2x_{2} & & & + & 3x_{4} & - & 5x_{5} & = & 3 \end{array}\)
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b. \(\mathbf{x} = \left[ \begin{array}{r} -3 \\ 0 \\ -1 \\ 0 \\ 0 \end{array} \right] + \left( s \left[ \begin{array}{r} 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} -5 \\ 0 \\ 2 \\ 0 \\ 1 \end{array} \right] \right).\)
Given vectors \(\mathbf{a}_{1} = \left[ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right]\),
\(\mathbf{a}_{2} = \left[ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right]\), and \(\mathbf{a}_{3} = \left[ \begin{array}{r} 0 \\ -1 \\ 1 \end{array} \right]\), find a vector \(\mathbf{b}\) that is not a linear combination of \(\mathbf{a}_{1}\), \(\mathbf{a}_{2}\), and \(\mathbf{a}_{3}\). Justify your answer. [Hint: Part (2) of Theorem \(\PageIndex{1}\).]
In each case either show that the statement is true, or give an example showing that it is false.
- \(\left[ \begin{array}{r} 3 \\ 2 \end{array} \right]\) is a linear combination of \(\left[ \begin{array}{r} 1 \\ 0 \end{array} \right]\) and \(\left[ \begin{array}{r} 0 \\ 1 \end{array} \right]\).
- If \(A\mathbf{x}\) has a zero entry, then \(A\) has a row of zeros.
- If \(A\mathbf{x} = \mathbf{0}\) where \(\mathbf{x} \neq \mathbf{0}\), then \(A = 0\).
- Every linear combination of vectors in \(\mathbb{R}^n\) can be written in the form \(A\mathbf{x}\).
- If \(A = \left[ \begin{array}{ccc} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} \end{array} \right]\) in terms of its columns, and if \(\mathbf{b} = 3\mathbf{a}_{1} - 2\mathbf{a}_{2}\), then the system \(A\mathbf{x} = \mathbf{b}\) has a solution.
- If \(A = \left[ \begin{array}{ccc} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} \end{array} \right]\) in terms of its columns, and if the system \(A\mathbf{x} = \mathbf{b}\) has a solution, then \(\mathbf{b} = s\mathbf{a}_{1} + t\mathbf{a}_{2}\) for some \(s\), \(t\).
- If \(A\) is \(m \times n\) and \(m < n\), then \(A\mathbf{x} = \mathbf{b}\) has a solution for every column \(\mathbf{b}\).
- If \(A\mathbf{x} = \mathbf{b}\) has a solution for some column \(\mathbf{b}\), then it has a solution for every column \(\mathbf{b}\).
- If \(\mathbf{x}_{1}\) and \(\mathbf{x}_{2}\) are solutions to \(A\mathbf{x} = \mathbf{b}\), then \(\mathbf{x}_{1} - \mathbf{x}_{2}\) is a solution to \(A\mathbf{x} = \mathbf{0}\).
- Let \(A = \left[ \begin{array}{ccc} \mathbf{a}_{1} & \mathbf{a}_{2} & \mathbf{a}_{3} \end{array}\right]\) in terms of its columns. If \(\mathbf{a}_{3} = s\mathbf{a}_{1} + t\mathbf{a}_{2}\), then \(A\mathbf{x} = \mathbf{0}\), where \(\mathbf{x} = \left[ \begin{array}{c} s \\ t \\ -1 \end{array} \right]\).
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- False. \(\left[ \begin{array}{rr} 1 & 2 \\ 2 & 4 \end{array} \right] \left[ \begin{array}{r} 2 \\ -1 \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \end{array} \right]\).
- True. The linear combination \(x_{1}\mathbf{a}_{1} + \cdots + x_{n}\mathbf{a}_{n}\) equals \(A\mathbf{x}\) where \(A = \left[ \begin{array}{ccc} \mathbf{a}_{1} & \cdots & \mathbf{a}_{n} \end{array} \right]\) by Theorem [thm:002684].
- False. If \(A = \left[ \begin{array}{rrr} 1 & 1 & -1 \\ 2 & 2 & 0 \end{array} \right]\) and \(\mathbf{x} = \left[ \begin{array}{r} 2 \\ 0 \\ 1 \end{array} \right]\), then \[A\mathbf{x} = \left[ \begin{array}{r} 1 \\ 4 \end{array} \right] \neq s \left[ \begin{array}{r} 1 \\ 2 \end{array} \right] + t \left[ \begin{array}{r} 1 \\ 2 \end{array} \right] \mbox{ for any } s \mbox{ and } t. \nonumber \]
- False. If \(A = \left[ \begin{array}{rrr} 1 & -1 & 1 \\ -1 & 1 & -1 \end{array} \right]\), there is a solution for \(\mathbf{b} = \left[ \begin{array}{r} 0 \\ 0 \end{array} \right]\) but not for \(\mathbf{b} = \left[ \begin{array}{r} 1 \\ 0 \end{array} \right]\).
Let \(T : \mathbb{R}^2 \to \mathbb{R}^2\) be a transformation. In each case show that \(T\) is induced by a matrix and find the matrix.
- \(T\) is a reflection in the \(y\) axis.
- \(T\) is a reflection in the line \(y = x\).
- \(T\) is a reflection in the line \(y = -x\).
- \(T\) is a clockwise rotation through \(\frac{\pi}{2}\).
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- Here \(T \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} y \\ x \end{array} \right] = \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right]\).
- Here \(T \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} y \\ -x \end{array} \right] = \left[ \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right]\).
The projection \(P : \mathbb{R}^3 \to \mathbb{R}^2\) is defined by \(P \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{c} x \\ y \end{array} \right]\) for all \(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\) in \(\mathbb{R}^3\). Show that \(P\) is induced by a matrix and find the matrix.
Let \(T : \mathbb{R}^3 \to \mathbb{R}^3\) be a transformation. In each case show that \(T\) is induced by a matrix and find the matrix.
- \(T\) is a reflection in the \(x-y\) plane.
- \(T\) is a reflection in the \(y-z\) plane.
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Here
b. \[T \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{c} -x \\ y \\ z \end{array} \right] = \left[ \begin{array}{rrr} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right], \nonumber \]
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Fix \(a > 0\) in \(\mathbb{R}\), and define \(T_{a} : \mathbb{R}^4 \to \mathbb{R}^4\) by \(T_{a}(\mathbf{x}) = a\mathbf{x}\) for all \(\mathbf{x}\) in \(\mathbb{R}^4\). Show that \(T\) is induced by a matrix and find the matrix. [\(T\) is called a dilation if \(a > 1\) and a contraction if \(a < 1\).]
Let \(A\) be \(m \times n\) and let \(\mathbf{x}\) be in \(\mathbb{R}^n\). If \(A\) has a row of zeros, show that \(A\mathbf{x}\) has a zero entry.
If a vector \(\mathbf{b}\) is a linear combination of the columns of \(A\), show that the system \(A\mathbf{x} = \mathbf{b}\) is consistent (that is, it has at least one solution.)
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Write \(A = \left[ \begin{array}{cccc} \mathbf{a}_{1} & \mathbf{a}_{2} & \cdots & \mathbf{a}_{n} \end{array} \right]\) in terms of its columns. If \(\mathbf{b} = x_{1}\mathbf{a}_{1} + x_{2}\mathbf{a}_{2} + \cdots + x_{n}\mathbf{a}_{n}\) where the \(x_{i}\) are scalars, then \(A\mathbf{x} = \mathbf{b}\) by Theorem [thm:002684] where \(\mathbf{x} = \left[ \begin{array}{cccc} x_{1} & x_{2} & \cdots & x_{n} \end{array} \right]^{T}\). That is, \(\mathbf{x}\) is a solution to the system \(A\mathbf{x} = \mathbf{b}\).
If a system \(A\mathbf{x} = \mathbf{b}\) is inconsistent (no solution), show that \(\mathbf{b}\) is not a linear combination of the columns of \(A\).
Let \(\mathbf{x}_{1}\) and \(\mathbf{x}_{2}\) be solutions to the homogeneous system \(A\mathbf{x} = \mathbf{0}\).
- Show that \(\mathbf{x}_{1} + \mathbf{x}_{2}\) is a solution to \(A\mathbf{x} = \mathbf{0}\).
- Show that \(t\mathbf{x}_{1}\) is a solution to \(A\mathbf{x} = \mathbf{0}\) for any scalar \(t\).
- Answer
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b. By Theorem [thm:002849], \(A(t\mathbf{x}_{1}) = t(A\mathbf{x}_{1}) = t \cdot \mathbf{0} = \mathbf{0}\); that is, \(t\mathbf{x}_{1}\) is a solution to \(A\mathbf{x} = \mathbf{0}\).
Suppose \(\mathbf{x}_{1}\) is a solution to the system \(A\mathbf{x} = \mathbf{b}\). If \(\mathbf{x}_{0}\) is any nontrivial solution to the associated homogeneous system \(A\mathbf{x} = \mathbf{0}\), show that \(\mathbf{x}_{1} + t\mathbf{x}_{0}\), \(t\) a scalar, is an infinite one parameter family of solutions to \(A\mathbf{x} = \mathbf{b}\). [Hint: Example:2.1.7].]
Let \(A\) and \(B\) be matrices of the same size. If \(\mathbf{x}\) is a solution to both the system \(A\mathbf{x} = \mathbf{0}\) and the system \(B\mathbf{x} = \mathbf{0}\), show that \(\mathbf{x}\) is a solution to the system \((A + B)\mathbf{x} = \mathbf{0}\).
If \(A\) is \(m \times n\) and \(A\mathbf{x} = \mathbf{0}\) for every \(\mathbf{x}\) in \(\mathbb{R}^n\), show that \(A = 0\) is the zero matrix. [Hint: Consider \(A\mathbf{e}_{j}\) where \(\mathbf{e}_{j}\) is the \(j\)th column of \(I_{n}\); that is, \(\mathbf{e}_{j}\) is the vector in \(\mathbb{R}^n\) with \(1\) as entry \(j\) and every other entry \(0\).]
Prove part (1) of Theorem \(\PageIndex{2}\).
- Answer
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If \(A\) is \(m \times n\) and \(\mathbf{x}\) and \(\mathbf{y}\) are \(n\)-vectors, we must show that \(A(\mathbf{x} + \mathbf{y}) = A\mathbf{x} + A\mathbf{y}\). Denote the columns of \(A\) by \(\mathbf{a}_{1}, \mathbf{a}_{2}, \dots, \mathbf{a}_{n}\), and write \(\mathbf{x} = \left[ \begin{array}{cccc} x_{1} & x_{2} & \cdots & x_{n} \end{array} \right]^{T}\) and \(\mathbf{y} = \left[ \begin{array}{cccc} y_{1} & y_{2} & \cdots & y_{n} \end{array} \right]^{T}\). Then \(\mathbf{x} + \mathbf{y} = \left[ \begin{array}{cccc} x_{1} + y_{1} & x_{2} + y_{2} & \cdots & x_{n} + y_{n} \end{array} \right]^{T}\), so Definition [def:002068] and Theorem [thm:002170] give \(A(\mathbf{x} + \mathbf{y}) = (x_{1} + y_{1})\mathbf{a}_{1} + (x_{2} + y_{2})\mathbf{a}_{2} + \cdots + (x_{n} + y_{n})\mathbf{a}_{n} = (x_{1}\mathbf{a}_{1} + x_{2}\mathbf{a}_{2} + \cdots + x_{n}\mathbf{a}_{n}) + (y_{1}\mathbf{a}_{1} + y_{2}\mathbf{a}_{2} + \cdots + y_{n}\mathbf{a}_{n}) = A\mathbf{x} + A\mathbf{y}\).
Prove part (2) of Theorem \(\PageIndex{2}\).