Chapter 1: Problem 22
Show that if \(A\) is a symmetric nonsingular matrix, then \(A^{-1}\) is also symmetric.
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Chapter 1: Problem 22
Show that if \(A\) is a symmetric nonsingular matrix, then \(A^{-1}\) is also symmetric.
These are the key concepts you need to understand to accurately answer the question.
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Explain why each of the following algebraic rules will not work in general when the real numbers \(a\) and \(b\) are replaced by \(n \times n\) matrices \(A\) and \(B\). (a) \((a+b)^{2}=a^{2}+2 a b+b^{2}\) (b) \((a+b)(a-b)=a^{2}-b^{2}\)
Prove that if \(A\) is row equivalent to \(B\), then \(B\) is row equivalent to \(A\)
Let \(C\) be a nonsymmetric \(n \times n\) matrix. For each of the following, determine whether the given matrix must be symmetric or could be nonsymmetric: (a) \(A=C+C^{T}\) (b) \(B=C-C^{T}\) (c) \(D=C^{T} C\) (d) \(E=C^{T} C-C C^{T}\) (e) \(F=(I+C)\left(I+C^{T}\right)\) (f) \(G=(I+C)\left(I-C^{T}\right)\)
For each of the choices of \(A\) and \(\mathbf{b}\) that follow, determine whether the system \(A \mathbf{x}=\mathbf{b}\) is consistent by examining how b relates to the column vectors of A. Explain your answers in each case. (a) \(A=\left[\begin{array}{rr}2 & 1 \\ -2 & -1\end{array}\right], \quad \mathbf{b}=\left[\begin{array}{l}3 \\ 1\end{array}\right]\) (b) \(A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right], \quad \mathbf{b}=\left[\begin{array}{l}5 \\ 5\end{array}\right]\) (c) \(A=\left[\begin{array}{lll}3 & 2 & 1 \\ 3 & 2 & 1 \\ 3 & 2 & 1\end{array}\right], \quad \mathbf{b}=\left[\begin{array}{r}1 \\ 0 \\\ -1\end{array}\right]\)
For each of the systems of equations that follow, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in row echelon form. Indicate whether the system is consistent. If the system is consistent and involves no free variables, use back substitution to find the unique solution. If the system is consistent and there are free variables, transform it to reduced row echelon form and find all solutions $$\begin{aligned} &\text { (a) } \quad x_{1}-2 x_{2}=3\\\ &2 x_{1}-x_{2}=9 \end{aligned}$$ $$\begin{aligned} &\text { (b) } \quad 2 x_{1}-3 x_{2}=5\\\ &-4 x_{1}+6 x_{2}=8 \end{aligned}$$ $$\begin{aligned} &\text { (c) } \quad x_{1}+x_{2}=0\\\ &\begin{array}{l} 2 x_{1}+3 x_{2}=0 \\ 3 x_{1}-2 x_{2}=0 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (d) } 3 x_{1}+2 x_{2}-x_{3}=4\\\ &\begin{array}{r} x_{1}-2 x_{2}+2 x_{3}=1 \\ 11 x_{1}+2 x_{2}+x_{3}=14 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (e) } 2 x_{1}+3 x_{2}+x_{3}=1\\\ &\begin{array}{r} x_{1}+x_{2}+x_{3}=3 \\ 3 x_{1}+4 x_{2}+2 x_{3}=4 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (f) } \quad x_{1}-x_{2}+2 x_{3}=4\\\ &\begin{array}{l} 2 x_{1}+3 x_{2}-x_{3}=1 \\ 7 x_{1}+3 x_{2}+4 x_{3}=7 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (g) } x_{1}+x_{2}+x_{3}+x_{4}=0\\\ &\begin{array}{l} 2 x_{1}+3 x_{2}-x_{3}-x_{4}=2 \\ 3 x_{1}+2 x_{2}+x_{3}+x_{4}=5 \\ 3 x_{1}+6 x_{2}-x_{3}-x_{4}=4 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (h) } \quad x_{1}-2 x_{2}=3\\\ &\begin{aligned} 2 x_{1}+x_{2} &=1 \\ -5 x_{1}+8 x_{2} &=4 \end{aligned} \end{aligned}$$ $$\begin{aligned} &\text { (i) } \quad-x_{1}+2 x_{2}-x_{3}=2\\\ &\begin{aligned} -2 x_{1}+2 x_{2}+x_{3} &=4 \\ 3 x_{1}+2 x_{2}+2 x_{3} &=5 \\ -3 x_{1}+8 x_{2}+5 x_{3} &=17 \end{aligned} \end{aligned}$$ $$\begin{aligned} &\text { (j) } \quad x_{1}+2 x_{2}-3 x_{3}+x_{4}=1\\\ &\begin{array}{r} -x_{1}-x_{2}+4 x_{3}-x_{4}=6 \\ -2 x_{1}-4 x_{2}+7 x_{3}-x_{4}=1 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (k) } x_{1}+3 x_{2}+x_{3}+x_{4}=3\\\ &\begin{array}{c} 2 x_{1}-2 x_{2}+x_{3}+2 x_{4}=8 \\ x_{1}-5 x_{2}+x_{4}=5 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (I) } \quad x_{1}-3 x_{2}+\quad x_{3}=1\\\ &\begin{aligned} 2 x_{1}+x_{2}-x_{3} &=2 \\ x_{1}+4 x_{2}-2 x_{3} &=1 \\ 5 x_{1}-8 x_{2}+2 x_{3} &=5 \end{aligned} \end{aligned}$$
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