Chapter 9: Problem 22
\(\left| \begin{array}{rr}{12} & {8} \\ {3} & {2}\end{array}\right|\)
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Chapter 9: Problem 22
\(\left| \begin{array}{rr}{12} & {8} \\ {3} & {2}\end{array}\right|\)
These are the key concepts you need to understand to accurately answer the question.
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Let $$\mathbf{A}=\left[ \begin{array}{rrr}{2} & {-1} & {1} \\ {-1} & {2} & {1} \\ {1} & {1} & {2}\end{array}\right].$$ (a) Show that \(\mathbf{A}\) is singular. (b) Show that \(\mathbf{A} \mathbf{x}=\left[ \begin{array}{c}{3} \\ {1} \\\ {3}\end{array}\right]\) has no solutions. (c) Show that \(\mathbf{A x}=\left[ \begin{array}{c}{3} \\ {0} \\\ {3}\end{array}\right]\) has infinitely many solutions.
\(\mathbf{A}=\left[ \begin{array}{rrr}{2} & {1} & {3} \\ {0} & {2} & {-1} \\\ {0} & {0} & {2}\end{array}\right]\)
\(\mathbf{A}(t)=\left[ \begin{array}{cc}{1} & {e^{-2 t}} \\ {3} & {e^{-2 t}}\end{array}\right], \quad \mathbf{B}(t)=\left[ \begin{array}{cc}{e^{-t}} & {e^{-t}} \\ {-e^{-t}} & {3 e^{-t}}\end{array}\right]\)
\(\left| \begin{array}{rrr}{1} & {0} & {2} \\ {0} & {3} & {-1} \\ {-1} & {2} & {1}\end{array}\right|\)
\(\mathbf{A}=\left[ \begin{array}{ll}{8} & {-4} \\ {4} & {-2}\end{array}\right], \quad \mathbf{f}(t)=\left[ \begin{array}{l}{t^{-2} / 2} \\ {t^{-2}}\end{array}\right]\)
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