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Chapter 2: Q9SE (page 93)

Suppose \[AB = \left[ {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{4}}\\{ - {\bf{2}}}&{\bf{3}}\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}{\bf{7}}&{\bf{3}}\\{\bf{2}}&{\bf{1}}\end{array}} \right]\]. Find A.

Short Answer

Expert verified

The matrix \[A = \left[ {\begin{array}{*{20}{c}}{ - 3}&{13}\\{ - 8}&{27}\end{array}} \right]\].

Step by step solution

01

Use the inverse formula

\[{\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]^{ - 1}} = \frac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]when \[ad - bc \ne 0\].

02

Find the inverse of B

\[\begin{array}{c}{B^{ - 1}} = {\left[ {\begin{array}{*{20}{c}}7&3\\2&1\end{array}} \right]^{ - 1}}\\ = \frac{1}{{7\left( 1 \right) - 3\left( 2 \right)}}\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\\ = \frac{1}{{7 - 6}}\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\\ = 1\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\\{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\end{array}\]

03

Find A

Right multiply \[{B^{ - 1}}\] on both sides of \[AB = \left[ {\begin{array}{*{20}{c}}7&3\\2&1\end{array}} \right]\]. Therefore,

\[\begin{array}{c}AB{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}5&4\\{ - 2}&3\end{array}} \right]{B^{ - 1}}\\A = \left[ {\begin{array}{*{20}{c}}5&4\\{ - 2}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{5 - 8}&{ - 15 + 28}\\{ - 2 - 6}&{6 + 21}\end{array}} \right]\\A = \left[ {\begin{array}{*{20}{c}}{ - 3}&{13}\\{ - 8}&{27}\end{array}} \right]\end{array}\]

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Most popular questions from this chapter

If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{9}}&{ - {\bf{1}}}&{\bf{3}}\\{ - {\bf{8}}}&{\bf{7}}&{ - {\bf{6}}}\\{ - {\bf{4}}}&{\bf{1}}&{\bf{8}}\end{aligned}} \right)\)

In exercises 11 and 12, mark each statement True or False. Justify each answer.

a. The definition of the matrix-vector product \(A{\bf{x}}\) is a special case of block multiplication.

b. If \({A_{\bf{1}}}\), \({A_{\bf{2}}}\), \({B_{\bf{1}}}\), and \({B_{\bf{2}}}\) are \(n \times n\) matrices, \[A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}\\{{A_{\bf{2}}}}\end{array}} \right]\] and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), then the product \(BA\) is defined, but \(AB\) is not.

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.

a. Verify that \({A^2} = I\) when \(A = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\).

b. Use partitioned matrices to show that \({M^2} = I\) when\(M = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\).
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