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Chapter 2: Q37Q (page 93)

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{1}}&{\bf{3}}\\{\bf{1}}&{\bf{5}}\end{aligned}} \right)\). Construct a \({\bf{2}} \times {\bf{3}}\) matrix \(C\) (by trial and error) using only \({\bf{1}}\), \( - {\bf{1}}\), and \({\bf{0}}\) as enteries, such that \(CA = {I_{\bf{2}}}\). Compute \(AC\) and note that \(AC \ne {I_{\bf{3}}}\).

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}1&1&{ - 1}\\{ - 1}&1&0\end{aligned}} \right)\)

Step by step solution

01

Construct matrix \(C\) using the elements \({\bf{1}}\), \( - {\bf{1}}\), and \({\bf{0}}\)

As the product \(CA\) is an identity matrix of \(2 \times 2\), the first row of \(C\) is \(\left( {\begin{aligned}{*{20}{c}}1&1&{ - 1}\end{aligned}} \right)\).

When \(A\) is multiplied by the first row of \(C\), the first row of \(CA\) becomes \(\left( {\begin{aligned}{*{20}{c}}1&0\end{aligned}} \right)\).

02

Construct matrix \(C\) using the elements \({\bf{1}}\), \( - {\bf{1}}\), and \({\bf{0}}\)

The second row of \(C\) is \(\left( {\begin{aligned}{*{20}{c}}{ - 1}&1&0\end{aligned}} \right)\).

When \(A\) is multiplied by the second row of \(C\), the second row of \(CA\) becomes \(\left( {\begin{aligned}{*{20}{c}}0&1\end{aligned}} \right)\).

So, matrix \(C\) is \(\left( {\begin{aligned}{*{20}{c}}1&1&{ - 1}\\{ - 1}&1&0\end{aligned}} \right)\).

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Most popular questions from this chapter

If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

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.

Let Ube the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of Ulists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).

Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.

Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?
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