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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

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

Prove Theorem 2(b) and 2(c). Use the row-column rule. The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\) or \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} \).

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).
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