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Chapter 2: Q8SE (page 93)

Find a matrix A such that the transformation \(x \mapsto Ax\) maps \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{3}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{7}}\end{aligned}} \right)\) into \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{1}}\end{aligned}} \right)\), respectively. (Hint: Write a matrix equation involving A, and solve for A.)

Short Answer

Expert verified

The matrix \(A = \left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\4&{ - 1}\end{aligned}} \right)\).

Step by step solution

01

Use the matrix multiplication

By definition of matrix multiplication, matrix A satisfies the following:

\(A\left( {\begin{aligned}{*{20}{c}}1&2\\3&7\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3\\1&1\end{aligned}} \right)\)

02

Find the inverse matrix

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

03

Find A

Right multiply \({\left( {\begin{aligned}{*{20}{c}}1&2\\3&7\end{aligned}} \right)^{ - 1}}\) on both sides of \(A\left( {\begin{aligned}{*{20}{c}}1&2\\3&7\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3\\1&1\end{aligned}} \right)\).

\(\begin{aligned}{c}A\left( {\begin{aligned}{*{20}{c}}1&2\\3&7\end{aligned}} \right){\left( {\begin{aligned}{*{20}{c}}1&2\\3&7\end{aligned}} \right)^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}1&3\\1&1\end{aligned}} \right){\left( {\begin{aligned}{*{20}{c}}1&2\\3&7\end{aligned}} \right)^{ - 1}}\\A = \left( {\begin{aligned}{*{20}{c}}1&3\\1&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}7&{ - 2}\\{ - 3}&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{7 - 9}&{ - 2 + 3}\\{7 - 3}&{ - 2 + 1}\end{aligned}} \right)\\A = \left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\4&{ - 1}\end{aligned}} \right)\end{aligned}\)

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Most popular questions from this chapter

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.

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

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1鈥�4.

1. \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\)

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\(M = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\).

Let \(A = \left[ {\begin{array}{*{20}{c}}B&{\bf{0}}\\{\bf{0}}&C\end{array}} \right]\), where \(B\) and \(C\) are square. Show that \(A\)is invertible if an only if both \(B\) and \(C\) are invertible.

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

37. Construct a random \({\bf{4}} \times {\bf{4}}\) matrix Aand test whether \(\left( {A + I} \right)\left( {A - I} \right) = {A^2} - I\). The best way to do this is to compute \(\left( {A + I} \right)\left( {A - I} \right) - \left( {{A^2} - I} \right)\) and verify that this difference is the zero matrix. Do this for three random matrices. Then test \(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^{\bf{2}}}\) the same way for three pairs of random \({\bf{4}} \times {\bf{4}}\) matrices. Report your conclusions.
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