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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

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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

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

Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).

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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