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Chapter 2: Q9Q (page 93)

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

Short Answer

Expert verified

\(k = 5\)

Step by step solution

01

Find the product \(AB\)

The product \(AB\) can be calculated as,

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}2&5\\{ - 3}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}4&{ - 5}\\3&k\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{2 \times 4 + 5 \times 3}&{2 \times \left( { - 5} \right) + 5 \times k}\\{\left( { - 3} \right) \times 4 + 1 \times 3}&{\left( { - 3} \right) \times \left( { - 5} \right) + 1 \times k}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{23}&{ - 10 + 5k}\\{ - 9}&{15 + k}\end{aligned}} \right)\end{aligned}\)

02

Find the product \(BA\)

The product \(BA\) can be calculated as,

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}4&{ - 5}\\3&k\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2&5\\{ - 3}&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 2 + \left( { - 5} \right) \times \left( { - 3} \right)}&{4 \times 5 + \left( { - 5} \right) \times 1}\\{3 \times 2 + k \times \left( { - 3} \right)}&{3 \times 5 + k \times 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{23}&{15}\\{6 - 3k}&{15 + k}\end{aligned}} \right)\end{aligned}\)

03

Comparison of products \(AB\) and \(BA\)

On comparing \(AB\) and \(BA\):

\(\left( {\begin{aligned}{*{20}{c}}{23}&{ - 10 + 5k}\\{ - 9}&{15 + k}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{23}&{15}\\{6 - 3k}&{15 + k}\end{aligned}} \right)\)

As the matrices are equal, therefore

\(\begin{aligned}{c} - 9 = 6 - 3k\\3k = 15\\k = 5\end{aligned}\)

And

\(\begin{aligned}{c} - 10 + 5k = 15\\5k = 25\\k = 5\end{aligned}\)

So, the value of \(k\) is 5.

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Most popular questions from this chapter

Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]

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

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.

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