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Chapter 2: Q1Q (page 93)

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\( - 2A\), \(B - 2A\), \(AC\), \(CD\).

Short Answer

Expert verified

\( - 2A = \left( {\begin{aligned}{*{20}{c}}{ - 4}&0&2\\{ - 8}&{10}&{ - 4}\end{aligned}} \right)\),

\(B - 2A = \left( {\begin{aligned}{*{20}{c}}3&{ - 5}&3\\{ - 7}&6&{ - 7}\end{aligned}} \right)\),

\(AC\)is not defined and

\(CD = \left( {\begin{aligned}{*{20}{c}}1&{13}\\{ - 7}&{ - 6}\end{aligned}} \right)\).

Step by step solution

01

Find the matrix \( - 2A\)

The value of \( - 2A\) can be calculated as follows:

\(\begin{aligned}{c} - 2A = - 2\left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 4}&0&2\\{ - 8}&{10}&{ - 4}\end{aligned}} \right)\end{aligned}\)

02

Find the matrix \(B - 2A\)

The value of \(B - 2A\) can be calculated as follows:

\(\begin{aligned}{c}B - 2A = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right) - 2\left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}4&0&{ - 2}\\8&{ - 10}&4\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}3&{ - 5}&3\\{ - 7}&6&{ - 7}\end{aligned}} \right)\end{aligned}\)

03

Find the matrix \(AC\)

The order of matrix \(A\) is \(2 \times 3\), and the order of matrix \(C\) is \(2 \times 2\). The number of column of \(A\) does not match with the number of rows of \(C\). Therefore, the matrix \(AC\) is not defined.

04

Find the matrix \(CD\)

The product \(CD\) can be calculated as follows:

\(\begin{aligned}{c}CD = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right) \times \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{1 \times 3 + 2 \times \left( { - 1} \right)}&{1 \times 5 + 2 \times 4}\\{\left( { - 2} \right) \times 3 + 1 \times \left( { - 1} \right)}&{\left( { - 2} \right) \times 5 + 1 \times 4}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{3 - 2}&{5 + 8}\\{ - 6 - 1}&{ - 10 + 4}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&{13}\\{ - 7}&{ - 6}\end{aligned}} \right)\end{aligned}\)

So, \( - 2A = \left( {\begin{aligned}{*{20}{c}}{ - 4}&0&2\\{ - 8}&{10}&{ - 4}\end{aligned}} \right)\), \(B - 2A = \left( {\begin{aligned}{*{20}{c}}3&{ - 5}&3\\{ - 7}&6&{ - 7}\end{aligned}} \right)\), \(AC\) is not defined and \(CD = \left( {\begin{aligned}{*{20}{c}}1&{13}\\{ - 7}&{ - 6}\end{aligned}} \right)\).

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Most popular questions from this chapter

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]

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.

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)

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

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. [Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)].
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