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Give a formula for \({\left( {ABx} \right)^T}\), where \({\bf{x}}\) is a vector and \(A\) and \(B\) are matrices of appropriate sizes.

Short Answer

Expert verified

The formula for \({\left( {ABx} \right)^T}\) is \({\left( {ABx} \right)^T} = {x^T}{B^T}{A^T}\).

Step by step solution

01

Write the transpose property

By the transpose property, \({\left( {AB} \right)^T} = {B^T}{A^T}\).

Here, A and B matrices are of appropriate sizes.

02

Consider x as a matrix

Since every vector is a column vector,

\(\begin{aligned}{c}{\left( {ABx} \right)^T} = {x^T}{\left( {AB} \right)^T}\\ = {x^T}{B^T}{A^T}.\end{aligned}\)

03

Draw of conclusion

Hence, the formula for \({\left( {ABx} \right)^T}\) is \({\left( {ABx} \right)^T} = {x^T}{B^T}{A^T}\).

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Most popular questions from this chapter

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

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

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

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 Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).

Suppose Ais an \(m \times n\) matrix and there exist \(n \times m\) matrices C and D such that \(CA = {I_n}\) and \(AD = {I_m}\). Prove that \(m = n\) and \(C = D\). (Hint: Think about the product CAD.)

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