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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 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.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{9}}&{ - {\bf{1}}}&{\bf{3}}\\{ - {\bf{8}}}&{\bf{7}}&{ - {\bf{6}}}\\{ - {\bf{4}}}&{\bf{1}}&{\bf{8}}\end{aligned}} \right)\)

Suppose \(AD = {I_m}\) (the \(m \times m\) identity matrix). Show that for any b in \({\mathbb{R}^m}\), the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) has a solution. (Hint: Think about the equation \(AD{\mathop{\rm b}\nolimits} = {\mathop{\rm b}\nolimits} \).) Explain why Acannot have more rows than columns.

If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.

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.)

Suppose a linear transformation \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) has the property that \(T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\) for some pair of distinct vectors u and v in \({\mathbb{R}^n}\). Can Tmap \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\)? Why or why not?

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