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Chapter 2: Q27Q (page 93)

Show that if ABis invertible, so is A. You cannot use Theorem 6(b), because you cannot assumethat Aand Bare invertible. (Hint:There is a matrix Wsuch that \(ABW = I\). Why?)

Short Answer

Expert verified

BothABand A are invertible.

Step by step solution

01

Write the algorithm for obtaining \({A^{ - 1}}\)

The inverse of an\(m \times m\)matrix A can be computed by using the augmented matrix\(\left( {\begin{array}{*{20}{c}}A&I\end{array}} \right)\), where\(I\)is the identity matrix. Matrix Ahas an inverse only if \(\left( {\begin{array}{*{20}{c}}A&I\end{array}} \right)\) is row equivalent to \(\left( {\begin{array}{*{20}{c}}I&{{A^{ - 1}}}\end{array}} \right)\).

02

Show that A is invertible

The product AB is invertible, so there should be an inverse of matrix AB. Let W be the inverse matrix of AB.

Then, it can be represented as shown below:

\(\begin{array}{c}\left( {AB} \right)W = I\\A\left( {BW} \right) = I\end{array}\)

The equation\(A\left( {BW} \right) = I\) shows that matrix\(BW\)is the inverse of matrix A.

Therefore, A is invertible.

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Most popular questions from this chapter

If A, B, and X are \(n \times n\) invertible matrices, does the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = {I_n}\) have a solution, X? If so, find it.

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 Ais a \(3 \times n\) matrix whose columns span \({\mathbb{R}^3}\). Explain how to construct an \(n \times 3\) matrix Dsuch that \(AD = {I_3}\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?
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