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Chapter 2: Q16Q (page 93)

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.)

Short Answer

Expert verified

It is proved that Ais invertible.

Step by step solution

01

Condition for an invertible matrix

Theorem 5states that Ais an invertible \(n \times n\) matrix, then for each b in \({\mathbb{R}^n}\), the equation \(Ax = b\) has a unique solution \(x = {A^{ - 1}}b\).

02

Show that A is invertible

Theorem 6states that Aand Bare \(n \times n\) invertible matrices; the inverse of ABis the product of the inverse of A and B in the reverse order. That is, \({\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\).

Consider \(C = AB\).

Multiply each side of the equation \(C = AB\) by \({B^{ - 1}}\):

\(\begin{aligned}{c}C{B^{ - 1}} = AB{B^{ - 1}}\\ = AI\\ = A\end{aligned}\)

According to theorem 6, Ais the product of invertible matrices; thus, Ais invertible.

Hence, it is proved that Ais invertible.

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Most popular questions from this chapter

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).

Show that if the columns of Bare linearly dependent, then so are the columns of AB.

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 \(AB\).

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

Show that \({I_n}A = A\) when \(A\) is \(m \times n\) matrix. (Hint: Use the (column) definition of \({I_n}A\).)

Suppose Ais \(n \times n\) and the equation \(A{\bf{x}} = {\bf{0}}\) has only the trivial solution. Explain why Ahas npivot columns and Ais row equivalent to \({I_n}\). By Theorem 7, this shows that Amust be invertible. (This exercise and Exercise 24 will be cited in Section 2.3.)
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