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Chapter 2: Q20Q (page 93)

Suppose A, B, and Care \(n \times n\) matrices with A, X, and \(A - AX\) invertible, and suppose

\({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) 鈥�(3)

  1. Explain why B is invertible.
  2. Solve (3) for X. If you need to invert a matrix, explain why that matrix is invertible.

Short Answer

Expert verified
  1. Bis invertible since it is the product of two invertible matrices.
  2. \(X = {\left( {A + {B^{ - 1}}} \right)^{ - 1}}A\).

Step by step solution

01

Explanation of B is invertible

(a)

Multiply each side of the equation \({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) by \(X\):

\(\begin{aligned}{c}{\left( {A - AX} \right)^{ - 1}}X = X{X^{ - 1}}B\\{A^{ - 1}}X - {X^{ - 1}}{A^{ - 1}}X = IB\\{A^{ - 1}}X - {X^{ - 1}}{A^{ - 1}}X = B\end{aligned}\)

It is the product of two invertible matrices. Thus, Bis invertible

02

Solve the equation \({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) for X

(b)

Use theorem 6 about the inverse of a product to invert each side of the equation \({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\):

\(\begin{aligned}{c}A - AX = {\left( {{X^{ - 1}}B} \right)^{ - 1}}\\ = {B^{ - 1}}{\left( {{X^{ - 1}}} \right)^{ - 1}}\\ = {B^{ - 1}}X\end{aligned}\)

Then,

\(\begin{aligned}{c}A - AX = {B^{ - 1}}X\\A = AX + {B^{ - 1}}X\\A = \left( {A + {B^{ - 1}}} \right)X\end{aligned}\)

The product \(\left( {A + {B^{ - 1}}} \right)X\) is invertible since Ais invertible. Xis invertible since the other factor \(\left( {A + {B^{ - 1}}} \right)\) is invertible. Therefore,

\(\begin{aligned}{c}{\left( {A + {B^{ - 1}}} \right)^{ - 1}}A = {\left( {A + {B^{ - 1}}} \right)^{ - 1}}\left( {A + {B^{ - 1}}} \right)X\\ = IX\\ = X\end{aligned}\)

Thus, \(X = {\left( {A + {B^{ - 1}}} \right)^{ - 1}}A\).

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Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

Suppose block matrix \(A\) on the left side of (7) is invertible and \({A_{{\bf{11}}}}\) is invertible. Show that the Schur component \(S\) of \({A_{{\bf{11}}}}\) is invertible. [Hint: The outside factors on the right side of (7) are always invertible. Verify this.] When \(A\) and \({A_{{\bf{11}}}}\) are invertible, (7) leads to a formula for \({A^{ - {\bf{1}}}}\), using \({S^{ - {\bf{1}}}}\) \(A_{{\bf{11}}}^{ - {\bf{1}}}\), and the other entries in \(A\).

In exercise 11 and 12, mark each statement True or False.Justify each answer.

a. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}&{{A_{\bf{2}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), with \({A_{\bf{1}}}\) and \({A_{\bf{2}}}\) the same sizes as \({B_{\bf{1}}}\) and \({B_{\bf{2}}}\), respectively then \(A + B = \left[ {\begin{array}{*{20}{c}}{{A_1} + {B_1}}&{{A_{\bf{2}}} + {B_{\bf{2}}}}\end{array}} \right]\).

b. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_{\bf{2}}}}\end{array}} \right]\), then the partitions of \(A\) and \(B\) are comfortable for block multiplication.

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5鈥�8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.
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