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Show that if ABis invertible, so is B.

Short Answer

Expert verified

Both AB and B 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

Product AB is invertible, so there should be an inverse of matrix AB. Let D be the inverse matrix of AB.

Then, it can be represented as shown below:

\(\begin{array}{c}D\left( {AB} \right) = I\\\left( {DA} \right)B = I\end{array}\)

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

Therefore, B is invertible.

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Most popular questions from this chapter

Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\)as\(n \times 1\)matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is a \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

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Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.

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2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

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