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Expert-verifiedFound in: Page 93

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Let A be an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).**

**If A is larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).**

The solution \(\begin{aligned}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}\) can be uniquely determined since *A *is an invertible matrix, and \(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{...}&{{b_p}}\end{aligned}} \right)\) must row reduced to \(\left( {\begin{aligned}{*{20}{c}}I&{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\).

It is known that \(A\)** **is a \(m \times n\) matrix. If *B *is a \(n \times p\) matrix with columns \({{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _p}\), the product *AB *is the \(m \times p\) matrix whose columns are \(A{{\mathop{\rm b}\nolimits} _1},...,A{{\mathop{\rm b}\nolimits} _p}\). That is, \(AB = A\left( {\begin{aligned}{*{20}{c}}{{b_1}}&{{b_2}}&{{b_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{A{b_1}}&{A{b_2}}&{A{b_3}}\end{aligned}} \right)\) .

Write the matrix \(B = \left( {\begin{aligned}{*{20}{c}}{{b_1}}&{...}&{{b_p}}\end{aligned}} \right)\) and \(X = \left( {\begin{aligned}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right)\). \(AX = \left( {\begin{aligned}{*{20}{c}}{A{{\mathop{\rm u}\nolimits} _1}}&{...}&{A{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right)\) according to the matrix multiplication. Therefore, the equation \(AX = B\) is equivalent to the *p* system.

\(A{{\mathop{\rm u}\nolimits} _1} = {{\mathop{\rm b}\nolimits} _1},...,A{{\mathop{\rm u}\nolimits} _p} = {{\mathop{\rm b}\nolimits} _p}\)

Since *A *is the coefficient matrix in each system, these systems can be solved simultaneously by placing the augmented columns of each system next to *A *such that \(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{...}&{{b_p}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\).

If *A *is an invertible \(n \times n\) matrix, for each b in \({\mathbb{R}^n}\), the equation \(Ax = b\) has the unique solution \(x = {A^{ - 1}}b\).

The solution \(\begin{aligned}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}\) can be uniquely determined since *A *is an invertible matrix, and \(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{...}&{{b_p}}\end{aligned}} \right)\)must row reduced to \(\left( {\begin{aligned}{*{20}{c}}I&{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\). Thus, \(X\) is a unique solution \({A^{ - 1}}B\) of the equation \(AX = B\).

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