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Expert-verified Found in: Page 93 ### Linear Algebra and its Applications

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).

See the step by step solution

## Step 1: The equation $$AX = B$$ is equivalent to the p systems

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).

## Step 2: The row reduced matrix of \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 $$AX = B$$. ### Want to see more solutions like these? 