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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 # 3. Find the inverse of the matrix \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{5}}\\{ - {\bf{7}}}&{ - {\bf{5}}}\end{aligned}} \right).

The inverse of \left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right) is \left( {\begin{aligned}{*{20}{c}}1&1\\{ - 1.4}&{ - 1.6}\end{aligned}} \right).

See the step by step solution

## Step 1: Check if the matrix is invertible

\begin{aligned}{c}\det \left( {\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)} \right) = 8\left( { - 5} \right) - 5\left( { - 7} \right)\\ = - 40 + 35\\\det \left( {\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)} \right) = - 5 \ne 0\end{aligned}

This implies that \left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right) is invertible.

## Step 2: Use the formula

{\left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{aligned}{*{20}{c}}d&{ - b}\\{ - c}&a\end{aligned}} \right) when $$ad - bc \ne 0$$.

## Step 3: Write the inverse matrix

\begin{aligned}{c}{\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)^{ - 1}} = \frac{1}{{ - 5}}\left( {\begin{aligned}{*{20}{c}}{ - 5}&{ - 5}\\7&8\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&1\\{ - \frac{7}{5}}&{ - \frac{8}{5}}\end{aligned}} \right)\\{\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}1&1\\{ - 1.4}&{ - 1.6}\end{aligned}} \right)\end{aligned} ### Want to see more solutions like these? 