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Found in: Page 165

### 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

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# Let $$A = \left[ {\begin{array}{*{20}{c}}3&1\\4&2\end{array}} \right]$$. Write $$5A$$. Is $$\det 5A = 5\det A$$?

$$\det 5A \ne 5\det A$$

See the step by step solution

## Step 1: Determine the matrix $$5A$$

Let $$A = \left[ {\begin{array}{*{20}{c}}3&1\\4&2\end{array}} \right]$$.

Compute the matrix $$5A$$ as shown below:

$$\begin{array}{c}5A = 5\left[ {\begin{array}{*{20}{c}}3&1\\4&2\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{15}&5\\{20}&{10}\end{array}} \right]\end{array}$$

## Step 2: Verify whether $$\det 5A = 5\det A$$

The determinant of matrix A is shown below:

$$\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}3&1\\4&2\end{array}} \right|\\ = 6 - 4\\ = 2\end{array}$$

The determinant of matrix $$5A$$ is shown below:

$\begin{array}{c}\det 5A = \left| {\begin{array}{*{20}{c}}{15}&5\\{20}&{10}\end{array}} \right|\\ = 150 - 100\\ = 50\end{array}$

Thus, $$\det 5A \ne 5\det A$$.

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