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Q5E

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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

# Find the determinants in Exercises 5-10 by row reduction to echelon form.$$\left| {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{ - {\bf{4}}}\\{ - {\bf{1}}}&{ - {\bf{4}}}&{\bf{5}}\\{ - {\bf{2}}}&{ - {\bf{8}}}&{\bf{7}}\end{array}} \right|$$

The value of the determinant is $$- 3$$.

See the step by step solution

## Step 1: Apply the row operation on the determinant

Apply the row operation to reduce the determinant into the echelon form.

At row 3, multiply row 1 by 2 and add it to row 3, i.e., $${R_3} \to {R_3} + 2{R_1}$$.

$$\left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\{ - 1}&{ - 4}&5\\0&2&{ - 1}\end{array}} \right|$$

## Step 2: Apply the row operation on the determinant

At row 2, add rows 1 and 2, i.e., $${R_2} \to {R_2} + {R_1}$$.

$$\left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\0&1&1\\0&2&{ - 1}\end{array}} \right|$$

## Step 3: Apply the row operation on the determinant

At row 3, multiply row 2 by 2 and subtract it from row 3.

$$\left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\0&1&1\\0&0&{ - 3}\end{array}} \right|$$

## Step 4: Find the value of the determinant

For a triangular matrix, the determinant is the product of diagonal elements.

$$\begin{array}{c}\det = \left( 1 \right)\left( 1 \right)\left( { - 3} \right)\\ = - 3\end{array}$$

So, the value of the determinant is $$- 3$$.