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Q5E

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Linear Algebra and its Applications
Found in: Page 165
Linear Algebra and its Applications

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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Short Answer

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 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 \( - 3\).

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