Question

Find the determinants in Exercises 5-10 by row reduction to echelon form.

\(\left| {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{3}}&{ - {\bf{3}}}\\{\bf{3}}&{\bf{4}}&{ - {\bf{4}}}\\{\bf{2}}&{ - {\bf{3}}}&{ - {\bf{5}}}\end{aligned}} \right|\)

Short answer

The value of the determinant is \( - 24\).

Step-by-step solution

Apply the row operation on the determinant

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

Divide row 1 by 3, i.e., \({R_1} \to \frac{1}{3}{R_1}\).

\(3\left| {\begin{aligned}{*{20}{c}}1&1&{ - 1}\\3&4&{ - 4}\\2&{ - 3}&{ - 5}\end{aligned}} \right|\)

At row 3, multiply row 1 by 2 and subtract it from row 3, i.e., \({R_3} \to {R_3} - 2{R_1}\).

\(3\left| {\begin{aligned}{*{20}{c}}1&1&{ - 1}\\3&4&{ - 4}\\0&{ - 5}&{ - 3}\end{aligned}} \right|\)

Apply the row operation on the determinant

At row 2, multiply row 1 by 3 and subtract it from row 2, i.e., \({R_2} \to {R_2} - 3{R_1}\).

\(3\left| {\begin{aligned}{*{20}{c}}1&1&{ - 1}\\0&1&{ - 1}\\0&{ - 5}&{ - 3}\end{aligned}} \right|\)

Apply the row operation on the determinant

At row 3, multiply row 2 by 5 and add it to row 3, i.e., \({R_3} \to {R_3} + 5{R_2}\).

\(3\left| {\begin{aligned}{*{20}{c}}1&1&{ - 1}\\0&1&{ - 1}\\0&0&{ - 8}\end{aligned}} \right|\)

Find the value of the determinant

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

\(\begin{aligned}{c}\det = 3\left( 1 \right)\left( 1 \right)\left( { - 8} \right)\\ = - 24\end{aligned}\)

So, the value of the determinant is \( - 24\).