91Ó°ÊÓ

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

At row 4, multiply row 1 by 3 and add it to row 4, i.e., \({R_4} \to {R_4} + 3{R_1}\).

\[\left| {\begin{aligned}{*{20}{c}}1&3&2&{ - 4}\\0&1&2&{ - 5}\\2&7&6&{ - 3}\\0&{ - 1}&{ - 1}&{ - 10}\end{aligned}} \right|\]

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

\[\left| {\begin{aligned}{*{20}{c}}1&3&2&{ - 4}\\0&1&2&{ - 5}\\0&1&2&5\\0&{ - 1}&{ - 1}&{ - 10}\end{aligned}} \right|\]

Apply the row operation \({R_3} \to {R_3} - {R_2}\).

\[\left| {\begin{aligned}{*{20}{c}}1&3&2&{ - 4}\\0&1&2&{ - 5}\\0&0&0&{10}\\0&{ - 1}&{ - 1}&{ - 10}\end{aligned}} \right|\]

At row 4, add rows 4 and 2, i.e., \({R_4} \to {R_4} + {R_2}\).

\[\left| {\begin{aligned}{*{20}{c}}1&3&2&{ - 4}\\0&1&2&{ - 5}\\0&0&0&{10}\\0&0&1&{ - 15}\end{aligned}} \right|\]

Interchange rows 3 and 4, i.e., \({R_3} \leftrightarrow {R_4}\).

\[ - \left| {\begin{aligned}{*{20}{c}}1&3&2&{ - 4}\\0&1&2&{ - 5}\\0&0&1&{ - 15}\\0&0&0&{10}\end{aligned}} \right|\]

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

\(\begin{aligned}{c}\det = - \left( 1 \right)\left( 1 \right)\left( 1 \right)\left( {10} \right)\\ = - 10\end{aligned}\)

So, the value of the determinant is \( - 10\).