Question

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

\(\left| {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{3}}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{5}}&{\bf{4}}\\{ - {\bf{1}}}&{\bf{0}}&{\bf{5}}&{\bf{3}}\\{\bf{3}}&{ - {\bf{3}}}&{ - {\bf{2}}}&{\bf{3}}\end{array}} \right|\)

Short answer

The value of the determinant is \( - 28\).

Step-by-step solution

Apply the row operation on the determinant

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

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

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

At row 3, add rows 1 and 3, i.e., \({R_3} \to {R_3} + {R_1}\).

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

Apply the row operation on the determinant

At row 3, add rows 2 and 3, i.e., \({R_3} \to {R_3} + {R_2}\).

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

Apply the row operation on the determinant

At row 4, subtract row 3 from row 4, i.e., \({R_4} \to {R_4} - {R_3}\).

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

Find the value of the determinant

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

\(\begin{array}{c}\det = \left( 1 \right)\left( 1 \right)\left( 7 \right)\left( { - 4} \right)\\ = - 28\end{array}\)

So, the value of the determinant is \( - 28\).