Question

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

\(\left| {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{3}}&{ - {\bf{1}}}&{\bf{0}}&{ - {\bf{2}}}\\{\bf{0}}&{\bf{2}}&{ - {\bf{4}}}&{ - {\bf{2}}}&{ - {\bf{6}}}\\{ - {\bf{2}}}&{ - {\bf{6}}}&{\bf{2}}&{\bf{3}}&{{\bf{10}}}\\{\bf{1}}&{\bf{5}}&{ - {\bf{6}}}&{\bf{2}}&{ - {\bf{3}}}\\{\bf{0}}&{\bf{2}}&{ - {\bf{4}}}&{\bf{5}}&{\bf{9}}\end{array}} \right|\)

Short answer

The value of the determinant is 6.

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, subtract row 1 from row 4, i.e., \({R_4} \to {R_4} - {R_1}\).

\[\left| {\begin{array}{*{20}{c}}1&3&{ - 1}&0&{ - 2}\\0&2&{ - 4}&{ - 2}&{ - 6}\\{ - 2}&{ - 6}&2&3&{10}\\0&2&{ - 5}&2&{ - 1}\\0&2&{ - 4}&5&9\end{array}} \right|\]

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

\[\left| {\begin{array}{*{20}{c}}1&3&{ - 1}&0&{ - 2}\\0&2&{ - 4}&{ - 2}&{ - 6}\\0&0&0&3&6\\0&2&{ - 5}&2&{ - 1}\\0&2&{ - 4}&5&9\end{array}} \right|\]

Apply the row operation on the determinant

Take 2 common from row 2.

\[2\left| {\begin{array}{*{20}{c}}1&3&{ - 1}&0&{ - 2}\\0&1&{ - 2}&{ - 1}&{ - 3}\\0&0&0&3&6\\0&2&{ - 5}&2&{ - 1}\\0&2&{ - 4}&5&9\end{array}} \right|\]

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

\[ - 2\left| {\begin{array}{*{20}{c}}1&3&{ - 1}&0&{ - 2}\\0&1&{ - 2}&{ - 1}&{ - 3}\\0&2&{ - 5}&2&{ - 1}\\0&0&0&3&6\\0&2&{ - 4}&5&9\end{array}} \right|\]

Apply the row operation on the determinant

At row 3, multiply row 2 by 2 and subtract it from row 3, i.e., \({R_3} \to {R_3} - 2{R_2}\).

\[ - 2\left| {\begin{array}{*{20}{c}}1&3&{ - 1}&0&{ - 2}\\0&1&{ - 2}&{ - 1}&{ - 3}\\0&0&{ - 1}&4&5\\0&0&0&3&6\\0&2&{ - 4}&5&9\end{array}} \right|\]

At row 5, multiply row 2 by 2 and subtract it from row 5, i.e., \({R_5} \to {R_5} - 2{R_2}\).

\[ - 2\left| {\begin{array}{*{20}{c}}1&3&{ - 1}&0&{ - 2}\\0&1&{ - 2}&{ - 1}&{ - 3}\\0&0&{ - 1}&4&5\\0&0&0&3&6\\0&0&0&7&{15}\end{array}} \right|\]

Divide row 4 by 3, i.e., \({R_4} \to \frac{{{R_4}}}{3}\).

\[ - 6\left| {\begin{array}{*{20}{c}}1&3&{ - 1}&0&{ - 2}\\0&1&{ - 2}&{ - 1}&{ - 3}\\0&0&{ - 1}&4&5\\0&0&0&1&2\\0&0&0&7&{15}\end{array}} \right|\]

Step 4: Apply the row operation on the determinant

Multiply row 3 by \( - 1\), i.e., \({R_3} \to - {R_3}\).

\[6\left| {\begin{array}{*{20}{c}}1&3&{ - 1}&0&{ - 2}\\0&1&{ - 2}&{ - 1}&{ - 3}\\0&0&1&{ - 4}&{ - 5}\\0&0&0&1&2\\0&0&0&7&{15}\end{array}} \right|\]

At row 5, multiply row 4 by 7 and subtract it from row 5, i.e., \({R_5} \to {R_5} - 7{R_4}\).

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

Find the value of the determinant

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

\(\begin{array}{c}\det = 6\left( 1 \right)\left( 1 \right)\left( 1 \right)\left( 1 \right)\left( 1 \right)\\ = 6\end{array}\)

So, the value of the determinant is 6.