Question

Find the determinant in Exercise 20, where \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{\bf{d}}&{\bf{e}}&{\bf{f}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{array}} \right| = {\bf{7}}\].

20. \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{{\bf{d}} + {\bf{3g}}}&{{\bf{e}} + {\bf{3h}}}&{{\bf{f}} + {\bf{3i}}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{array}} \right|\]

Short answer

Hence \[\left| {\begin{array}{*{20}{c}}a&b&c\\{d + 3g}&{e + 3h}&{f + 3i}\\g&h&i\end{array}} \right| = 7\].

Step-by-step solution

Reduce the given determinant

At row 2, add \[ - 3\] times row 3 to row 2 to obtain:

\[\left| {\begin{array}{*{20}{c}}a&b&c\\{d + 3g}&{e + 3h}&{f + 3i}\\g&h&i\end{array}} \right| = \left| {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right|\]

Use the given statement

The given statement is \[\left| {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right| = 7\].

Conclusion

Therefore,

\[\left| {\begin{array}{*{20}{c}}a&b&c\\{d + 3g}&{e + 3h}&{f + 3i}\\g&h&i\end{array}} \right| = 7\]