Question

Find the value of det(3A) for the following matrix:

\({\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 4&7&1\\ { - 1}&3&2\\ { - 2}&0&5 \end{array}} \right]\)

1. 1458
2. 81
3. 27
4. 1971
5. None of these

Answer 1

Correct Answer - Option 4 : 1971

Concept:

1. Determinant of a 3 × 3 matrix:

• Let A be a 3 × 3 matrix given by:

\({\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{b}}&{\rm{c}}\\ {\rm{f}}&{\rm{e}}&{\rm{d}}\\ {\rm{g}}&{\rm{h}}&{\rm{i}} \end{array}} \right]\)

then the value of |A| also written as det(A) is:

det (A) = a (ei - dh) – b (fi - dg) + c (fh - eg)

2. Property of determinant of a matrix:

• Let A be a matrix of order n × n and det(A) = k. Then for a scaler c, the following property holds:

          det(cA) = cⁿ det(A)

Calculation:

First evaluate the determinant of the given matrix:

det(A) = 4(15 - 0) – 7(-5 + 4) + 1(0 + 6)

= 4(15) -7(-1) + 1(6)

= 60 + 7 + 6

= 73

Now using the property the value of det(3A) is:

det(3A) = 3³ det(A)

= 27 × 73

= 1971