Question

Find the value of det(2²A) for the following matrix:

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

1. 1340
2. -1282
3. -1088
4. 1542

Answer 1

Correct Answer - Option 3 : -1088

Concept:

Property of determinant of a matrix:

Let A be a matrix of order n × n then det(kA) = kn det(A)

Calculation:

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

Here the order of the matrix is 3.

Now,

det(A) = 2(0 - 2) - 3(-2 + 6) + 1(-1 + 0)

= 2(-2) - 3(4) + 1(-1)

= - 4 - 12 - 1

= -17

Now using the property the value of det(2²A) is:

det(2²A) = det(4A) = 43 det(A)

= 64 × -17

= -1088