Question

If \(A = \left[ {\begin{array}{*{20}{c}} {11}&{20}\\ {31}&{41} \end{array}} \right]\) and |3A| = k then find the value of k?
1. 1600
2. 2560
3. -1521
4. None of these

Answer 1

Correct Answer - Option 3 : -1521

CONCEPT:

• If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)
• If A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

CALCULATION:

Given: \(A = \left[ {\begin{array}{*{20}{c}} {11}&{20}\\ {31}&{41} \end{array}} \right]\) and |3A| = k |A|

As we know that, if \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)

⇒ |A| = (41 × 11) - (20 × 31) = 451 - 620 = - 169

As we know that, if A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

⇒ |3A| = 3² × - 169 = - 1521

Hence, the correct option is 3.