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If A=[203113040]A = \begin{bmatrix} 2 & 0 & 3 \\ -1 & 1 & 3 \\ 0 & -4 & 0 \end{bmatrix}, then the value of det (2A) is

Solution

Correct Option: 3

For any n×nn \times n matrix: det(kA)=kn×det(A)\det(kA) = k^n \times \det(A)

Since AA is a 3×33 \times 3 matrix:

det(2A)=23×det(A)\det(2A) = 2^3 \times \det(A)

det(2A)=8×det(A)\det(2A) = 8 \times \det(A)


Given: A=[203113040]A = \begin{bmatrix} 2 & 0 & 3 \\ -1 & 1 & 3 \\ 0 & -4 & 0 \end{bmatrix}

Expanding along Row 3 (which contains the most zeros):

det(A)=0C31(4)C32+0C33\det(A) = 0 \cdot C_{31} - (-4) \cdot C_{32} + 0 \cdot C_{33}

det(A)=4M32\det(A) = 4 \cdot M_{32}

where M32M_{32} is the minor obtained by removing Row 3 and Column 2.


M32=2313M_{32} = \begin{vmatrix} 2 & 3 \\ -1 & 3 \end{vmatrix}

M32=(2)(3)(3)(1)M_{32} = (2)(3) - (3)(-1)

M32=6(3)M_{32} = 6 - (-3)

M32=6+3M_{32} = 6 + 3

M32=9M_{32} = 9


det(A)=4×9\det(A) = 4 \times 9

det(A)=36\det(A) = 36


det(2A)=8×det(A)\det(2A) = 8 \times \det(A)

det(2A)=8×36\det(2A) = 8 \times 36

det(2A)=288\det(2A) = 288

Therefore, the value of det(2A)=288\det(2A) = 288.

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