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Let A=[aij]3×3A = [a_{ij}]_{3 \times 3} such that ∣A∣=−5|A| = -5. Then the value of det⁡(5A)\det(5A) is equal to

Solution

✅ Correct Option: 4

Given matrix AA is a 3×33 \times 3 matrix with ∣A∣=−5|A| = -5.

For an n×nn \times n matrix, when multiplied by a scalar kk:

det⁡(kA)=kn×det⁡(A)\det(kA) = k^n \times \det(A)


Since AA is a 3×33 \times 3 matrix, n=3n = 3.

Applying the formula:

det⁡(5A)=53×det⁡(A)\det(5A) = 5^3 \times \det(A)


Calculate 535^3:

53=5×5×5=1255^3 = 5 \times 5 \times 5 = 125


Substitute det⁡(A)=−5\det(A) = -5:

det⁡(5A)=125×(−5)\det(5A) = 125 \times (-5)

det⁡(5A)=−625\det(5A) = -625

Therefore, det⁡(5A)=−625\det(5A) = -625.

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