Given matrix A is a 3×3 matrix with ∣A∣=−5.
For an n×n matrix, when multiplied by a scalar k:
det(kA)=kn×det(A)
Since A is a 3×3 matrix, n=3.
Applying the formula:
det(5A)=53×det(A)
Calculate 53:
53=5×5×5=125
Substitute det(A)=−5:
det(5A)=125×(−5)
det(5A)=−625
Therefore, det(5A)=−625.