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If A is a square matrix of order 3 such that A(adjA)=[−2000−2000−2]A( {adj } A) = \begin{bmatrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix}, then ∣A∣|A| is equal to

Solution

✅ Correct Option: 4

For any square matrix AA of order nn:

A×(adj A)=∣A∣×IA \times (\text{adj } A) = |A| \times I

where ∣A∣|A| is the determinant of matrix AA and II is the identity matrix.


Given that:

A(adj A)=[−2000−2000−2]A(\text{adj } A) = \begin{bmatrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix}

The matrix on the right has −2-2 on all diagonal elements and 00 elsewhere.

[−2000−2000−2]=−2×[100010001]\begin{bmatrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix} = -2 \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

A(adj A)=−2IA(\text{adj } A) = -2I


Using the property A×(adj A)=∣A∣×IA \times (\text{adj } A) = |A| \times I:

∣A∣×I=−2I|A| \times I = -2I

Therefore:

∣A∣=−2|A| = -2

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