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If A=[3−2321−14−32]A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}, then the matrix (adj A)A is equal to

Solution

✅ Correct Option: 2

The property of adjoint matrices states that:

(adj A)A = |A|I

where |A| is the determinant of matrix A and I is the identity matrix.

Therefore, only the determinant of A is needed.


Given: A=[3−2321−14−32]A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}

Using first row expansion:

∣A∣=3∣1−1−32∣−(−2)∣2−142∣+3∣214−3∣|A| = 3 \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} - (-2) \begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix}


For the first term:

3∣1−1−32∣=3(1×2−(−1)×(−3))3 \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} = 3(1 \times 2 - (-1) \times (-3))

=3(2−3)= 3(2 - 3)

=3(−1)= 3(-1)

=−3= -3


For the second term:

2∣2−142∣=2(2×2−(−1)×4)2 \begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} = 2(2 \times 2 - (-1) \times 4)

=2(4+4)= 2(4 + 4)

=2(8)= 2(8)

=16= 16


For the third term:

3∣214−3∣=3(2×(−3)−1×4)3 \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} = 3(2 \times (-3) - 1 \times 4)

=3(−6−4)= 3(-6 - 4)

=3(−10)= 3(-10)

=−30= -30


∣A∣=−3+16+(−30)|A| = -3 + 16 + (-30)

=−17= -17


Using the property (adj A)A = |A|I:

(adj A)A =−17I= -17I

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