CUET UG 2025 — Mathematics Arithmetic

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.

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Given: $A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}$A=​324​−21−3​3−12​​

Using first row expansion:

$$|A| = 3 \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} - (-2) \begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix}$$∣A∣=3​1−3​−12​​−(−2)​24​−12​​+3​24​1−3​​

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For the first term:

$$3 \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} = 3(1 \times 2 - (-1) \times (-3))$$3​1−3​−12​​=3(1×2−(−1)×(−3))

$$= 3(2 - 3)$$=3(2−3)

$$= 3(-1)$$=3(−1)

$$= -3$$=−3

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For the second term:

$$2 \begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} = 2(2 \times 2 - (-1) \times 4)$$2​24​−12​​=2(2×2−(−1)×4)

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

$$= 2(8)$$=2(8)

$$= 16$$=16

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For the third term:

$$3 \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} = 3(2 \times (-3) - 1 \times 4)$$3​24​1−3​​=3(2×(−3)−1×4)

$$= 3(-6 - 4)$$=3(−6−4)

$$= 3(-10)$$=3(−10)

$$= -30$$=−30

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$$|A| = -3 + 16 + (-30)$$∣A∣=−3+16+(−30)

$$= -17$$=−17

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

(adj A)A $= -17I$=−17I