CUET UG 2025 — Mathematics Algebra

Given $A = \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix}$A=[10​32​] and $B = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}$B=[10​−12​], we need to find matrix $X$X such that $AX = B$AX=B.

From $AX = B$AX=B:

$$X = A^{-1}B$$X=A−1B

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For a 2×2 matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$[ac​bd​], the inverse is:

$$A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$A−1=ad−bc1​[d−c​−ba​]

For matrix $A$A where $a = 1, b = 3, c = 0, d = 2$a=1,b=3,c=0,d=2:

$$ad - bc = (1)(2) - (3)(0)$$ad−bc=(1)(2)−(3)(0)

$$= 2$$=2

Therefore:

$$A^{-1} = \frac{1}{2}\begin{bmatrix} 2 & -3 \\ 0 & 1 \end{bmatrix}$$A−1=21​[20​−31​]

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Now calculate $X = A^{-1}B$X=A−1B:

$$X = \frac{1}{2}\begin{bmatrix} 2 & -3 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}$$X=21​[20​−31​][10​−12​]

Position (1,1): $2(1) + (-3)(0) = 2$2(1)+(−3)(0)=2

Position (1,2): $2(-1) + (-3)(2) = -2 - 6 = -8$2(−1)+(−3)(2)=−2−6=−8

Position (2,1): $0(1) + 1(0) = 0$0(1)+1(0)=0

Position (2,2): $0(-1) + 1(2) = 2$0(−1)+1(2)=2

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$$X = \frac{1}{2}\begin{bmatrix} 2 & -8 \\ 0 & 2 \end{bmatrix}$$X=21​[20​−82​]