CUET UG 2025 — Mathematics Algebra

$$A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}, \quad 2A^{-1} = KI - A$$A=[2−4​−37​],2A−1=KI−A

---

$$\det(A) = (2)(7) - (-3)(-4) = 14 - 12 = 2$$det(A)=(2)(7)−(−3)(−4)=14−12=2

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

$$\text{adj}(A) = \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$$adj(A)=[74​32​]

$$A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)$$A−1=det(A)1​⋅adj(A)

$$= \frac{1}{2}\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$$=21​[74​32​]

---

$$2A^{-1} = 2 \times \frac{1}{2}\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$$2A−1=2×21​[74​32​]

$$= \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$$=[74​32​]

---

$$KI - A = K\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$$KI−A=K[10​01​]−[2−4​−37​]

$$= \begin{bmatrix} K-2 & 3 \\ 4 & K-7 \end{bmatrix}$$=[K−24​3K−7​]

---

Setting $2A^{-1} = KI - A$2A−1=KI−A:

$$\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} K-2 & 3 \\ 4 & K-7 \end{bmatrix}$$[74​32​]=[K−24​3K−7​]

From position (1,1): $7 = K - 2 \implies K = 9$7=K−2⟹K=9

From position (2,2): $2 = K - 7 \implies K = 9$2=K−7⟹K=9

---

Therefore, $K = 9$K=9