CUET UG 2025 — Mathematics Algebra

Given $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$A=[25​3−2​] and $A^{-1} = KA$A−1=KA.

For any matrix $A$A, we have $A \cdot A^{-1} = I$A⋅A−1=I where $I$I is the identity matrix.

Substituting $A^{-1} = KA$A−1=KA:

$$A \cdot (KA) = I$$A⋅(KA)=I

$$K(A \cdot A) = I$$K(A⋅A)=I

$$KA^2 = I$$KA2=I

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Calculate $A^2$A2:

$$A^2 = A \times A$$A2=A×A

$$A^2 = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$$A2=[25​3−2​]×[25​3−2​]

For position (1,1): $(2)(2) + (3)(5) = 4 + 15 = 19$(2)(2)+(3)(5)=4+15=19

For position (1,2): $(2)(3) + (3)(-2) = 6 - 6 = 0$(2)(3)+(3)(−2)=6−6=0

For position (2,1): $(5)(2) + (-2)(5) = 10 - 10 = 0$(5)(2)+(−2)(5)=10−10=0

For position (2,2): $(5)(3) + (-2)(-2) = 15 + 4 = 19$(5)(3)+(−2)(−2)=15+4=19

$$A^2 = \begin{bmatrix} 19 & 0 \\ 0 & 19 \end{bmatrix}$$A2=[190​019​]

$$A^2 = 19\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$A2=19[10​01​]

$$A^2 = 19I$$A2=19I

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From $KA^2 = I$KA2=I and $A^2 = 19I$A2=19I:

$$K(19I) = I$$K(19I)=I

$$19KI = I$$19KI=I

$$K = \frac{1}{19}$$K=191​

Therefore, the value of $K$K is $\frac{1}{19}$191​.