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Question 27 of 47

Q.If A=[24−32]A = \begin{bmatrix} 2 & 4 \\ -3 & 2 \end{bmatrix} then, find A−1A^{-1}.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2022Subjective· 2mImportance★★★★★
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Compute ∣A∣=16|A|=16 and the adjoint, then A−1=116[2−432]A^{-1}=\dfrac{1}{16}\begin{bmatrix}2&-4\\3&2\end{bmatrix}.

Given A=[24−32]A = \begin{bmatrix} 2 & 4 \\ -3 & 2 \end{bmatrix}.

Step 1 — determinant:

∣A∣=(2)(2)−(4)(−3)=4+12=16≠0|A| = (2)(2) - (4)(-3) = 4 + 12 = 16 \neq 0

since ∣A∣eq0|A| eq0, the inverse exists.

Step 2 — adjoint. For [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} the adjoint is [d−b−ca]\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}:

adj(A)=[2−432]\text{adj}(A) = \begin{bmatrix} 2 & -4 \\ 3 & 2 \end{bmatrix}

Step 3 — inverse:

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