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Q.If A=[231−4]A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} and B=[1−2−13]B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}, then verify that (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.

Madhya Pradesh MpbseMP Board Higher Secondary 2026Subjective· 4mImportance★★★★★
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Compute (AB)−1(AB)^{-1} and B−1A−1B^{-1}A^{-1} separately; they match.

AB=[231−4][1−2−13]=[−155−14]AB=\begin{bmatrix}2&3\\1&-4\end{bmatrix}\begin{bmatrix}1&-2\\-1&3\end{bmatrix}=\begin{bmatrix}-1&5\\5&-14\end{bmatrix}, det⁡(AB)=14−25=−11\det(AB)=14-25=-11.

(AB)−1=1−11[−14−5−5−1]=111[14551].(AB)^{-1}=\dfrac{1}{-11}\begin{bmatrix}-14&-5\\-5&-1\end{bmatrix}=\dfrac{1}{11}\begin{bmatrix}14&5\\5&1\end{bmatrix}.

A−1=1−11[−4−3−12]=111[431−2]A^{-1}=\dfrac{1}{-11}\begin{bmatrix}-4&-3\\-1&2\end{bmatrix}=\dfrac{1}{11}\begin{bmatrix}4&3\\1&-2\end{bmatrix} (since det⁡A=−11\det A=-11); B−1=[3211]B^{-1}=\begin{bmatrix}3&2\\1&1\end{bmatrix} (since det⁡B=1\det B=1).

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