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Q.Answer both

(a) and
(b) :
(a) [2 marks] If A=(3175)A=\begin{pmatrix}3 & 1 \\ 7 & 5\end{pmatrix}, I=(1001)I=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}, then find xx and yy such that A2+xI=yAA^{2}+xI=yA.
(b) [2 marks] If AA and BB are invertible matrices of the same order, then prove that (AB)−1=B−1A−1(AB)^{-1}=B^{-1}A^{-1}.
Assam AhsecAHSEC Higher Secondary (HS) Final Examination 2026Subjective· 4mImportance★★★★★
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(a) A2+xI=yA⇒x=8,y=8A^2+xI=yA\Rightarrow x=8,y=8; (b) (AB)−1=B−1A−1(AB)^{-1}=B^{-1}A^{-1} proved. (OR: (c) skew-symmetric diagonals are zero; (d) x=5x=5 or −3-3.)

Main (a). With A=(3175)A=\begin{pmatrix}3&1\\7&5\end{pmatrix},

A2=(3175)(3175)=(1685632).A^{2}=\begin{pmatrix}3&1\\7&5\end{pmatrix}\begin{pmatrix}3&1\\7&5\end{pmatrix}=\begin{pmatrix}16&8\\56&32\end{pmatrix}.

Then A2+xI=(16+x85632+x)A^{2}+xI=\begin{pmatrix}16+x&8\\56&32+x\end{pmatrix} and yA=(3yy7y5y)yA=\begin{pmatrix}3y&y\\7y&5y\end{pmatrix}.

Comparing the (1,2)(1,2) entries: 8=y8=y. Comparing (1,1)(1,1): 16+x=3y=24⇒x=816+x=3y=24\Rightarrow x=8. (Check (2,2)(2,2): 32+8=40=5⋅832+8=40=5\cdot 8.) So x=8, y=8x=8,\ y=8.

Main (b). To prove (AB)−1=B−1A−1(AB)^{-1}=B^{-1}A^{-1}, verify it is the inverse of ABAB:

(AB)(B−1A−1)=A(BB−1)A−1=A I A−1=AA−1=I,(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=A\,I\,A^{-1}=AA^{-1}=I,

(B−1A−1)(AB)=B−1(A−1A)B=B−1IB=I.(B^{-1}A^{-1})(AB)=B^{-1}(A^{-1}A)B=B^{-1}IB=I.

Since B−1A−1B^{-1}A^{-1} multiplies ABAB to II on both sides, (AB)−1=B−1A−1(AB)^{-1}=B^{-1}A^{-1}.

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