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Miscellaneous Exercise 2(A) · Q66

Q.Find the inverse of [123115247]\begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix} by adjoint method.

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Step 1: A=[123115247]A=\begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix}.

Step 2: Cofactors: A11=(1)(7)−(5)(4)=−13A_{11}=(1)(7)-(5)(4)=-13; wait -- computing carefully: A11=∣1547∣=7−20=−13A_{11}=\begin{vmatrix}1&5\\4&7\end{vmatrix}=7-20=-13; A12=−∣1527∣=−(7−10)=3A_{12}=-\begin{vmatrix}1&5\\2&7\end{vmatrix}=-(7-10)=3; A13=∣1124∣=4−2=2A_{13}=\begin{vmatrix}1&1\\2&4\end{vmatrix}=4-2=2; A21=−∣2347∣=−(14−12)=−2A_{21}=-\begin{vmatrix}2&3\\4&7\end{vmatrix}=-(14-12)=-2; A22=∣1327∣=7−6=1A_{22}=\begin{vmatrix}1&3\\2&7\end{vmatrix}=7-6=1; A23=−∣1224∣=−(4−4)=0A_{23}=-\begin{vmatrix}1&2\\2&4\end{vmatrix}=-(4-4)=0; A31=∣2315∣=10−3=7A_{31}=\begin{vmatrix}2&3\\1&5\end{vmatrix}=10-3=7; A32=−∣1315∣=−(5−3)=−2A_{32}=-\begin{vmatrix}1&3\\1&5\end{vmatrix}=-(5-3)=-2; A33=∣1211∣=1−2=−1A_{33}=\begin{vmatrix}1&2\\1&1\end{vmatrix}=1-2=-1. …

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