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Question 30 of 37

Q.Find the adjoint of the following matrix : A=[1−1230−2103]A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}

ChseodishaCHSE Odisha Plus Two (Class 12) Commerce Board 2022Subjective· 5mImportance★★★★★est
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adj(A)(A) = transpose of the cofactor matrix of AA.

For A=[1−1230−2103]A=\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}, compute each cofactor Cij=(−1)i+jMijC_{ij}=(-1)^{i+j}M_{ij}.

C11=+∣0−203∣=0,C12=−∣3−213∣=−(9+2)=−11,C13=+∣3010∣=0.C_{11}=+\begin{vmatrix}0&-2\\0&3\end{vmatrix}=0,\quad C_{12}=-\begin{vmatrix}3&-2\\1&3\end{vmatrix}=-(9+2)=-11,\quad C_{13}=+\begin{vmatrix}3&0\\1&0\end{vmatrix}=0.

C21=−∣−1203∣=−(−3)=3,C22=+∣1213∣=3−2=1,C23=−∣1−110∣=−(0+1)=−1.C_{21}=-\begin{vmatrix}-1&2\\0&3\end{vmatrix}=-(-3)=3,\quad C_{22}=+\begin{vmatrix}1&2\\1&3\end{vmatrix}=3-2=1,\quad C_{23}=-\begin{vmatrix}1&-1\\1&0\end{vmatrix}=-(0+1)=-1.

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