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Exercise 2.2 · Q13

Q.Find the matrix of co-factors for the following matrix. [102−21303−5]\begin{bmatrix} 1 & 0 & 2 \\ -2 & 1 & 3 \\ 0 & 3 & -5 \end{bmatrix}

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Step 1: A=[102−21303−5]A=\begin{bmatrix} 1 & 0 & 2 \\ -2 & 1 & 3 \\ 0 & 3 & -5 \end{bmatrix}.

Step 2: M11=∣133−5∣=−5−9=−14⇒A11=−14M_{11}=\begin{vmatrix}1&3\\3&-5\end{vmatrix}=-5-9=-14\Rightarrow A_{11}=-14. M12=∣−230−5∣=10−0=10⇒A12=−10M_{12}=\begin{vmatrix}-2&3\\0&-5\end{vmatrix}=10-0=10\Rightarrow A_{12}=-10. M13=∣−2103∣=−6−0=−6⇒A13=−6M_{13}=\begin{vmatrix}-2&1\\0&3\end{vmatrix}=-6-0=-6\Rightarrow A_{13}=-6.

Step 3: M21=∣023−5∣=0−6=−6⇒A21=(−1)3(−6)=6M_{21}=\begin{vmatrix}0&2\\3&-5\end{vmatrix}=0-6=-6\Rightarrow A_{21}=(-1)^3(-6)=6. M22=∣120−5∣=−5−0=−5⇒A22=−5M_{22}=\begin{vmatrix}1&2\\0&-5\end{vmatrix}=-5-0=-5\Rightarrow A_{22}=-5. M23=∣1003∣=3−0=3⇒A23=(−1)5(3)=−3M_{23}=\begin{vmatrix}1&0\\0&3\end{vmatrix}=3-0=3\Rightarrow A_{23}=(-1)^5(3)=-3. …

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