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Exercise 4.1 · Q8

Q.Find the minor and cofactor of element of the determinant D=∣2−1312−1572∣D = \begin{vmatrix} 2 & -1 & 3 \\ 1 & 2 & -1 \\ 5 & 7 & 2 \end{vmatrix}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Step 1: D=∣2−1312−1572∣D = \begin{vmatrix} 2 & -1 & 3 \\ 1 & 2 & -1 \\ 5 & 7 & 2 \end{vmatrix}, so a11=2, a12=−1, a13=3, a21=1, a22=2, a23=−1, a31=5, a32=7, a33=2a_{11}=2,\ a_{12}=-1,\ a_{13}=3,\ a_{21}=1,\ a_{22}=2,\ a_{23}=-1,\ a_{31}=5,\ a_{32}=7,\ a_{33}=2.

Step 2 (Row 1): M11=∣2−172∣=4+7=11M_{11}=\begin{vmatrix}2 & -1\\ 7 & 2\end{vmatrix}=4+7=11, so C11=(+1)(11)=11C_{11}=(+1)(11)=11.

M12=∣1−152∣=2+5=7M_{12}=\begin{vmatrix}1 & -1\\ 5 & 2\end{vmatrix}=2+5=7, so C12=(−1)(7)=−7C_{12}=(-1)(7)=-7.

M13=∣1257∣=7−10=−3M_{13}=\begin{vmatrix}1 & 2\\ 5 & 7\end{vmatrix}=7-10=-3, so C13=(+1)(−3)=−3C_{13}=(+1)(-3)=-3.

Step 3 (Row 2): M21=∣−1372∣=−2−21=−23M_{21}=\begin{vmatrix}-1 & 3\\ 7 & 2\end{vmatrix}=-2-21=-23, so C21=(−1)(−23)=23C_{21}=(-1)(-23)=23.

M22=∣2352∣=4−15=−11M_{22}=\begin{vmatrix}2 & 3\\ 5 & 2\end{vmatrix}=4-15=-11, so C22=(+1)(−11)=−11C_{22}=(+1)(-11)=-11.

M23=∣2−157∣=14+5=19M_{23}=\begin{vmatrix}2 & -1\\ 5 & 7\end{vmatrix}=14+5=19, so C23=(−1)(19)=−19C_{23}=(-1)(19)=-19. …

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