Question 11 of 58beginner💡 UnderstandShort Answer2 marks

Write Minors and Cofactors of the elements of following determinants: (i) ∣2−403∣\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix} (ii) ∣acbd∣\begin{vmatrix} a & c \\ b & d \end{vmatrix}

(i)
Answer

For the determinant ∣2−403∣\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}, the Minors are M11=3M_{11} = 3, M12=0M_{12} = 0, M21=−4M_{21} = -4, and M22=2M_{22} = 2. The Cofactors are calculated using Aij=(−1)i+jMijA_{ij} = (-1)^{i+j} M_{ij}. Thus, A11=(−1)1+1(3)=3A_{11} = (-1)^{1+1}(3) = 3, A12=(−1)1+2(0)=0A_{12} = (-1)^{1+2}(0) = 0, A21=(−1)2+1(−4)=4A_{21} = (-1)^{2+1}(-4) = 4, and A22=(−1)2+2(2)=2A_{22} = (-1)^{2+2}(2) = 2.

(ii)
Answer

For the determinant ∣acbd∣\begin{vmatrix} a & c \\ b & d \end{vmatrix}, the Minors are M11=dM_{11} = d, M12=bM_{12} = b, M21=cM_{21} = c, and M22=aM_{22} = a. The Cofactors are A11=(−1)1+1(d)=dA_{11} = (-1)^{1+1}(d) = d, A12=(−1)1+2(b)=−bA_{12} = (-1)^{1+2}(b) = -b, A21=(−1)2+1(c)=−cA_{21} = (-1)^{2+1}(c) = -c, and A22=(−1)2+2(a)=aA_{22} = (-1)^{2+2}(a) = a.

Explanation

The solution follows Definition 1 and Definition 2 provided in the text. The Minor (MijM_{ij}) is found by deleting the respective row and column. The Cofactor (AijA_{ij}) is calculated by multiplying the minor by (−1)i+j(-1)^{i+j}. This method is consistent with Example 9 in the provided context.

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