If the cofactors of the elements 3,7 and 6 of | vedclass

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(C) Let the matrix be $A = \begin{bmatrix} 1 & 2 & 3 \ 4 & -1 & 7 \ 2 & 4 & 6 \end{bmatrix}$.Elements $3, 7, 6$ are in the third column $(C_3)$.Cofa

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(C) Let the matrix be $A = \begin{bmatrix} 1 & 2 & 3 \ 4 & -1 & 7 \ 2 & 4 & 6 \end{bmatrix}$.Elements $3, 7, 6$ are in the third column $(C_3)$.Cofactor $a$ of element $3$ $(A_{13})$: $a = (-1)^{1+3} \begin{vmatrix} 4 & -1 \ 2 & 4 \end{vmatrix} = (16 - (-2)) = 18$.Cofactor $b$ of element $7$ $(A_{23})$: $b = (-1)^{2+3} \begin{vmatrix} 1 & 2 \ 2 & 4 \end{vmatrix} = -(4 - 4) = 0$.Cofactor $c$ of element $6$ $(A_{33})$: $c = (-1)^{3+3} \begin{vmatrix} 1 & 2 \ 4 & -1 \end{vmatrix} = (-1 - 8) = -9$.We need to calculate $\begin{bmatrix} a & b & c \end{bmatrix} \begin{bmatrix} 1 \ 4 \ 2 \end{bmatrix} + \begin{bmatrix} a & b & c \end{bmatrix} \begin{bmatrix} 3 \ 7 \ 6 \end{bmatrix}$.This is equal to $\begin{bmatrix} a & b & c \end{bmatrix} \left( \begin{bmatrix} 1 \ 4 \ 2 \end{bmatrix} + \begin{bmatrix} 3 \ 7 \ 6 \end{bmatrix} ight) = \begin{bmatrix} 18 & 0 & -9 \end{bmatrix} \begin{bmatrix} 4 \ 11 \ 8 \end{bmatrix}$.$= (18 	imes 4) + (0 	imes 11) + (-9 	imes 8) = 72 + 0 - 72 = 0$.

Element	Co-factor
$A$. $-1$	$(1)$ $-2$
$B$. $1$	$(2)$ $32$
$C$. $3$	$(3)$ $4$
$D$. $6$	$(4)$ $6$
	$(5)$ $-6$

Similar Questions

Let $A = [a_{ij}]_{n \times n}$ be a square matrix and let $c_{ij}$ be the cofactor of $a_{ij}$ in $A$. If $C = [c_{ij}]$,then which of the following is true?

Using cofactors of elements of the second row,evaluate $\Delta = \left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$.

Match the following elements of the matrix $A = \left[\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 4 & 2 \\ 3 & -4 & 6 \end{array}\right]$ with their co-factors and choose the correct answer.
Element Co-factor
$A$. $-1$ $(1)$ $-2$
$B$. $1$ $(2)$ $32$
$C$. $3$ $(3)$ $4$
$D$. $6$ $(4)$ $6$
$(5)$ $-6$

Write the minors and cofactors of the elements of the following determinant: $\left|\begin{array}{ccc}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$

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