If Delta = begin{vmatrix} a_1 & b_1 & c_1 a_ | vedclass

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(B) Let $\Delta'$ be the given determinant of co-factors.We know that the product of a determinant and its co-factor matrix determinant is given by $\

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$\Delta^2$

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(B) Let $\Delta'$ be the given determinant of co-factors.We know that the product of a determinant and its co-factor matrix determinant is given by $\Delta \cdot \Delta' = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} \cdot \begin{vmatrix} A_1 & B_1 & C_1 \ A_2 & B_2 & C_2 \ A_3 & B_3 & C_3 \end{vmatrix}$.By the property of multiplication of determinants,this equals $\begin{vmatrix} a_1A_1 + b_1B_1 + c_1C_1 & a_1A_2 + b_1B_2 + c_1C_2 & a_1A_3 + b_1B_3 + c_1C_3 \ a_2A_1 + b_2B_1 + c_2C_1 & a_2A_2 + b_2B_2 + c_2C_2 & a_2A_3 + b_2B_3 + c_2C_3 \ a_3A_1 + b_3B_1 + c_3C_1 & a_3A_2 + b_3B_2 + c_3C_2 & a_3A_3 + b_3B_3 + c_3C_3 \end{vmatrix}$.Using the property that the sum of the product of elements of a row with their corresponding co-factors is $\Delta$,and the sum of the product of elements of a row with co-factors of another row is $0$,we get:$\Delta \cdot \Delta' = \begin{vmatrix} \Delta & 0 & 0 \ 0 & \Delta & 0 \ 0 & 0 & \Delta \end{vmatrix}$.Evaluating this diagonal determinant,we get $\Delta \cdot \Delta' = \Delta^3$.Therefore,$\Delta' = \Delta^2$ (assuming $\Delta 
eq 0$).

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

If $\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ and $A_1, B_1, C_1$ denote the co-factors of $a_1, b_1, c_1$ respectively,then the value of the determinant $\begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}$ is

Similar Questions

The sum of the minor and the cofactor of the element $7$ in the determinant $\left|\begin{array}{ccc}2 & 3 & 5 \\ 1 & 0 & 7 \\ -1 & -2 & 4\end{array}\right|$ is . . . . . .

If $A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & -3 \\ -1 & 2 & 3 \end{bmatrix}$,then find the value of $A_{31} + A_{32} + A_{33}$,where $A_{ij}$ denotes the cofactor of the element $a_{ij}$ of matrix $A$.

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Let $A = [a_{ij}] = \begin{bmatrix} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{bmatrix}$. If $A_{ij}$ is the cofactor of $a_{ij}$,$C_{ij} = \sum_{k=1}^2 a_{ik} A_{jk}$,$1 \leq i, j \leq 2$,and $C = [C_{ij}]$,then $8|C|$ is equal to:

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
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$B$. $1$ $(2)$ $32$
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$D$. $6$ $(4)$ $6$
$(5)$ $-6$

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