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Question: If \({{A}_{ij}}\) is the cofactor of the element \({{a}_{ij}}\)of the determinant \(\left[ \begin{ma...

If Aij{{A}_{ij}} is the cofactor of the element aij{{a}_{ij}}of the determinant [2−35 604 15−7 ]\left[ \begin{matrix} 2 & -3 & 5 \\\ 6 & 0 & 4 \\\ 1 & 5 & -7 \\\ \end{matrix} \right] then write the value of a32.A32{{a}_{32}}.{{A}_{32}}

Explanation

Solution

A cofactor is a number you get when you remove the column and row of a designated element in a matrix.
For a given matrix of order 3×33\times 3; say A=[a11a12a13 a21a22a23 a31a32a33 ]A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right],
Cofactor of any element aij{{a}_{ij}} is given as Aij=(−1)i+jMij{{A}_{ij}}={{\left( -1 \right)}^{i+j}}{{M}_{ij}}
Where, Mij{{M}_{ij}} is the matrix formed by removing the ith{{i}^{th}} row and jth{{j}^{th}} column of the matrix. Use the above formulas to find the cofactor of a32{{a}_{32}}i.e. A32{{A}_{32}} and get the value of a32.A32{{a}_{32}}.{{A}_{32}}

Complete step-by-step solution:
Since we have a matrix of order 3×33\times 3; A=[2−35 604 15−7 ]A=\left[ \begin{matrix} 2 & -3 & 5 \\\ 6 & 0 & 4 \\\ 1 & 5 & -7 \\\ \end{matrix} \right]
So, element a32{{a}_{32}} is 5.
Now, we need to find M32{{M}_{32}}first.
So, M32=[25 64 ]......(1){{M}_{32}}=\left[ \begin{matrix} 2 & 5 \\\ 6 & 4 \\\ \end{matrix} \right]......(1)
Now, solve the equation (1), we get:
⇒M32=[8−30] ⇒M32=−22 \begin{aligned} & \Rightarrow {{M}_{32}}=\left[ 8-30 \right] \\\ & \Rightarrow {{M}_{32}}=-22 \\\ \end{aligned}
Now, we need to calculate value of A32{{A}_{32}}
By using the formula Aij=(−1)i+jMij{{A}_{ij}}={{\left( -1 \right)}^{i+j}}{{M}_{ij}}, where i=3 and j = 2, we get:
⇒A32=(−1)3+2(−22) ⇒A32=−1×−22 ⇒A32=22 \begin{aligned} & \Rightarrow {{A}_{32}}={{\left( -1 \right)}^{3+2}}\left( -22 \right) \\\ & \Rightarrow {{A}_{32}}=-1\times -22 \\\ & \Rightarrow {{A}_{32}}=22 \\\ \end{aligned}
Now substitute the values of a32{{a}_{32}}and A32{{A}_{32}} and get the value of a32.A32{{a}_{32}}.{{A}_{32}}.
We get:
a32.A32=5×22 =110\begin{aligned} & {{a}_{32}}.{{A}_{32}}=5\times 22 \\\ & =110 \end{aligned}
Hence, the value of a32.A32{{a}_{32}}.{{A}_{32}} is 110.

Note: Always remember that while finding the co-factor of an element, find the matrix formed by removing the elements of row and column of that element first. Elements of the matrix can be misplaced in hurry. So, be careful while finding Mij{{M}_{ij}}. And to find the value of Mij{{M}_{ij}}, expand the matrix formed to get a finite value.
Also, some might confuse with the definition of co-factor. So, instead of using the given formula to find co-factorAij=(−1)i+jMij{{A}_{ij}}={{\left( -1 \right)}^{i+j}}{{M}_{ij}}, they might write Mij{{M}_{ij}} as co-factor of the element directly. It is an incomplete answer. Always follow the given formula to find the co-factor.