Question

Adj$\left[ {\begin{array}{*{20}{c}}1&0&2\\{ - 1}&1&{ - 2}\\0&2&1\end{array}}
\right] = \left[ {\begin{array}{*{20}{c}}5&a&{ - 2}\\1&1&0\\{ - 2}&{ - 2}&b\end{array}}
\right] \Rightarrow \left[ {\begin{array}{*{20}{c}}a&b\end{array}} \right] = $
E. $\left[ {\begin{array}{*{20}{c}}{ - 4}&1\end{array}} \right]$
F. $\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 1}\end{array}} \right]$
G. $\left[ {\begin{array}{*{20}{c}}4&1\end{array}} \right]$
H. $\left[ {\begin{array}{*{20}{c}}4&{ - 1}\end{array}} \right]$

Answer 1

Hint: In this problem, first we have to find the cofactors of the given matrix by using the formula
${\left( { - 1} \right)^{i + j}}{m_{ij}}$, where ${m_{ij}}$ represents the minor of the matrix.
After getting the cofactor matrix we have to take transpose of that matrix. This will give us the
adjoin of the given matrix. Now we have to compare the calculated adjoin matrix with the given
adjoin matrix.
It is known that the cofactors of a matrix can be determined by ${\left( { - 1} \right)^{i +
j}}{m_{ij}}$.
Now, for the minor ${m_{11}} = \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\2&1\end{array}}
\right]$
For the cofactor ${c_{11}}$,
$\begin{array}{c}{c_{11}} = {\left( { - 1} \right)^{1 + 1}}\left[ {\begin{array}{*{20}{c}}1&{
- 2}\\2&1\end{array}} \right]\\ = 1 \times \left( {1 + 4} \right)\\ = 5\end{array}$

For the minor ${m_{12}} = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\0&1\end{array}}
\right]$
For the cofactor ${c_{12}}$,
$\begin{array}{c}{c_{12}} = {\left( { - 1} \right)^{1 + 2}}\left[ {\begin{array}{*{20}{c}}{ -
1}&{ - 2}\\0&1\end{array}} \right]\\ = - 1 \times - 1\\ = 1\end{array}$

For the minor ${m_{13}} = \left[ {\begin{array}{*{20}{c}}{ - 1}&1\\0&2\end{array}} \right]$
For the cofactor ${c_{13}}$,
$\begin{array}{c}{c_{13}} = {\left( { - 1} \right)^{1 + 3}}\left[ {\begin{array}{*{20}{c}}{ -
1}&1\\0&2\end{array}} \right]\\ = 1 \times - 2\\ = - 2\end{array}$

For the minor ${m_{21}} = \left[ {\begin{array}{*{20}{c}}0&2\\2&1\end{array}} \right]$
For the cofactor ${c_{21}}$,
$\begin{array}{c}{c_{21}} = {\left( { - 1} \right)^{2 + 1}}\left[
{\begin{array}{*{20}{c}}0&2\\2&1\end{array}} \right]\\ = - 1 \times - 4\\ = 4\end{array}$

For the minor ${m_{22}} = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right]$
For the cofactor ${c_{22}}$,
$\begin{array}{c}{c_{22}} = {\left( { - 1} \right)^{2 + 2}}\left[
{\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right]\\ = 1 \times 1\\ = 1\end{array}$

For the minor ${m_{23}} = \left[ {\begin{array}{*{20}{c}}1&0\\0&2\end{array}} \right]$
For the cofactor ${c_{23}}$,
$\begin{array}{c}{c_{23}} = {\left( { - 1} \right)^{2 + 3}}\left[
{\begin{array}{*{20}{c}}1&0\\0&2\end{array}} \right]\\ = - 1 \times 2\\ = - 2\end{array}$

For the minor ${m_{31}} = \left[ {\begin{array}{*{20}{c}}0&2\\1&{ - 2}\end{array}} \right]$
For the cofactor ${c_{31}}$,
$\begin{array}{c}{c_{31}} = {\left( { - 1} \right)^{3 + 1}}\left[
{\begin{array}{*{20}{c}}0&2\\1&{ - 2}\end{array}} \right]\\ = 1 \times - 2\\ = -
2\end{array}$

For the minor ${m_{32}} = \left[ {\begin{array}{*{20}{c}}1&2\\{ - 1}&{ - 2}\end{array}}
\right]$
For the cofactor ${c_{32}}$,
$\begin{array}{c}{c_{32}} = {\left( { - 1} \right)^{3 + 2}}\left[
{\begin{array}{*{20}{c}}1&2\\{ - 1}&{ - 2}\end{array}} \right]\\ = - 1 \times 0\\ =
0\end{array}$

For the minor ${m_{33}} = \left[ {\begin{array}{*{20}{c}}1&0\\{ - 1}&1\end{array}} \right]$
For the cofactor ${c_{33}}$,

$\begin{array}{c}{c_{33}} = {\left( { - 1} \right)^{3 + 3}}\left[
{\begin{array}{*{20}{c}}1&0\\{ - 1}&1\end{array}} \right]\\ = 1 \times 1\\ = 1\end{array}$

Therefor cofactor matrix is $\left[ {{c_{ij}}} \right]$.
$\left[ {{c_{ij}}} \right] = \left[ {\begin{array}{*{20}{c}}5&1&{ - 2}\\4&1&{ - 2}\\{ -
2}&0&1\end{array}} \right]$
It is known that the adjoint of a matrix is equal to the transpose of the cofactor of the matrix.
Now, the adjoin matrix Adj ${a_{ij}}$ can be written as
$\begin{array}{c}{\rm{Adj}}\left( {{a_{ij}}} \right) = {\left[ {{c_{ij}}} \right]^T}\\ = \left[
{\begin{array}{*{20}{c}}5&4&{ - 2}\\1&1&0\\{ - 2}&{ - 2}&1\end{array}}
\right]\end{array}$

$\therefore $Adj$\left[ {\begin{array}{*{20}{c}}1&0&2\\{ - 1}&1&{ -
2}\\0&2&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&4&{ - 2}\\1&1&0\\{ -
2}&{ - 2}&1\end{array}} \right]$
Given adjoin matrix is $\left[ {\begin{array}{*{20}{c}}5&a&{ - 2}\\1&1&0\\{ - 2}&{ -
2}&b\end{array}} \right]$.
Now, comparing the calculated adjoin matrix with the given adjoin matrix.
$\left[ {\begin{array}{*{20}{c}}5&4&{ - 2}\\1&1&0\\{ - 2}&{ - 2}&1\end{array}} \right] =
\left[ {\begin{array}{*{20}{c}}5&a&{ - 2}\\1&1&0\\{ - 2}&{ - 2}&b\end{array}} \right]$
By comparing we get $a = 4$ and $b = 1$
Therefore, in matrix form $\left[ {\begin{array}{*{20}{c}}a&b\end{array}} \right] = \left[
{\begin{array}{*{20}{c}}4&1\end{array}} \right]$
Hence, the correct option is C.

Note: The minors ${m_{ij}}$ of a matrix can be formed by removing ${i^{th}}$ row and
${j^{th}}$ column of the matrix. Here we have to find the value of $a$ and $b$ in the given
matrix. Since the original matrix and its adjoin matrix are given, so we need to calculate the
cofactors of the matrix. From that we will get us adjoint matrix. By comparing the adjoin matrix
with the given adjoin matrix from which we get our required values of $a$ and $b$.