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# If ${\rm A} = \left[ {\begin{array}{*{20}{c}} {5a}&{ - b} \\ 3&2 \end{array}} \right]$ and ${\rm A}(adj{\rm A}) = {\rm A}{{\rm A}^{\rm T}}$ , then $5a + b$ is equal toA. $- 1$ B. $5$ C. $4$ D. $13$

Last updated date: 24th Mar 2023
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Hint: In order to determine the adjoint , ${\rm A}(adj{\rm A}) = {\rm A}{{\rm A}^{\rm T}}$ can be defined as the form of ${\rm A}(adj{\rm A}) = \left| {\rm A} \right|{{\rm I}_n}$ from the matrix of ${\rm A}$ be a square matrix of order, $2 \times 2$ . The adjoint of a matrix ${\rm A}$ is the transpose of the cofactor matrix of ${\rm A}$ . An adjoint matrix is also called an adjugate matrix. The transpose of a matrix is a new matrix whose rows are the columns of the original. We need to find out $5a + b$ is equal to the required option.

In the given problem,
We have the $2 \times 2$ matrix ${\rm A}$ . we need to solve this $5a + b$ by the adjacent matrix ${\rm A}(adj{\rm A}) = {\rm A}{{\rm A}^{\rm T}}$ , The adjoint of ${\rm A}$ can be determined from the form of ${\rm A}(adj{\rm A}) = \left| {\rm A} \right|{{\rm I}_n}$
An identity matrix is a square matrix having 1s on the main diagonal, and 0s everywhere else. For example, the $2 \times 2$ identity matrices are ${{\rm I}_2} = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$ .These are called identity matrices because, when you multiply them with a compatible matrix, you get back the same matrix. 
We need to find out the ${\rm A}(adj{\rm A}) = {\rm A}{{\rm A}^{\rm T}}$ , where the matrix ${\rm A} = \left[ {\begin{array}{*{20}{c}} {5a}&{ - b} \\ 3&2 \end{array}} \right]$
Since, ${\rm A}(adj{\rm A}) = {\rm A}{{\rm A}^{\rm T}}$
If ${\rm A}$ is the square matrix of order $n$ , then ${\rm A}(adj{\rm A}) = \left| {\rm A} \right|{{\rm I}_n}$ . Let us consider the $2 \times 2$ identity matrices are ${{\rm I}_2} = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$ . Here the matrix : ${\rm A} = \left[ {\begin{array}{*{20}{c}} {5a}&{ - b} \\ 3&2 \end{array}} \right]$
LHS: ${\rm A}(adj{\rm A}) = \left| {\rm A} \right|{{\rm I}_n}$
Now, Substitute all the matrix values into the formula to solve in further, we get
${\rm A}(adj{\rm A}) = \left| {\begin{array}{*{20}{c}} {5a}&{ - b} \\ 3&2 \end{array}} \right|{{\rm I}_2}$
By performing multiplication of the modulus of ${\rm A}$ , we get
${\rm A}(adj{\rm A}) = 10a + 3b\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$ 
${\rm A}(adj{\rm A}) = \left[ {\begin{array}{*{20}{c}} {10a + 3b}&0 \\ 0&{10a + 3b} \end{array}} \right]$ ---------(1)
Here, we need to find the RHS of ${\rm A}(adj{\rm A}) = {\rm A}{{\rm A}^{\rm T}}$ where ${\rm A} = \left[ {\begin{array}{*{20}{c}} {5a}&{ - b} \\ 3&2 \end{array}} \right]$
Find out the Transpose of ${\rm A}$ makes the columns of the new matrix the rows of the original. Here is a matrix ${\rm A} = \left[ {\begin{array}{*{20}{c}} {5a}&{ - b} \\ 3&2 \end{array}} \right]$ and its transpose: ${\rm A} = \left[ {\begin{array}{*{20}{c}} {5a}&3 \\ { - b}&2 \end{array}} \right]$
RHS:
${\rm A}{{\rm A}^{\rm T}} = \left[ {\begin{array}{*{20}{c}} {5a}&{ - b} \\ 3&2 \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {5a}&3 \\ { - b}&2 \end{array}} \right]$
By simplify in further, we can get
${\rm A}{{\rm A}^{\rm T}} = \left[ {\begin{array}{*{20}{c}} {25{a^2} + {b^2}}&{15a - 2b} \\ {15a - 2b}&{9 + 4} \end{array}} \right]$
${\rm A}{{\rm A}^{\rm T}} = \left[ {\begin{array}{*{20}{c}} {25{a^2} + {b^2}}&{15a - 2b} \\ {15a - 2b}&{13} \end{array}} \right]$ -----------(2)
From the equation (1) and (2),form two equations for finding the value of $a$ and $b$ .
$15a - 2b = 0$ --------(3)
$10a + 3b = 13$ --------(4)
Expanding the equation (3) of LHS to RHS, we can get
$\Rightarrow 15a = 2b \\ a = \dfrac{{2b}}{{15}} \;$
On substituting the value of $a$ into the equation (4), we can get
$\Rightarrow 10\left( {\dfrac{{2b}}{{15}}} \right) + 3b = 13$
Expanding the brackets and take LCM on RHS, we can solve it in further
$\left( {\dfrac{{20b}}{{15}}} \right) + 3b = 13 \\ \dfrac{{20b + 45b}}{{15}} = 13 \;$
By simplify in further to get the value of $b$ , we get
$\dfrac{{65b}}{{15}} = 13 \\ \dfrac{{13b}}{3} = 13 \\ b = 3 \;$
By substituting the value $b$ into the equation (3)
$\Rightarrow 15a - 2b = 0 \\ 15a - 2(3) = 0 \\ 15a = 6 \\ a = \dfrac{2}{5} \;$
Therefore, the value of $a = \dfrac{2}{5}$ and $b = 3$
From the question $5a + b$ ------(5)
Now, substituting the value of $a$ and $b$ into the equation (5), we get
$\Rightarrow 5a + b = 5\left( {\dfrac{2}{5}} \right) + 3 = 5$
Therefore, $5a + b = 5$
Hence, the option (B) $5$ is the correct answer.
So, the correct answer is “Option B”.

Note: We note that we use the formula to solve ${\rm A}(adj{\rm A}) = {\rm A}{{\rm A}^{\rm T}}$ . Here, The transpose of a matrix ${{\rm A}^{\rm T}}$ simply interchange the rows and columns of the matrix i.e. write the elements of the rows as columns and write the elements of a column as rows. Finally we can find out the appropriate value.