Question

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 to
A. $- 1$
B. $5$
C. $4$
D. $13$

Answer 1

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.

Complete step-by-step answer :
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

{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.