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
 \[{\rm A}(adj{\rm A}) = \left| {\begin{array}{*{20}{c}}
  {5a}&{ - b} \\
  3&2
\end{array}} \right|{{\rm I}_2}\]
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.