# 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. \[13\]

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.

**:**

__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.

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