Question

For any $ 2 \times 2 $ matrix, if $ {\rm{A}}\left( {{\rm{adj A}}} \right){\rm{ = }}\left[ {\begin{array}{*{20}{c}}
{15}&0\\
0&{15}
\end{array}} \right] $ , then $ \left| {\rm{A}} \right| $ is equal to
A.20
B.625
C.15

Answer 1

Hint: As per the basic rules of matrix, identity matrix of the order $ 2 \times 2 $ can be given by \[{\rm{I = }}\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]\].
Also, the value of inverse of a matrix can be calculated by taking the ratio of adjoint of the matrix A to the determinant of the matrix A. The product of the matrix A and its inverse can be given by an identity matrix, i.e. $ {\rm{A}} \cdot {{\rm{A}}^{ - 1}} = {\rm{I}} $ .

Complete step-by-step answer:
As per the question, it is given that $ {\rm{A}}\left( {{\rm{adj A}}} \right){\rm{ = }}\left[ {\begin{array}{*{20}{c}}
{15}&0\\
0&{15}
\end{array}} \right] $
Now, we can take 15 common from the left-hand side of the equation.
 $ {\rm{A}}\left( {{\rm{adj A}}} \right){\rm{ = 15 }}\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right] $
 $ \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right] $
is the identity matrix denoted by I.
 $ {\rm{A}}\left( {{\rm{adj A}}} \right){\rm{ = 15I}}......\left( 1 \right) $
We know that the inverse of matrix A can be found out by taking the ratio of adjoint of matrix A to the determinant of matrix A.
 $ {A^{ - 1}} = \dfrac{{{\rm{adj A}}}}{{\left| {\rm{A}} \right|}} $
Now, multiply by A on both sides of the equation.
 $\Rightarrow A \cdot {A^{ - 1}} = A \cdot \dfrac{{{\rm{adj A}}}}{{\left| {\rm{A}} \right|}} $
Now, we should multiply by matrix $ \left| {\rm{A}} \right| $ on both sides of the equation $\Rightarrow \left| {\rm{A}} \right|{A^{ - 1}} = {\rm{adj A}} $ .
$\Rightarrow \left| {\rm{A}} \right|A \cdot {A^{ - 1}} = {\rm{A}} \cdot \left( {{\rm{adj A}}} \right) $
We know that the product of matrix A and its inverse can be given by the identity matrix denoted by I.
Substitute \[{\rm{A}} \cdot {{\rm{A}}^{{\rm{ - 1}}}}{\rm{ = I}}\] in the equation $ \left| {\rm{A}} \right|A \cdot {A^{ - 1}} = {\rm{A}} \cdot \left( {{\rm{adj A}}} \right) $ .
After substituting, we get, $ \left| {\rm{A}} \right|{\rm{I}} = \left( {{\rm{adj A}}} \right) \cdot {\rm{A}}.......\left( 2 \right) $
Now, from equations (1) and (2), we can conclude that $ \left| A \right|{\rm{I = 15I}} $
$\Rightarrow \left| A \right|{\rm{I = 15I}} $ means that $ \left| {\rm{A}} \right| = 15 $
Therefore, the determinant of the matrix A i.e. $ \left| {\rm{A}} \right| $ is equal to 15.
So, the correct answer is “Option C”.

Note: While solving the question for matrix, please keep in mind that the multiplication, addition, subtraction operation in matrix is different from that of the normal algebraic equations. In matrix equations students often divide the equation by any variable (provided it is greater than zero) without thinking much. But this can’t happen in the matrix equation. In such situations, when we want to divide a matrix equation by a matrix B, we should multiply the equation by matrix $ {{\rm{B}}^{ - 1}} $ . This would help in reducing the complications of the question.