Question

If $A = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)$ and $A\left( {{\text{Adj A}}} \right) = KI$ , then the value of $K$ is:

Answer 1

Hint: In order to find the value of $K$, find the adjoint matrix of the given matrix. Then multiply the Matrix with the adjoint founded according to the relation given $A\left( {{\text{Adj A}}} \right) = KI$. Then equate the result obtained with the equation given, compare the values and get the result for the unknown value $K$.

Complete step-by-step answer:
We are given with a 2x2 matrix $A = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)$ and an equation $A\left( {{\text{Adj A}}} \right) = KI$.
From the equation $A\left( {{\text{Adj A}}} \right) = KI$ we can see that it is the product of the matrix $A$ and its adjoint.
So, we need to find the adjoint of the matrix first.
We know that for a 2x2 matrix $A = \left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right)$, the adjoint matrix obtained is ${\text{Adj A}} = \left( {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right)$.
So, comparing $A = \left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right)$ with our given matrix $A = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)$, the adjoint matrix becomes same as the adjoint matrix of $A = \left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right)$.
Therefore, from ${\text{Adj A}} = \left( {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right)$ , we get the adjoint matrix for $A = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)$ as ${\text{Adj A}} = \left( {\begin{array}{*{20}{c}}
  4&{ - 2} \\
  { - 3}&1
\end{array}} \right)$.
Now, as per the given equation, we need to multiply the matrix $A$ and its adjoint matrix.
So, multiplying both of the matrix and it’s adjoint matrix, we get:
${\text{A}}\left( {{\text{Adj A}}} \right) = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  4&{ - 2} \\
  { - 3}&1
\end{array}} \right)$
For multiplication of two matrices, we know that the element of the new matrix is the sum of the elements of the row of the first matrix with the same position in the second matrix. Same to be done for all the other elements.
Following these steps, we get:
${\text{A}}\left( {{\text{Adj A}}} \right) = \left( {\begin{array}{*{20}{c}}
  {1 \times 4 + 2\left( { - 3} \right)}&{1 \times \left( { - 2} \right) + 2 \times 1} \\
  {3 \times 4 + 4\left( { - 3} \right)}&{3 \times \left( { - 2} \right) + 4 \times 1}
\end{array}} \right)$
Solving the elements:
${\text{A}}\left( {{\text{Adj A}}} \right) = \left( {\begin{array}{*{20}{c}}
  {4 + \left( { - 6} \right)}&{\left( { - 2} \right) + 2} \\
  {12 + \left( { - 12} \right)}&{\left( { - 6} \right) + 4}
\end{array}} \right)$
${\text{A}}\left( {{\text{Adj A}}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 2}&0 \\
  0&{ - 2}
\end{array}} \right)$
Taking $ - 2$ common from the matrix, we get:
${\text{A}}\left( {{\text{Adj A}}} \right) = \left( { - 2} \right)\left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)$
Since, we know that $\left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)$ is nothing but an Identity matrix which is represented as $I$.
${\text{A}}\left( {{\text{Adj A}}} \right) = \left( { - 2} \right)I$

Comparing this obtained relation ${\text{A}}\left( {{\text{Adj A}}} \right) = \left( { - 2} \right)I$ with the given relation $A\left( {{\text{Adj A}}} \right) = KI$ , we get: $K = \left( { - 2} \right)$
Therefore, the value of $K$ is $ - 2$.
So, the correct answer is “ $ - 2$”.

Note: The adjoint matrix is nothing but the Transverse of the Cofactors of the given matrix, which is represented as ${\text{Adj A = }}{\left( C \right)^T}$.
An Identity matrix is a matrix having $1$ in its diagonal and $0$ on the remaining parts and it is represented as $\left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)$.
When multiplying two matrices the number of columns of the first matrix must be equal to the number of rows to the second matrix, they can only be multiplied otherwise not.