Question

If A=\[\left[{\begin{array}{*{20}{c}}1&{ - 1}\\2&3\end{array}} \right]\],
then adj A is equal to
A. \[\left[
{\begin{array}{*{20}{c}}{ - 3}&{ - 1}\\2&{ - 1}\end{array}} \right]\]
B. \[\left[
{\begin{array}{*{20}{c}}3&1\\{ - 2}&1\end{array}} \right]\]
C. \[\left[
{\begin{array}{*{20}{c}}3&{ - 2}\\1&1\end{array}} \right]\]
D. \[\left[
{\begin{array}{*{20}{c}}3&{ - 1}\\{ - 2}&1\end{array}} \right]\]

Answer 1

Hint:
In this question, we have to find adj A of the given matrix. So for
that we have know about adj A. so the adjoint of a square matrix A = \[{\left[ {{a_{ij}}}\right]_{n \times n}}\] can be obtained by changing the diagonal elements and changing the sign of off-diagonal elements.


Formula Used:

The formula for the adjoint of a matrix can be derived using the cofactor and transpose of a matrix. However, it is easy to find the adj of a matrix for a 2 x 2 matrix.
Let A be the 2 x 2 matrix and is given by:
\[A=\left[{\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}}\right]\]
Now, the adj A will be calculated by interchanging\[{a_{11}}\] and\[{a_{22}}\] and by changing signs of \[{a_{12}}\] and \[{a_{21}}\]
This can be shown as:
adjA of \[\left[{\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}}\right]\]=\[\left[ {\begin{array}{*{20}{c}}{{a_{22}}}&{- {a_{12}}}\\{ - {a_{21}}}&{{a_{11}}}\end{array}} \right]\]



Complete Step-by-Step Solution:

Now we are given a matrix A= \[\left[{\begin{array}{*{20}{c}}1&{ - 1}\\2&3\end{array}} \right]\]
And we have to find its adjoint of this matrix
Now we will interchange the elements \[{a_{11}}\] and \[{a_{22}}\]as per the formula of the adj A and changing signs of \[{a_{12}}\] and\[{a_{21}}\].
So adj A= \[\left[{\begin{array}{*{20}{c}}3&1\\{ - 2}&1\end{array}} \right]\]
Therefore, option B is correct.


Note:
Students have to remember the formula of adj A for a 2×2 matrix.
As many students make mistakes in interchanging elements and changing signs of
elements. The adj A also has some properties which students have to remember
while solving such questions.
If A be any given square matrix of order n, we can define the
following:
 A (adj A) = (adj A) A = A I, where I is the identity
matrix of order n
For a zero matrix 0, adj (0) = 0
For an identity matrix I, adj (I) = I
For any scalar k, adj(kA) \[ = {k^{n - 1}}\] adj(A)
\[adj({A^T}) = {(adjA)^T}\]