Question

What is the adjoint of the matrix $\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& 1\\ 2& 3& 4\end{array}\right]$?

• A. (a)
• B. (b)
• C. (c)
• D. (d)

Answer 1

The correct option is B

(b)

Calculation of adjoint is a step by step process. You need to know what minor and cofactor of an element first. We first find the cofactors of all elements to create a cofactor matrix. Simply taking the transpose of it will give the adjoint matrix

From the video you saw that minor of an element $a_{ij}$ is given by the determinant of matrix formed by excluding the row and column of that particular element. ${\alpha }_{12}$

If minor of an element $a_{ij}$ is $M_{ij}$, then its Cofactor $A_{ij}$ is given by $(-1{)}^2$

$A_{ij}$ = $(-1{)}^{i+j}$ $M_{ij}$

Lets take each element of the give matrix and find their cofactors.

$Cofactorof{\alpha }_{11}=(-1{)}^2\left|\begin{array}{cc}1& 1\\ 3& 4\end{array}\right|=1.(4-3)=1Cofactorof{\alpha }_{12}=(-1{)}^3\left|\begin{array}{cc}1& 1\\ 2& 4\end{array}\right|=(-1)\left(2\right)=-2$

$Cofactorof{\alpha }_{13}=(-1{)}^4\left|\begin{array}{cc}1& 1\\ 2& 3\end{array}\right|=1Cofactorof{\alpha }_{21}=(-1{)}^3\left|\begin{array}{cc}2& 3\\ 3& 4\end{array}\right|=(-1)(-1)=+1$

$Cofactorof{\alpha }_{22}=(-1{)}^4\left|\begin{array}{cc}1& 3\\ 2& 4\end{array}\right|=-2Cofactorof{\alpha }_{23}=(-1{)}^5\left|\begin{array}{cc}1& 2\\ 2& 3\end{array}\right|=(-1)(-1)=1$

$Cofactorof{\alpha }_{31}=(-1{)}^4\left|\begin{array}{cc}2& 3\\ 1& 1\end{array}\right|=(-1)$

$Cofactorof{\alpha }_{32}=(-1{)}^5\left|\begin{array}{cc}1& 3\\ 1& 1\end{array}\right|=(-1\left)\right(-2)=2$

$Cofactorof{\alpha }_{33}=(-1{)}^6\left|\begin{array}{cc}1& 2\\ 1& 1\end{array}\right|=1-1=-1$

Cofactor matrix =

$\left[\begin{array}{ccc}1& -2& 1\\ 1& -2& 1\\ -1& 2& -1\end{array}\right]$

Now take the transpose of cofactor matrix to get Adjoint of the Matrix.

$\therefore Adj\left(A\right)=\left[\begin{array}{ccc}1& 1& -1\\ -2& -2& 2\\ 1& 1& -1\end{array}\right]$

Hence the option (b)