Here you will learn how to find adjoint of the matrix 2×2 and 3×3, cofactors and its properties with examples.
Let’s begin –
Adjoint of the Matrix
Let A = \([a_{ij}]\) be a square matrix of order n and let \(C_{ij}\) be a cofactor of \(a_{ij}\) in A. Then the transpose of the matrix of cofactors of elements of A is called adjoint of A and is denoted by adj A.
Thus, adj A = \([C{ij}]^T\) \(\implies\) \((adj A)_{ij}\) = \(C_{ij}\) = Cofactor of \(a_{ij}\) in A.
If A = \(\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}\) then,
adj A = \({\begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23}\\ C_{31} & C_{32} & C_{33} \end{bmatrix}}^T\) = \(\begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32}\\ C_{13} & C_{23} & C_{33} \end{bmatrix}\)
where \(C_{ij}\) denotes cofactor of \(a_{ij}\) in A.
How to find Cofactors and Adjoint for 2×2 Matrix :
Example : Let A = \([a_{ij}]\) = \(\begin{bmatrix} p & q \\ r & s \end{bmatrix}\)
then, cofactor of \(a_{11}\) = s
and cofactor of \(a_{12}\) = -r
cofactor of \(a_{21}\) = -q
cofactor of \(a_{22}\) = p
\(\therefore\) adj A = \({\begin{bmatrix} s & -r \\ -q & p \end{bmatrix}}^T\) = \(\begin{bmatrix} s & -q \\ -r & p \end{bmatrix}\)
Rule : It is evident from this example that the adjoint of a square matrix of order 2 can be easily obtained by interchanging the diagonal elements and changing signs of off-diagonal elements.
If A = \(\begin{bmatrix} -2 & 3 \\ -5 & 4 \end{bmatrix}\) then by the above rule, we obtain adj A \(\begin{bmatrix} 4 & -3 \\ 5 & -2 \end{bmatrix}\)
How to find Cofactors and Adjoint for 3×3 Matrix :
Example : Let A = \([a_{ij}]\) = \(\begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & -3 \\ -1 & 2 & 3 \end{bmatrix}\)
Let \(C_{ij}\) be cofactor of \(a_{ij}\) in A. Then the cofactors of elements of A are given by
\(C_{11}\) = \(\begin{vmatrix} 1 & -3 \\ 2 & 3 \end{vmatrix}\) = 9, \(C_{12}\) = -\(\begin{vmatrix} 2 & -3 \\ -1 & 3 \end{vmatrix}\) = -3, \(C_{13}\) = \(\begin{vmatrix} 2 & 1 \\ -1 & 2 \end{vmatrix}\) = 5
\(C_{21}\) = -\(\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix}\) = -1, \(C_{22}\) = \(\begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix}\) = 4, \(C_{23}\) = -\(\begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix}\) = -3
\(C_{31}\) = \(\begin{vmatrix} 1 & 1 \\ 1 & -3 \end{vmatrix}\) = -4, \(C_{32}\) = -\(\begin{vmatrix} 1 & 1 \\ 2 & -3 \end{vmatrix}\) = 5, \(C_{33}\) = \(\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\) = -1
\(\therefore\) adj A = \({\begin{bmatrix} 9 & -3 & 5 \\ -1 & 4 & -3 \\ -4 & 5 & -1 \end{bmatrix}}^T\) = \(\begin{bmatrix} 9 & -1 & -4 \\ -3 & 4 & 5 \\ 5 & -3 & -1 \end{bmatrix}\)
Note : Let A be a square matrix of order n. Then, A (adj A) = |A| \(I_n\) = (adj A) A.
