Adjoint of the Matrix (2×2 & 3×3) – Properties, Examples

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.

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