{"id":4998,"date":"2021-09-04T00:21:01","date_gmt":"2021-09-03T18:51:01","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4998"},"modified":"2021-11-26T20:41:06","modified_gmt":"2021-11-26T15:11:06","slug":"adjoint-of-the-matrix","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/adjoint-of-the-matrix\/","title":{"rendered":"Adjoint of the Matrix (2×2 & 3×3) – Properties, Examples"},"content":{"rendered":"
Here you will learn how to find adjoint of the matrix 2×2 and 3×3, cofactors and its properties with examples.<\/p>\n
Let’s begin –<\/p>\n
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<\/strong> and is denoted by adj A<\/strong>.<\/p>\n Thus, adj A = \\([C{ij}]^T\\) \\(\\implies\\) \\((adj A)_{ij}\\) = \\(C_{ij}\\) = Cofactor of \\(a_{ij}\\) in A.<\/p>\n If A = \\(\\begin{bmatrix} a_{11} & a_{12} &\u00a0 a_{13} \\\\ a_{21} & a_{22} & a_{23}\\\\\u00a0 a_{31} & a_{32} & a_{33} \\end{bmatrix}\\) then,<\/p>\n adj A = \\({\\begin{bmatrix} C_{11} & C_{12} &\u00a0 C_{13} \\\\ C_{21} & C_{22} & C_{23}\\\\\u00a0 C_{31} & C_{32} & C_{33} \\end{bmatrix}}^T\\) = \\(\\begin{bmatrix} C_{11} & C_{21} &\u00a0 C_{31} \\\\ C_{12} & C_{22} & C_{32}\\\\\u00a0 C_{13} & C_{23} & C_{33} \\end{bmatrix}\\)<\/p>\n<\/blockquote>\n where \\(C_{ij}\\) denotes cofactor of \\(a_{ij}\\) in A.<\/p>\n Example<\/span><\/strong> : Let A = \\([a_{ij}]\\) = \\(\\begin{bmatrix} p & q \\\\ r & s\u00a0 \\end{bmatrix}\\)<\/p>\n then, cofactor of \\(a_{11}\\) = s<\/p>\n and cofactor of \\(a_{12}\\) = -r<\/p>\n cofactor of \\(a_{21}\\) = -q<\/p>\n cofactor of \\(a_{22}\\) = p<\/p>\n \\(\\therefore\\)\u00a0 adj A = \\({\\begin{bmatrix} s & -r \\\\ -q & p\u00a0 \\end{bmatrix}}^T\\) = \\(\\begin{bmatrix} s & -q \\\\ -r & p\u00a0 \\end{bmatrix}\\)<\/p>\n Rule :\u00a0<\/strong>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.<\/p>\n If A = \\(\\begin{bmatrix} -2 & 3 \\\\\u00a0 -5 & 4\u00a0 \\end{bmatrix}\\) then by the above rule, we obtain adj A \\(\\begin{bmatrix} 4 & -3 \\\\\u00a0 5 & -2\u00a0 \\end{bmatrix}\\)<\/p>\n Example : <\/span><\/strong>Let A = \\([a_{ij}]\\) = \\(\\begin{bmatrix} 1 & 1 &\u00a0 1 \\\\\u00a0 2 & 1 & -3 \\\\\u00a0 -1 & 2 & 3 \\end{bmatrix}\\)<\/p>\n Let \\(C_{ij}\\) be cofactor of \\(a_{ij}\\) in A. Then the cofactors of elements of A are given by<\/p>\n \\(C_{11}\\) = \\(\\begin{vmatrix} 1 & -3 \\\\\u00a0 2 & 3\u00a0 \\end{vmatrix}\\) = 9, \\(C_{12}\\) = -\\(\\begin{vmatrix} 2 & -3 \\\\\u00a0 -1 & 3\u00a0 \\end{vmatrix}\\) = -3,\u00a0 \\(C_{13}\\) = \\(\\begin{vmatrix} 2 & 1 \\\\\u00a0 -1 & 2\u00a0 \\end{vmatrix}\\) = 5<\/p>\n \\(C_{21}\\) = -\\(\\begin{vmatrix} 1 & 1 \\\\\u00a0 2 & 3\u00a0 \\end{vmatrix}\\) = -1, \\(C_{22}\\) = \\(\\begin{vmatrix} 1 & 1 \\\\\u00a0 -1 & 3\u00a0 \\end{vmatrix}\\) = 4,\u00a0 \\(C_{23}\\) = -\\(\\begin{vmatrix} 1 & 1 \\\\\u00a0 -1 & 2\u00a0 \\end{vmatrix}\\) = -3<\/p>\n \\(C_{31}\\) = \\(\\begin{vmatrix} 1 & 1 \\\\\u00a0 1 & -3\u00a0 \\end{vmatrix}\\) = -4, \\(C_{32}\\) = -\\(\\begin{vmatrix} 1 & 1 \\\\\u00a0 2 & -3\u00a0 \\end{vmatrix}\\) = 5,\u00a0 \\(C_{33}\\) = \\(\\begin{vmatrix} 1 & 1 \\\\\u00a0 2 & 1\u00a0 \\end{vmatrix}\\) = -1<\/p>\n \\(\\therefore\\)\u00a0 \u00a0adj A = \\({\\begin{bmatrix} 9 & -3 &\u00a0 5 \\\\\u00a0 -1 & 4 & -3 \\\\\u00a0 -4 & 5 & -1 \\end{bmatrix}}^T\\) = \\(\\begin{bmatrix} 9 & -1 &\u00a0 -4 \\\\\u00a0 -3 & 4 & 5 \\\\\u00a0 5 & -3 & -1 \\end{bmatrix}\\)<\/p>\n Note :\u00a0<\/strong>Let A be a square matrix of order n. Then, A (adj A) = |A| \\(I_n\\) = (adj A) A.<\/p>\n\n\n\n
How to find Cofactors and Adjoint for 2×2 Matrix :<\/strong><\/h4>\n
How to find Cofactors and Adjoint for 3×3 Matrix :<\/strong><\/h4>\n