# Find the adjoint of matrix A=$\begin{bmatrix} 3 & -3 & 4 \\2 & -3 & 4 \\0 & -1 & 1 \end{bmatrix}$ and verify the result $A\;(adj A)=(adj A)\;A=$|$A$|$.I$

Toolbox:
• Let $A = [ a_{ij} ]$ be a square matrix x. Let $A_{ij}$ be the cofactor of $a_{ij}$. Then $[ A_{ij}]$ is the matrix of cofactors and $adj\: A$ ( or adjoint of the matrix A) is given by $adj\: A=[A_{ij}]^T$
• Multiplication of two matrices A and B is possible when the number of columns in the first matrix equals the number of rows in the second matrix B. If A is of type $m$ x $p$ and B is of type $p$ x $x$ then AB is of type $m$ x $n$
• $AB = [ c_{ij}]$ where $c_{ij} = \sum^p_{\substack{k=1}}a_{ik}b_{kj}$
• A determinant can be expanded by using the elements of any row or column.
Step 1
$A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$
To find $adj \: A$
$[A_{ij}] = \begin{bmatrix} (-3+4) & -(2-0) & (-2-0) \\ -(-3+4) & (3-0) & -(-3-0) \\ (-12+12) & -(12-8) & (-9+6) \end{bmatrix} = \begin{bmatrix} 1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3 \end{bmatrix}$
$adj\: A = [ A_{ij}]^T = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$
Step 2
To find $A(adj\: A)$ and $(adj\: A) A$
$A(adj\: A) = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$
$= \begin{bmatrix} (3)(1)+(-3)(-2)+(4)(-2) & (3)(-1)+(-3)(3)+(4)(3) & 3(0)+(-3)(-4)+(4)(-3) \\ (2)(1)+(-3)(-2)+(4)(-2) & (2)(-1)+(-3)(3)+(4)(3) & 2(0)+(-3)(-4)+(4)(-3) \\ 0+(-1)(-2)+(1)(-2) & 0+(-1)(3)+(1)(3) & 0+(-1)(-4)+(1)(-3) \end{bmatrix}$
$= \begin{bmatrix} 3+6-8 & -3-9+12 & 0+12-12 \\ 2+6-8 & -2-9+12 & 0+12-12 \\ 0+2-2 & 0-3+3 & 0+4-3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$
$(adj\: A )A = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$
$= \begin{bmatrix} (1)(3)+(-1)(2)+0 & (1)(-3)+(-1)(-3)+0 & (1)(4)+(-1)(4)+0 \\ (-2)(3)+(3)(2)+0 & (-2)(-3)+(3)(-3)+(-4)(-1) & (-2)(4)+(3)(4)+(-4)(1) \\ (-2)(3)+(3)(2)+0 & (-2)(-3)+(3)(-3)+(-3)(-1) & (-2)(4)+(3)(4)+(-3)(1) \end{bmatrix}$
$= \begin{bmatrix} 3-2 & -3+3 & 4-4 \\ -6+6 & 6-9+4 & -8+12-4 \\ -6+6 & 6-9+3 & -8+12-3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$
It can be seen that $A ( adj\: A ) = (adj\: A) A$
Step 3
To find $| A | \: I$
$|A| = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = 3(-3+4)\: -2(-3+4)$
$= 3-2=1$
( The determinant was expanded by $c_1$ )
$|A|\: I = 1\: I = I$
Hence from step 2 $A(adj\: A)=(adj\: A)A=|A|\: I$ Thus verified.

edited May 17, 2013