Inverse matrix: Determinant method

Invent a $3\times3$ matrix and calculate its inverse.

Why can it not be any matrix? What condition must be satisfied?

We define the matrix $A=\left( \begin{array}{ccc} 1 & 0 & -2 \ -1 & 1 & 1 \ 0 & -1 & -2 \end{array} \right)$

We are going to use the second method

Calculating the determinant (using the rule of Sarrus):

$$|A|=\left| \begin{array}{ccc} 1 & 0 & -2 \\ -1 & 1 & 1 \\ 0 & -1 & -2 \end{array} \right|= (1\cdot1\cdot-2)+(-1\cdot-1\cdot-2)+(0)-0-(1\cdot-1\cdot1)-0=-3$$

The adjoint matrix is

$$Adj(A)=\left(\begin{array}{ccc} +\left|\begin{array}{cc} 1 & 1 \\ -1 & -2 \end{array} \right| & -\left|\begin{array}{cc} -1 & 1 \\ 0 & -2 \end{array} \right| & +\ldots \\ -\ldots & +\ldots & -\ldots \\ +\ldots & -\ldots & +\left|\begin{array}{cc} 1 & 0 \\ -1 & 1 \end{array} \right| \end{array} \right)= \left( \begin{array}{ccc} -1 & -2 & 1 \\ -2 & -2 & 1 \\ 2 & 1 & 1 \end{array} \right)$$

We transpose the adjoint matrix

$$(Adj(A))^t=\left( \begin{array}{ccc} -1 & -2 & 2 \\ -2 & -2 & 1 \\ 1 & 1 & 1 \end{array} \right)$$

Finally, the inverse matrix is:

$$A^{-1}=\dfrac{1}{|A|}\cdot(Adj(A))^t=\dfrac{1}{-3}\left( \begin{array}{ccc} -1 & -2 & 2 \\ -2 & -2 & 1 \\ 1 & 1 & 1 \end{array} \right)=\dfrac{1}{3}\left( \begin{array}{ccc} 1 & 2 & -2 \\ 2 & 2 & -1 \\ -1 & -1 & -1 \end{array} \right)$$

$$A^{-1}=\dfrac{1}{3}\left( \begin{array}{ccc} 1 & 2 & -2 \\ 2 & 2 & -1 \\ -1 & -1 & -1 \end{array} \right)$$

When dividing by the determinant, it is necessary that the determinant of $A$ is nonempty.

That is to say, the condition that the matrix $A$ should satisfy so as to be considered invertible.

Back to topic