Q.If $A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$ then $A^{-1} =$

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

so the whole system becomes $AX=B$.

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$. The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$:

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$, so:

• $\det A\neq0$: the system is consistent with the unique solution $X=A^{-1}B$.
• $\det A=0$: no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$, and

Then …