A 2 by 2 matrix can be inverted by a formula: swap the two diagonal entries, change the signs of the other two, and divide by the determinant. The division is exactly where things break when the determinant is zero.

To turn dollars back into euros you divide by the rate you multiplied by. The 2 by 2 inverse divides by the determinant, and a rate of zero is one no division can undo.

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ with $ad-bc\neq0$, the matrix below is the inverse: multiplying it by $A$ in either order gives $I$. If $ad-bc=0$ the formula would divide by zero, and indeed no inverse exists, because an invertible matrix has a non-zero determinant.

• Swap the two entries on the main diagonal, change the signs of the two off it, and divide every entry by the determinant.
• Check by multiplying: $\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=(ad-bc)I$.
• A $2\times2$ matrix is invertible exactly when $ad-bc\neq0$: the formula gives an inverse when it is not zero, and the last chapter showed that none exists when it is.
• The system $A\mathbf{x}=\mathbf{b}$ then has exactly one solution, $\mathbf{x}=A^{-1}\mathbf{b}$.

Multiply $A$ by the matrix with the diagonal swapped and the other two entries negated. Both diagonal entries of the product are $ad-bc$ and the others cancel, so the product is $(ad-bc)I$. Dividing by the non-zero determinant leaves $I$, and the product in the other order works out the same way.

Multiply $A$ by the adjusted matrix, row against column.

Row 1 against column 1 gives $ad-bc$, row 1 against column 2 gives $-ab+ba$, and so on.

The off-diagonal entries cancel, and both diagonal entries are the determinant.

Divide by the determinant, which is allowed because it is not zero.

In the other order the product has the same entries arranged the same way, so the formula undoes $A$ from both sides.

• Swap the two diagonal entries.
• Change the signs of the two off-diagonal entries.
• Divide every entry by the determinant.

If $A$ is invertible, multiply both sides of $A\mathbf{x}=\mathbf{b}$ by $A^{-1}$ on the left: $\mathbf{x}=A^{-1}\mathbf{b}$, the one and only solution.

• $A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ for $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$.
• Swap the diagonal, negate the other two entries, divide by the determinant.
• A $2\times2$ matrix is invertible exactly when $ad-bc\neq0$.
• $A\mathbf{x}=\mathbf{b}$ then has the one solution $\mathbf{x}=A^{-1}\mathbf{b}$.

Great work! You now know how to:

• invert a $2\times2$ matrix by the formula;
• check an inverse by multiplying;
• tell from $ad-bc$ whether an inverse exists;
• solve a $2\times2$ system with the inverse.