13.3 Cramer's Rule for Inverse of a 2x2 System

Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: 13.1 Inverse of a Square Matrix

Properties: N/A
Sufficiencies: N/A
Questions: N/A

Cramer's Rule for Inverse of a

2 \times 2

System

Consider the $2 \times 2$ matrix with real entries given by

A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

Then, the inverse of matrix $A$ is given by

A^{- 1} = \frac{1}{δ} [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

where $δ = a_{11} a_{22} - a_{21} a_{12}$ .

\begin{proof}
Suppose matrix $A \in R^{2 \times 2}$ is nonsingular. Let

A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

To produce the inverse of $A$ , we will transform $A$ into the 8.6 Identity Matrix $I_{2}$ using elementary matrices.

Let $δ = a_{11} a_{22} - a_{21} a_{12}$ .
$δ$ is known as the determinant of $A$ and $δ \neq 0$ iff $a_{11} \neq 0$ or $a_{21} \neq 0$ .

We will assume $a_{11} \neq 0$ and reduce $A$ to $I_{2}$ in the following series of equations.

S_{21} (- \frac{a_{21}}{a_{11}}) \cdot A = [\begin{matrix} 1 & 0 \\ - \frac{a_{21}}{a_{11}} & 1 \end{matrix}] [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] = [\begin{matrix} a_{11} & a_{12} \\ 0 & \frac{δ}{a_{11}} \end{matrix}]

Now, consider

D_{2} (\frac{a_{11}}{δ}) \cdot D_{1} (\frac{1}{a_{11}}) = [\begin{matrix} \frac{1}{a_{11}} & 0 \\ 0 & \frac{a_{11}}{δ} \end{matrix}]

We can multiply our product by this diagonal matrix to find

[\begin{matrix} \frac{1}{a_{11}} & 0 \\ 0 & \frac{a_{11}}{δ} \end{matrix}] [\begin{matrix} a_{11} & a_{12} \\ 0 & \frac{δ}{a_{11}} \end{matrix}] = [\begin{matrix} 1 & \frac{a_{12}}{a_{11}} \\ 0 & 1 \end{matrix}]

We finish converting our matrix to $I_{2}$ by multiplying this entire product on the left-hand side to annihilate $\frac{a_{12}}{a_{11}}$ .

[\begin{matrix} 1 & - \frac{a_{12}}{a_{11}} \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & \frac{a_{12}}{a_{11}} \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

Thus, we see that $A^{- 1} \cdot A = I_{2}$ with

A^{- 1} = S_{12} (- \frac{a_{12}}{a_{11}}) \cdot D_{2} (\frac{a_{11}}{δ}) \cdot D_{1} (\frac{1}{a_{11}}) \cdot S_{21} (- \frac{a_{21}}{a_{11}})

The reduction of matrix $A$ into $I_{2}$ gives us a format to specifically calculate $A^{- 1}$ . We can do this by first finding the product

(D_{2} (\frac{a_{11}}{δ}) \cdot D_{1} (\frac{1}{a_{11}})) \cdot S_{21} (- \frac{a_{21}}{a_{11}})

given as

[\begin{matrix} \frac{1}{a_{11}} & 0 \\ 0 & \frac{a_{11}}{δ} \end{matrix}] [\begin{matrix} 1 & 0 \\ - \frac{a_{21}}{a_{11}} & 1 \end{matrix}] = [\begin{matrix} \frac{1}{a_{11}} & 0 \\ - \frac{a_{21}}{δ} & \frac{a_{11}}{δ} \end{matrix}]

Using our product, we can calculate $A^{- 1}$ .

[\begin{matrix} 1 & - \frac{a_{12}}{a_{11}} \\ 0 & 1 \end{matrix}] [\begin{matrix} \frac{1}{a_{11}} & 0 \\ - \frac{a_{21}}{δ} & \frac{a_{11}}{δ} \end{matrix}] = \frac{1}{δ} [\begin{matrix} a_{22} & - a_{12} \\ - a_{21} & a_{11} \end{matrix}]

\end{proof}