# How do you find the inverse of A=((5, 2), (-1,a))?

Feb 27, 2016

The matrix is invertible as long as $5 a + 2 \setminus \ne 0$,
${A}^{- 1} = \left(\begin{matrix}a & - 2 \\ 1 & 5\end{matrix}\right)$.

#### Explanation:

Consider a matrix of the form $A = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ and let its inverse be of the form ${A}^{- 1} = \left(\begin{matrix}\setminus \alpha & \setminus \beta \\ \setminus \gamma & \setminus \delta\end{matrix}\right)$.

Our job is to solve for $\setminus \alpha , \setminus \beta , \setminus \gamma$ and $\setminus \delta$ in terms of $a , b , c$ and $d$.

Use the fact that $A \setminus \times {A}^{- 1} = I$ to get the four equations to solve the four unknowns.

$\left(\begin{matrix}a & b \\ c & d\end{matrix}\right) . \left(\begin{matrix}\setminus \alpha & \setminus \beta \\ \setminus \gamma & \setminus \delta\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$
$\left(\begin{matrix}a \setminus \alpha + b \setminus \gamma & a \setminus \beta + b \setminus \delta \\ c \setminus \alpha + d \setminus \gamma & c \setminus \beta + d \setminus \delta\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$

Comparing component wise, we get four equations required to solve the four unknowns

$a \setminus \alpha + b \setminus \gamma = 1$ ....... (1)$\setminus \quad \setminus q \quad$ $a \setminus \beta + b \setminus \delta = 0$ ........ (2)
$c \setminus \alpha + d \setminus \gamma = 0$ ....... (3) $\setminus q \quad$ $c \setminus \beta + d \setminus \delta = 1$ ......... (4)

Step 1: Use (3) to eliminate $\setminus \gamma$ in (1) and solve for $\setminus \alpha$.
From (3), $\setminus \gamma = - \frac{c}{d} \setminus \alpha$, substitute this in (1)
$\setminus \alpha \left(a - b \frac{c}{d}\right) = 1 \setminus q \quad \setminus \rightarrow \setminus \alpha = \frac{d}{a d - b c}$

$\setminus \gamma = - \frac{c}{d} \setminus \alpha = - \frac{c}{a d - b c}$
Step 2: Use (2) to eliminate $\setminus \delta$ in (4) and solve for $\setminus \beta$.
From (2), $\setminus \delta = - \frac{a}{b} \setminus \beta$, substitute this in (4)
$\setminus \beta \left(c - d \frac{a}{b}\right) = 1 \setminus q \quad \setminus \rightarrow \setminus \beta = - \frac{b}{a d - b c}$
$\setminus \delta = - \frac{a}{b} \setminus \beta = \frac{a}{a d - b c}$

Step 3: Recognise that $\left(a d - b c\right)$ is the determinant value of the matrix $A$
$| A | = | \left(a , b\right) , \left(c , d\right) | = a d - b c$

\alpha = d/|A|; \qquad \beta = -b/|A|; \qquad \gamma = -c/|A|; \qquad \delta = a/|A|.

Therefore, ${A}^{- 1} = \frac{1}{|} A | \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$.
You can see that this works only as long as $| A | \setminus \ne 0$

If $| A | = 0$ then the matrix is said to be singular and non-invertible.