# Find the inverse of Matrix $A$

Find the inverse of Matrix $A$.

$$A=\begin{pmatrix}a+c & b+a \\ d-a & b \end{pmatrix}, A∈\mathbb{R^{2,2}}$$

I tried to use the formula $F \cdot F^{-1}=I$ and reduce it using the row reduced echelon concept but I don't know what values I should put in the matrix for $F$.

Hint: if $$A=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$$ then $$A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d & -b \\ -c & a \end{pmatrix}$$

• Using there determinant doesn't show the answer as elegantly as the Gaussian elimination and the answer comes out to be very long. – Ski Mask Jun 5 '18 at 20:24

Let consider

$$A=\begin{pmatrix}x & y \\ z & w \end{pmatrix}$$

then by gaussian elimination we have

$$\left(\begin{array}{cc|cc} x & y & 1 & 0 \\ z & w & 0 & 1\end{array}\right)$$

$$\left(\begin{array}{cc|cc} xz & yz & z & 0 \\ xz & xw & 0 & x\end{array}\right)$$

$$\left(\begin{array}{cc|cc} xz & yz & z & 0 \\ 0 & xw-yz & -z & x\end{array}\right)$$

$$\left(\begin{array}{cc|cc} xz( xw-yz) & yz( xw-yz) & z( xw-yz) & 0 \\ 0 & yz(xw-yz) & -z(yz) & x(yz)\end{array}\right)$$

$$\left(\begin{array}{cc|cc} xz( xw-yz) & 0 & zxw & -xyz \\ 0 & yz(xw-yz) & -yz^2 & xyz\end{array}\right)$$

$$\left(\begin{array}{cc|cc} 1 & 0 & \frac{zxw}{xz( xw-yz)} & \frac{-xyz}{xz( xw-yz)} \\ 0 & 1 & \frac{-yz^2}{yz(xw-yz)} & \frac{xyz}{yz(xw-yz)}\end{array}\right)$$

$$\left(\begin{array}{cc|cc} 1 & 0 & \frac{w}{( xw-yz)} & \frac{-y}{( xw-yz)} \\ 0 & 1 & \frac{-z}{(xw-yz)} & \frac{x}{(xw-yz)}\end{array}\right)$$

and thus the inverse matrix is

$$A^{-1}=\frac{1}{xw-yz}\begin{pmatrix}w & -y \\ -z & x \end{pmatrix}$$

then consider

• $x=a+c$
• $y=b+a$
• $z=d-a$
• $w=b$

and then the inverse is

$$A^{-1}=\frac{1}{(a+c)b-(b+a)(d-a)}\begin{pmatrix}b & -(b+a) \\ -(d-a) & (a+c) \end{pmatrix}$$

• I want to use gaussian elimination but I can't set $x=c-d$ rather I'd have to put numerical values in the matrix. My question is what would these values be? – Ski Mask Jun 5 '18 at 20:23
• @SkiMask I've added the derivation by gaussian elimination. – gimusi Jun 5 '18 at 21:32