# Thread: Calculating the inverse matrix

1. ## Calculating the inverse matrix

Hello everyone,

In my textbook I am asked to calculate the inverse matrix of $\begin{pmatrix}2 & -1\\1
& 3
\end{pmatrix}$
by a system of equations.
This is no problem at all, I quickly found that a = 3/7, b = 1/7, c = -1/7, d = 2/7 is the fitting set of solutions. It is then quickly proven that this inverse matrix can be written as $1/7\begin{pmatrix}3 & 1\\-1
& 2
\end{pmatrix}$
.

What I'm stuck with is proving that the inverse matrix for any matrix $\begin{pmatrix}p & q\\r
& s
\end{pmatrix}$
equals $1/(ps-qr)\begin{pmatrix}s & -q\\-r
& p
\end{pmatrix}$

If I create the system:
ap + br = 1
cp + dr = 0
aq + bs = 0
cq + ds = 1

I am stuck with the variables a, b, c and d. Has anyone got a clue how to do this proof?

Note: Please excuse me for the random <br/> in the matrices, I am not yet very good at LaTeX.

2. ## Re: Calculating the inverse matrix

Hi tomkoolen!

What do you get if you multiply your matrix with its supposed inverse matrix?
If you get the identity matrix, you're done.

As for the <br/>, they come from a deficiency in the latex engine of this site.
They occur when you put a new line in a latex expression.
The workaround is to put the entire latex expression on one line.
Or otherwise, if you latex expression is suitable for it, splitting up the latex expression in multiple latex expressions that each consist of one line.

3. ## Re: Calculating the inverse matrix

suppose ps - qr ≠ 0.

multiplying your first equation by q, and your third equation by p we get:

apq + bqr = q
apq + bps = 0

subtracting the top from the bottom we get:

b(ps - qr) = -q and since ps - qr ≠ 0:

b = -q/(ps - qr) (*)

similarly, multiplying your second equation by q, and your fourth equation by p gives:

cpq + dqr = 0
cpq + dps = p

subtracting the top from the bottom gives:

d(ps - qr) = p so:

d = p/(ps - qr) (**)

finally, substituting (*) back in:

ap + br = 1 yields:

ap - qr/(ps - qr) = 1

ap = (ps - qr)/(ps - qr) + qr/(ps - qr) = ps/(ps - qr), so if p ≠ 0:

a = s/(ps - qr)

and substituting (**) back in:

cq + ds = 1 gives:

cq + ps/(ps - qr) = 1

cq = (ps - qr)/(ps - qr) - ps/(ps - qr) = -qr(ps - qr), so if q ≠ 0:

c = -r/(ps - qr).

note that since we are assuming ps - qr ≠ 0, we can't have BOTH p,q = 0, or p,r = 0, or s,q = 0, or s,r = 0.

so let's look at what happens if p = 0. in this case, ps - qr = -qr ≠ 0, and your original equation 1 is now:

br = 1, so that b = 1/r = -q/(-qr) = -q/(ps - qr) (since ps = 0). and your original equation 3 is now:

aq + bs = 0, so that a = -bs/q (remember, if p = 0, q cannot be)

= (-1/r)(s/q) = s/(-qr) = s/(ps - qr) as before.

we have a similar situation when q = 0, it turns out then that c = -r/(ps) = -r/(ps - qr) (since then qr = 0).

so in all cases when ps - qr ≠ 0, solving the system of equations:

ap + br = 1
cp + dr = 0
aq + bs = 0
cq + ds = 1

leads to the solution:

a = s/(ps - qr)
b = -q/(ps - qr)
c = -r/(ps - qr)
d = p/(ps - qr)

if we call the quantity ps - qr = D, we can pretty this up a bit:

a = s/D
b = -q/D
c = -r/D
d = p/D

or, in matrix form:

$\begin{bmatrix}p&q\\r&s \end{bmatrix}^{-1} = \frac{1}{D}\begin{bmatrix}s&-q\\-r&p\end{bmatrix} = \frac{1}{ps - qr}\begin{bmatrix}s&-q\\-r&p\end{bmatrix}$