# Using linear equations to solve unknowns in matrices

• January 5th 2007, 03:47 AM
Trapper Dave
Using linear equations to solve unknowns in matrices
Question 87

Find the solution of the following set of linear equations.

[ 1 2 1 4 ] {x1} = { 13 }
[ 2 0 4 3 ] {x2} = {28.3}
[ 4 2 2 1 ] {x3} = {20.9}
[ -3 1 3 2 ] {x4} = {6.5 }

Just to be clear there are three matrices, of the orders 4x4, 4x1 and 4x1. I've managed to split them into the linear equations, but just looking at them make me sweat with fear.

x1 + 2 x2 + x3 + 4 x4 = 13
2 x1 + 4 x3 + 3 x4 = 28.3
4 x1 + 2 x2 + 2 x3 + x4 = 20.9
-3 x1 + x2 + 3 x3 + 2 x4 = 6.5
• January 5th 2007, 04:52 AM
Soroban
Hello, Trapper Dave!

You didn't explain what method you're supposed to use . . .

Quote:

Question 87

Find the solution of the following set of linear equations:

. . $\begin{array}{cccc}w + 2x + y + 4z & = & 13 \\ 2w \qquad + 4y + 3z & = & 28.3 \\ 4w + 2x + 2y + z & = & 20.9 \\ \text{-}3w + x + 3y + 2z & = & 6.5\end{array}$

$\begin{pmatrix}1&2&1&4\\2 &0&4&3\\4&2&2&1\\\text{-}3&1&3&2\end{pmatrix}\begin{pmatrix}w\\x\\y\\z\end {pmatrix} \;=\;\begin{pmatrix}13 \\ 28.3 \\ 20.9 \\ 6.5\end{pmatrix}$

We can solve the system using "normal" algebra . . . lots of it!

If we want a matrix solution, we need the inverse of the 4-by-4 matrix.

We'd work with: . $\begin{vmatrix}1&2&1&4&|&1&0 &0&0\: \\ 2&0&4&3&|&0&1&0&0\: \\ 4&2&2&1&|&0&0&1&0\: \\ \text{-}3&1&3&2&|&0&0&0&1\:\end{vmatrix}$

• January 5th 2007, 02:31 PM
Trapper Dave
So that would give me;

http://www.trapperdave.com/images/003.JPG

What should I do next? Any method at all will do.
• January 5th 2007, 03:28 PM
Plato
Quote:

Originally Posted by Trapper Dave
What should I do next? Any method at all will do.

$CV = D\quad \Rightarrow \quad V = C^{ - 1} D$
• January 6th 2007, 12:51 AM
CaptainBlack
Quote:

Originally Posted by Trapper Dave
So that would give me;

http://www.trapperdave.com/images/003.JPG

What should I do next? Any method at all will do.

If you did the inversion using the Gauss-Jordan method you would have
saved some work by doing it with:

. $\left( \begin{array}{cccc|c}1&2&1&4&13 \\ 2&0&4&3&28.3 \\ 4&2&2&1&20.9 \\ \text{-}3&1&3&2&6.5\end{array} \right)$

as this will solve the equations directly without the need for a further matrix
multiplication.

RonL