# matrices

• Dec 2nd 2007, 03:23 PM
DINOCALC09
matrices
X is a 2 by 2 matric

2 -3
4 -1

multiplies by unkonw matrix X

=

-12 -9
1 3

Find the matrix X
• Dec 2nd 2007, 04:18 PM
jabroni1212
Make the unknow matrix and label each part, eg.

$\displaystyle \begin{bmatrix} {2}\;\; {-3}\\ {4}\;\;{-1} \end{bmatrix} \cdot\begin{bmatrix} {a}\;\; {b}\\ {c}\;\;{d} \end{bmatrix}=\begin{bmatrix} {-12}\;\; {-9}\\ {1}\;\;\;{3} \end{bmatrix}$
Perform the multiplication and solve for the variables, eg.
$\displaystyle 2a-3c=-12$
$\displaystyle 4a-c=1$
• Dec 2nd 2007, 04:24 PM
Soroban
Hello, DINOCALC09!

Quote:

$\displaystyle X$ is a 2x2 matrix such that: .$\displaystyle \begin{pmatrix}2 & \text{-}3 \\ 4 & \text{-}1\end{pmatrix}\,X \;=\;\begin{pmatrix}\text{-}12 & \text{-}9 \\ 1 & 3 \end{pmatrix}$

Find the matrix $\displaystyle X$

Let $\displaystyle X \:=\:\begin{pmatrix}w & x \\ y & z\end{pmatrix}$

Then we have: .$\displaystyle \begin{pmatrix}2 & \text{-}3 \\ 4 & \text{-}1\end{pmatrix}\,\begin{pmatrix}w & x \\ y & z \end{pmatrix} \;=\;\begin{pmatrix}\text{-}12 & \text{-}9 \\ 1 & 3\end{pmatrix}$

. . . . $\displaystyle \begin{pmatrix}2w-3y & 2x-3z \\ 4w - y & 4x-z\end{pmatrix} \;=\;\begin{pmatrix}-12 & -9 \\ 1 & 3\end{pmatrix}$

We have two systems of equations: .$\displaystyle \begin{Bmatrix}2w - 3y & = & \text{-}12 \\ 4w - y &=&1\end{Bmatrix}\quad\begin{Bmatrix}2x-3z &=&\text{-}9 \\ 4x-z &=&3\end{Bmatrix}$

Solve them . . .

You should get: .$\displaystyle X \;=\;\begin{pmatrix}\frac{3}{2} & \frac{9}{5} \\ 5 & \frac{21}{5}\end{pmatrix}$
• Dec 2nd 2007, 04:28 PM
topsquark
Quote:

Originally Posted by DINOCALC09
X is a 2 by 2 matric

2 -3
4 -1

multiplies by unkonw matrix X

=

-12 -9
1 3

Find the matrix X

Solve
$\displaystyle \left ( \begin{matrix} 2 & -3 \\ 4 & -1 \end{matrix} \right ) X = \left ( \begin{matrix} -12 & -9 \\ 1 & 3 \end{matrix} \right )$

Multiply both sides by the inverse of the coefficient matrix:
$\displaystyle \left ( \begin{matrix} 2 & -3 \\ 4 & -1 \end{matrix} \right ) ^{-1} = \frac{1}{2 \cdot -1 - (-3) \cdot 4} \left ( \begin{matrix} -1 & 3 \\ -4 & 2 \end{matrix} \right )$

So
$\displaystyle \frac{1}{10} \left ( \begin{matrix} -1 & 3 \\ -4 & 2 \end{matrix} \right ) \left ( \begin{matrix} 2 & -3 \\ 4 & -1 \end{matrix} \right ) X = \frac{1}{10} \left ( \begin{matrix} -1 & 3 \\ -4 & 2 \end{matrix} \right ) \left ( \begin{matrix} -12 & -9 \\ 1 & 3 \end{matrix} \right )$

$\displaystyle X = \frac{1}{10} \left ( \begin{matrix} 15 & 18 \\ 50 & 42 \end{matrix} \right )$

-Dan