# Thread: Matrix equations solving help needed?

1. ## Matrix equations solving help needed?

Can anyone solve this matrix question please:

Solve the matrix equation for A.
A 2x2 matrix of
1 2 multiplied by A
3 5

=

1 0
0 1

2. Originally Posted by alpha
Can anyone solve this matrix question please:

Solve the matrix equation for A.
A 2x2 matrix of
1 2 multiplied by A
3 5

=

1 0
0 1
Are you saying

$\displaystyle \left[\begin{matrix} 1 & 2 \\ 3 & 5 \end{matrix}\right]\left[\begin{matrix} a & b\\ c & d \end{matrix}\right] = \left[\begin{matrix} 1 & 0\\ 0 & 1 \end{matrix}\right]$?

Premultiply both sides of the equation by the inverse matrix.

3. a b = A
c d

a=-5
b= 2
c= 3
d= -1

or another way of writing it

[1,2;3,5] x [-5,2;3,-1] = [1,0;0,1]

sorry cant do a better way of showing it on this computer

4. Correct.

BTW, the matrix you were trying to write is $\displaystyle \left[\begin{matrix}-5 & 2\\ 3 & -1\end{matrix}\right]$.

5. provided we interpreted the question correctly

6. im new to this forum, how did you write the matrix in your post?

7. Use "LaTex" code. There is a tutorial here: http://www.mathhelpforum.com/math-help/latex-help/.

You can see the code used for
$\displaystyle \begin{bmatrix}1 & 2 \\ 3 & 5\end{bmatrix}A= \begin{bmatrix}1 & 0 \\0 & 1\end{bmatrix}$, or any formula, by clicking on it.

Since $\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ is the identity matrix, as other have told you, A is just the "inverse matrix to $\displaystyle \begin{bmatrix}1 & 2 \\ 3 & 5\end{bmatrix}$.

There are a number of different ways to find the inverse matrix to a given matrix and, of course, we don't know which you have learned or are expected to use.

One very fundamental way would be to write that matrix equations as four linear equations: Writing A as $\displaystyle \begin{bmatrix} a & b \\ c & d\end{bmatrix}$ we can write that matrix equation as
$\displaystyle \begin{bmatrix}1 & 2 \\ 3 & 5\end{bmatrix}A$$\displaystyle = \begin{bmatrix}1 & 0 \\0 & 1\end{bmatrix}\begin{bmatrix} a & b \\ c & d\end{bmatrix}$$\displaystyle = \begin{bmatrix}a+ 2c & b+ 2d \\ 3a+ 5c & 3b+ 5d\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ and, setting the same components equal, a+ 2c= 1, b+ 2d= 0, 3a+ 5c= 0, 3b+ 4d= 1, four equations in four values. (Actually, they can be separated- you really have two sets of two equation for two values which is a little easier.) Solve those for a, b, c, and d.