# Matrix multiplication problem.

1. Apr 6, 2005

### theCandyman

I know how to solve basic problems like this, but I have no clue where to start with one of the first parts in this example. I am given the following information about C, which is a 2 X 2 matrix.
$C \left[ \begin{array}{cc} 1\\ 2 \end{array} \right] = \left[ \begin{array}{cc} 2\\ 1 \end{array} \right]$ and $C^2 \left[ \begin{array}{cc} 1\\ 2 \end{array} \right] = \left[ \begin{array}{cc} -1\\ 1 \end{array} \right]$

The question asks for 2 X 2 matrices A and B so that CA = B, then solve for C.

My problem is finding what the B matrix is. A is $\left[ \begin{array}{cc} 1 & 2\\ 2 & 1 \end{array} \right]$, so how can I find B if C is squared?

2. Apr 7, 2005

### xanthym

SOLUTION HINTS:
From the problem statement, we have:

$$1: \ \ \ \ C \left[ \begin{array}{cc} 1\\ 2 \end{array} \right] \ = \ \left[ \begin{array}{cc} 2\\ 1 \end{array} \right]$$

$$2: \ \ \ \ C^2 \left[ \begin{array}{cc} 1\\ 2 \end{array} \right] \ = \ C \cdot C \cdot \left[ \begin{array}{cc} 1\\ 2 \end{array} \right] \ = \ \left[ \begin{array}{cc} -1\\ 1 \end{array} \right]$$

Thus, the following is also known from Eq #2 (together with Eq #1):

$$3: \ \ \ \ \color{red} C \cdot \color{blue} \left ( C \cdot \left[ \begin{array}{cc} 1\\ 2 \end{array} \right] \right ) \ = \ \color{red} C \cdot \color{blue} \left[ \begin{array}{cc} 2\\ 1 \end{array} \right] \ = \ \color{red} \left[ \begin{array}{cc} -1\\ 1 \end{array} \right]$$

Hence, from Eq #1 & #3, we can now write:

$$4: \ \ \ \ C \cdot \left[ \begin{array}{cc} 1 & 2\\ 2 & 1 \end{array} \right] \ = \ \left[ \begin{array}{cc} 2 & -1\\ 1 & 1 \end{array} \right ]$$

Solve for "C" by finding the INVERSE of the matrix shown below and multiplying both sides of the last equation (Eq #4) from the RIGHT:

$$5: \ \ \ \ \left[ \begin{array}{cc} 1 & 2\\ 2 & 1 \end{array} \right]$$

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Last edited: Apr 7, 2005