How to Determine Matrix B in Matrix Multiplication Problem?

[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[<br /> \begin{array}{cc}<br /> 1\\<br /> 2<br /> \end{array}<br /> \right] = \left[<br /> \begin{array}{cc}<br /> 2\\<br /> 1<br /> \end{array}<br /> \right] and C^2 \left[<br /> \begin{array}{cc}<br /> 1\\<br /> 2<br /> \end{array}<br /> \right] = \left[<br /> \begin{array}{cc}<br /> -1\\<br /> 1<br /> \end{array}<br /> \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[<br /> \begin{array}{cc}<br /> 1 &amp; 2\\<br /> 2 &amp; 1<br /> \end{array}<br /> \right], so how can I find B if C is squared?

 

[xanthym]

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[<br /> \begin{array}{cc}<br /> 1\\<br /> 2<br /> \end{array}<br /> \right] = \left[<br /> \begin{array}{cc}<br /> 2\\<br /> 1<br /> \end{array}<br /> \right] and C^2 \left[<br /> \begin{array}{cc}<br /> 1\\<br /> 2<br /> \end{array}<br /> \right] = \left[<br /> \begin{array}{cc}<br /> -1\\<br /> 1<br /> \end{array}<br /> \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[<br /> \begin{array}{cc}<br /> 1 &amp; 2\\<br /> 2 &amp; 1<br /> \end{array}<br /> \right], so how can I find B if C is squared?

SOLUTION HINTS:
From the problem statement, we have:

1: \ \ \ \ C \left[<br /> \begin{array}{cc}<br /> 1\\<br /> 2<br /> \end{array}<br /> \right] \ = \ \left[<br /> \begin{array}{cc}<br /> 2\\<br /> 1<br /> \end{array}<br /> \right]

2: \ \ \ \ C^2 \left[<br /> \begin{array}{cc}<br /> 1\\<br /> 2<br /> \end{array}<br /> \right] \ = \ C \cdot C \cdot \left[<br /> \begin{array}{cc}<br /> 1\\<br /> 2<br /> \end{array}<br /> \right]<br /> \ = \ \left[<br /> \begin{array}{cc}<br /> -1\\<br /> 1<br /> \end{array}<br /> \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[<br /> \begin{array}{cc}<br /> 1\\<br /> 2<br /> \end{array}<br /> \right] \right ) \ = \ \color{red} C \cdot \color{blue} \left[<br /> \begin{array}{cc}<br /> 2\\<br /> 1<br /> \end{array}<br /> \right] \ = \ \color{red} \left[<br /> \begin{array}{cc}<br /> -1\\<br /> 1<br /> \end{array}<br /> \right]

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

4: \ \ \ \ C \cdot \left[<br /> \begin{array}{cc}<br /> 1 &amp; 2\\<br /> 2 &amp; 1<br /> \end{array}<br /> \right] \ = \ \left[<br /> \begin{array}{cc}<br /> 2 &amp; -1\\<br /> 1 &amp; 1<br /> \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[<br /> \begin{array}{cc}<br /> 1 &amp; 2\\<br /> 2 &amp; 1<br /> \end{array}<br /> \right]


~~

 

[thepatientmental]

To solve this problem, we need to first understand the properties of matrix multiplication. In general, matrix multiplication is not commutative, meaning that AB is not always equal to BA. However, it is associative, meaning that (AB)C = A(BC). This means that we can rearrange the order of the matrices in a multiplication problem as long as we maintain the order of operations.

In this problem, we are given the information about C and asked to find matrices A and B so that CA = B. We know that C is a 2x2 matrix and we are given the results of multiplying C by itself (C^2). This means that C is a square matrix and we can use the associative property to rearrange the order of the matrices in the multiplication problem. So instead of finding B, we can find A and then use the associative property to find B.

To find A, we can use the given information that C multiplied by the column vector [1,2] equals [2,1]. This means that the first column of A must be [1,2] and the second column must be [2,1]. Therefore, A is \left[
\begin{array}{cc}
1 & 2\\
2 & 1
\end{array}
\right].

Now, to find B, we can use the associative property to rearrange the order of the matrices. Since we know that CA = B, we can also say that C^2A = CB. We are given the result of C^2 multiplied by the column vector [1,2], which is [-1,1]. This means that the first column of B must be [-1,1] and the second column must be [1,-1]. Therefore, B is \left[
\begin{array}{cc}
-1 & 1\\
1 & -1
\end{array}
\right].

Finally, to solve for C, we can use the associative property again to rearrange the order of the matrices. Since we know that CA = B, we can also say that CAA^{-1} = BA^{-1}. We know that A is invertible because it has a non-zero determinant, so we can find its inverse. The inverse of A is \frac{1}{3} \left[
\begin{array}{cc}
-1 & 2\\
2 & -