If `A=[(1,2,2),(2,2,3),(1,-1,3)], C=[(2,1,1),(2,2,1),(1,1,1)], D=[(10),(13),(9)]`, and `CB=D`. Solve the equation `AX=B`.

Explanation

To solve the equation \( AX = B \) given the matrices \( A \), \( C \), and \( D \), we will follow these steps:

### Step 1: Write down the equation
We start with the equation:
\[ AX = B \]

### Step 2: Pre-multiply by matrix \( C \)
To eliminate \( A \) from the left side, we pre-multiply both sides by \( C \):
\[ CAX = CB \]
Since we are given that \( CB = D \), we can substitute:
\[ CAX = D \]

### Step 3: Find \( CA \)
Next, we need to calculate the product \( CA \). Given:
\[ C = \begin{pmatrix} 2 & 1 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 1 \end{pmatrix}, \quad A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 2 & 3 \\ 1 & -1 & 3 \end{pmatrix} \]
We perform the matrix multiplication \( CA \):
\[ CA = \begin{pmatrix} 2 & 1 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 2 \\ 2 & 2 & 3 \\ 1 & -1 & 3 \end{pmatrix} \]

Calculating each element:
- First row:
- \( 2*1 + 1*2 + 1*1 = 2 + 2 + 1 = 5 \)
- \( 2*2 + 1*2 + 1*(-1) = 4 + 2 - 1 = 5 \)
- \( 2*2 + 1*3 + 1*3 = 4 + 3 + 3 = 10 \)

- Second row:
- \( 2*1 + 2*2 + 1*1 = 2 + 4 + 1 = 7 \)
- \( 2*2 + 2*2 + 1*(-1) = 4 + 4 - 1 = 7 \)
- \( 2*2 + 2*3 + 1*3 = 4 + 6 + 3 = 13 \)

- Third row:
- \( 1*1 + 1*2 + 1*1 = 1 + 2 + 1 = 4 \)
- \( 1*2 + 1*2 + 1*(-1) = 2 + 2 - 1 = 3 \)
- \( 1*2 + 1*3 + 1*3 = 2 + 3 + 3 = 8 \)

Thus, we have:
\[ CA = \begin{pmatrix} 5 & 5 & 10 \\ 7 & 7 & 13 \\ 4 & 3 & 8 \end{pmatrix} \]

### Step 4: Find \( (CA)^{-1} \)
Next, we need to find the inverse of \( CA \). We can use the formula for the inverse of a 3x3 matrix or apply row reduction. For simplicity, we will calculate the inverse directly.

The determinant of \( CA \) can be calculated, and if it is non-zero, we can find the inverse.

After calculating, we find:
\[ (CA)^{-1} = \begin{pmatrix} -\frac{17}{5} & 2 & 1 \\ \frac{4}{5} & 0 & -1 \\ \frac{7}{5} & -1 & 0 \end{pmatrix} \]

### Step 5: Multiply \( (CA)^{-1} \) by \( D \)
Now, we multiply \( (CA)^{-1} \) by \( D \):
\[ D = \begin{pmatrix} 10 \\ 13 \\ 9 \end{pmatrix} \]
So we compute:
\[ X = (CA)^{-1}D \]

Calculating each element:
- First row:
- \( -\frac{17}{5}*10 + 2*13 + 1*9 = -34 + 26 + 9 = 1 \)

- Second row:
- \( \frac{4}{5}*10 + 0*13 - 1*9 = 8 - 9 = -1 \)

- Third row:
- \( \frac{7}{5}*10 - 1*13 + 0*9 = 14 - 13 = 1 \)

Thus, we find:
\[ X = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \]

### Final Solution
The solution to the equation \( AX = B \) is:
\[ X = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \]