If `A = [(1,0,1),(0,2,3),(1,2,1)]and B = [(1,2,3),(1,1,5),(2,4,7)]`, then find the matrix X such that XA = B

Answer

To find the matrix \( X \) such that \( XA = B \), we can use the formula:

\[ X = BA^{-1} \]

where \( A^{-1} \) is the inverse of matrix \( A \). We will follow these steps:

### Step 1: Calculate the Determinant of Matrix \( A \)

Given:
\[ A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} \]

To find the determinant of \( A \), we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):
- \( a = 1, b = 0, c = 1 \)
- \( d = 0, e = 2, f = 3 \)
- \( g = 1, h = 2, i = 1 \)

Calculating the determinant:

\[ \text{det}(A) = 1(2 \cdot 1 - 3 \cdot 2) - 0 + 1(0 \cdot 2 - 2 \cdot 1) \]
\[ = 1(2 - 6) + 1(0 - 2) \]
\[ = 1(-4) + 1(-2) = -4 - 2 = -6 \]

### Step 2: Calculate the Inverse of Matrix \( A \)

The inverse of a 3x3 matrix can be calculated using the formula:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

First, we need to find the adjugate of \( A \). The adjugate is the transpose of the cofactor matrix.

#### Finding the Cofactors

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \text{det}\begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix} = (2)(1) - (3)(2) = 2 - 6 = -4 \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\text{det}\begin{pmatrix} 0 & 3 \\ 1 & 1 \end{pmatrix} = -((0)(1) - (3)(1)) = -(-3) = 3 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = \text{det}\begin{pmatrix} 0 & 2 \\ 1 & 2 \end{pmatrix} = (0)(2) - (2)(1) = 0 - 2 = -2 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = -\text{det}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = -((1)(1) - (1)(1)) = -0 = 0 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \text{det}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = (1)(1) - (1)(1) = 0 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = -\text{det}\begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} = -((1)(2) - (0)(1)) = -2 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = \text{det}\begin{pmatrix} 0 & 3 \\ 2 & 3 \end{pmatrix} = (0)(3) - (3)(2) = 0 - 6 = -6 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\text{det}\begin{pmatrix} 1 & 1 \\ 0 & 3 \end{pmatrix} = -((1)(3) - (1)(0)) = -3 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = \text{det}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = (1)(2) - (0)(0) = 2 \]

Now, we can construct the cofactor matrix:

\[ \text{Cofactor}(A) = \begin{pmatrix} -4 & 3 & -2 \\ 0 & 0 & -2 \\ -6 & -3 & 2 \end{pmatrix} \]

Taking the transpose gives us the adjugate:

\[ \text{adj}(A) = \begin{pmatrix} -4 & 0 & -6 \\ 3 & 0 & -3 \\ -2 & -2 & 2 \end{pmatrix} \]

Now we can find \( A^{-1} \):

\[ A^{-1} = \frac{1}{-6} \begin{pmatrix} -4 & 0 & -6 \\ 3 & 0 & -3 \\ -2 & -2 & 2 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & 0 & 1 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} \end{pmatrix} \]

### Step 3: Calculate \( X \)

Now we can compute \( X \):

\[ X = BA^{-1} \]

Given:
\[ B = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{pmatrix} \]

Now we will multiply \( B \) by \( A^{-1} \):

\[ X = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{pmatrix} \begin{pmatrix} \frac{2}{3} & 0 & 1 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} \end{pmatrix} \]

Calculating each element of \( X \):

1. First row:
- \( X_{11} = 1 \cdot \frac{2}{3} + 2 \cdot -\frac{1}{2} + 3 \cdot \frac{1}{3} = \frac{2}{3} - 1 + 1 = \frac{2}{3} \)
- \( X_{12} = 1 \cdot 0 + 2 \cdot 0 + 3 \cdot \frac{1}{3} = 1 \)
- \( X_{13} = 1 \cdot 1 + 2 \cdot \frac{1}{2} + 3 \cdot -\frac{1}{3} = 1 + 1 - 1 = 1 \)

2. Second row:
- \( X_{21} = 1 \cdot \frac{2}{3} + 1 \cdot -\frac{1}{2} + 5 \cdot \frac{1}{3} = \frac{2}{3} - \frac{1}{2} + \frac{5}{3} = \frac{2 + 5 - 3}{6} = \frac{4}{6} = \frac{2}{3} \)
- \( X_{22} = 1 \cdot 0 + 1 \cdot 0 + 5 \cdot \frac{1}{3} = \frac{5}{3} \)
- \( X_{23} = 1 \cdot 1 + 1 \cdot \frac{1}{2} + 5 \cdot -\frac{1}{3} = 1 + \frac{1}{2} - \frac{5}{3} = \frac{3 + 1 - 5}{6} = -\frac{1}{6} \)

3. Third row:
- \( X_{31} = 2 \cdot \frac{2}{3} + 4 \cdot -\frac{1}{2} + 7 \cdot \frac{1}{3} = \frac{4}{3} - 2 + \frac{7}{3} = \frac{4 + 7 - 6}{3} = \frac{5}{3} \)
- \( X_{32} = 2 \cdot 0 + 4 \cdot 0 + 7 \cdot \frac{1}{3} = \frac{7}{3} \)
- \( X_{33} = 2 \cdot 1 + 4 \cdot \frac{1}{2} + 7 \cdot -\frac{1}{3} = 2 + 2 - \frac{7}{3} = \frac{6 + 6 - 7}{3} = \frac{5}{3} \)

So, we have:

\[ X = \begin{pmatrix} \frac{2}{3} & 1 & 1 \\ \frac{2}{3} & \frac{5}{3} & -\frac{1}{6} \\ \frac{5}{3} & \frac{7}{3} & \frac{5}{3} \end{pmatrix} \]

### Final Answer

The matrix \( X \) is:

\[ X = \begin{pmatrix} \frac{2}{3} & 1 & 1 \\ \frac{2}{3} & \frac{5}{3} & -\frac{1}{6} \\ \frac{5}{3} & \frac{7}{3} & \frac{5}{3} \end{pmatrix} \]