If `[[1, 1, 1], [1, -2, -2], [1, 3, 1]][[x], [y], [z]]=[[0], [3], [4]]`, then `[[x], [y], [z]]=`

Options

A. `[[1], [1], [1]]`
B. `[[1], [-2], [1]]`
C. `[[1], [-2], [3]]`
D. `[[1], [2], [-3]]`

Answer

To solve the equation given in matrix form:

\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 4 \end{bmatrix} \]

we need to find the values of \(x\), \(y\), and \(z\). We can represent this as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. To find \(X\), we can use the formula:

\[ X = A^{-1}B \]

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

The first step is to calculate the determinant of matrix \(A\):

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 3 & 1 \end{bmatrix} \]

The determinant \(|A|\) can be calculated using the formula for a 3x3 matrix:

\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

\[ |A| = 1((-2)(1) - (-2)(3)) - 1(1(1) - (-2)(1)) + 1(1(3) - (-2)(1)) \]

Calculating each term:

- First term: \(1(-2 + 6) = 1 \cdot 4 = 4\)
- Second term: \(-1(1 + 2) = -1 \cdot 3 = -3\)
- Third term: \(1(3 + 2) = 1 \cdot 5 = 5\)

Thus,

\[ |A| = 4 - 3 + 5 = 6 \]

### Step 2: Find the Adjoint of Matrix \(A\)

Next, we need to find the cofactor matrix and then the adjoint of \(A\).

The cofactor matrix \(C\) is calculated by finding the determinant of the 2x2 matrices formed by removing the row and column of each element, applying the sign convention:

1. For element \(a_{11} = 1\): \(C_{11} = (-2)(1) - (-2)(3) = -2 + 6 = 4\)
2. For element \(a_{12} = 1\): \(C_{12} = -((1)(1) - (-2)(1)) = -1 + 2 = 1\) (with a negative sign, so \(-1\))
3. For element \(a_{13} = 1\): \(C_{13} = (1)(1) - (-2)(1) = 1 + 2 = 3\)

Continuing this for all elements, we get:

\[ C = \begin{bmatrix} 4 & -1 & 3 \\ -3 & 1 & -5 \\ 2 & -1 & -3 \end{bmatrix} \]

Now, we find the adjoint \(A^*\) by taking the transpose of the cofactor matrix:

\[ A^* = C^T = \begin{bmatrix} 4 & -3 & 2 \\ -1 & 1 & -1 \\ 3 & -5 & -3 \end{bmatrix} \]

### Step 3: Find the Inverse of Matrix \(A\)

Using the formula for the inverse:

\[ A^{-1} = \frac{1}{|A|} A^* \]

Substituting the values we found:

\[ A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & -3 & 2 \\ -1 & 1 & -1 \\ 3 & -5 & -3 \end{bmatrix} \]

### Step 4: Multiply \(A^{-1}\) with \(B\)

Now we multiply \(A^{-1}\) with matrix \(B\):

\[ B = \begin{bmatrix} 0 \\ 3 \\ 4 \end{bmatrix} \]

Calculating \(X = A^{-1}B\):

\[ X = \frac{1}{6} \begin{bmatrix} 4 & -3 & 2 \\ -1 & 1 & -1 \\ 3 & -5 & -3 \end{bmatrix} \begin{bmatrix} 0 \\ 3 \\ 4 \end{bmatrix} \]

Calculating each element:

1. First row: \(4(0) + (-3)(3) + 2(4) = 0 - 9 + 8 = -1\)
2. Second row: \(-1(0) + 1(3) + (-1)(4) = 0 + 3 - 4 = -1\)
3. Third row: \(3(0) + (-5)(3) + (-3)(4) = 0 - 15 - 12 = -27\)

Thus,

\[ X = \frac{1}{6} \begin{bmatrix} -1 \\ -1 \\ -27 \end{bmatrix} = \begin{bmatrix} -\frac{1}{6} \\ -\frac{1}{6} \\ -\frac{27}{6} \end{bmatrix} \]

### Final Result

The values of \(x\), \(y\), and \(z\) are:

\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -\frac{1}{6} \\ -\frac{1}{6} \\ -\frac{27}{6} \end{bmatrix} \]