Using matrix-inversion method, solve the following system of equations: x + y + z = 6; 3x - y + 3z = 10; 5x + 5y - 4z = 3.

Steps to Solve

1. Set up the equations in matrix form

We start with the system of equations:

[ \begin{align*}

1. & \quad x + y + z = 6 \
2. & \quad 3x - y + 3z = 10 \
3. & \quad 5x + 5y - 4z = 3 \ \end{align*} ]

This can be represented in matrix form as $AX = B$, where

[ A = \begin{pmatrix} 1 & 1 & 1 \ 3 & -1 & 3 \ 5 & 5 & -4 \end{pmatrix}, \quad X = \begin{pmatrix} x \ y \ z \end{pmatrix}, \quad B = \begin{pmatrix} 6 \ 10 \ 3 \end{pmatrix} ]

2. Find the inverse of matrix A

To solve for $X$, we need to find the inverse of matrix $A$. The formula for the inverse of a 3x3 matrix is given by:

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

First, calculate the determinant of $A$:

[ \text{det}(A) = 1(-1 \cdot -4 - 3 \cdot 5) - 1(3 \cdot -4 - 3 \cdot 5) + 1(3 \cdot 5 - -1 \cdot 5) ]

Calculating this yields:

[ \text{det}(A) = 1(4 - 15) - 1(-12 - 15) + 1(15 + 5) = -11 + 27 + 20 = 36 ]

3. Calculate the adjugate of A

Now we calculate the adjugate of $A$, which is the transpose of the cofactor matrix. The adjugate matrix is:

[ \text{adj}(A) = \begin{pmatrix} (-1)(-4) - (3)(5) & -((1)(-4) - (3)(5)) & (1)(5) - (1)(-1) \ -(1 \cdot -4 - 3 \cdot 1) & (1 \cdot -4 - 1 \cdot 1) & -(1 \cdot 1 - (-1)(3)) \ (1)(5) - (1)(-1) & -((1)(3) - (1)(3)) & (1)(-1) - (3)(1) \end{pmatrix} ]

Simplifying gives us:

[ \text{adj}(A) = \begin{pmatrix} -11 & 1 & 6 \ -1 & -5 & 2 \ -8 & 0 & -2 \end{pmatrix} ]

4. Calculate the inverse of A

Now we calculate $A^{-1}$:

[ A^{-1} = \frac{1}{36}{\begin{pmatrix} -11 & 1 & 6 \ -1 & -5 & 2 \ -8 & 0 & -2 \end{pmatrix}} ]

Thus,

[ A^{-1} = \begin{pmatrix} -\frac{11}{36} & \frac{1}{36} & \frac{1}{6} \ -\frac{1}{36} & -\frac{5}{36} & \frac{1}{18} \ -\frac{2}{9} & 0 & -\frac{1}{18} \end{pmatrix} ]

5. Multiply A^{-1} by B to find X

Now we find $X$:

[ X = A^{-1}B ]

Calculating this gives us the values of $x$, $y$, and $z$.

6. Final Result Calculation

After performing the matrix multiplication, we find:

[ X = \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} ]