Question

Solve the following system of linear equations by matrix method: \[ 5x + y - 3z = -10, \quad 3x - 2y + z = -3, \quad x + 3y + z = 4 \]

Hint:

In matrix methods, always write the system in the form \( AX = B \) and solve for \( X \) using \( X = A^{-1}B \).

Answer 1

Step 1: Writing the system in matrix form.
We can write the system of equations as a matrix equation \( AX = B \), where: \[ A = \begin{bmatrix} 5 & 1 & -3
3 & -2 & 1
1 & 3 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} -10
-3
4 \end{bmatrix} \] Step 2: Solving for \( X \) using the inverse of \( A \).
The solution is given by: \[ X = A^{-1}B \] First, find the inverse of matrix \( A \). The inverse of a 3x3 matrix is given by: \[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \] Find the determinant of \( A \): \[ \text{det}(A) = 5 \begin{vmatrix} -2 & 1
3 & 1 \end{vmatrix} - 1 \begin{vmatrix} 3 & 1
1 & 1 \end{vmatrix} + (-3) \begin{vmatrix} 3 & -2
1 & 3 \end{vmatrix} \] Simplifying the 2x2 determinants: \[ \text{det}(A) = 5((-2)(1) - (1)(3)) - 1((3)(1) - (1)(1)) + (-3)((3)(3) - (1)(-2)) \] \[ \text{det}(A) = 5(-2 - 3) - 1(3 - 1) + (-3)(9 + 2) = 5(-5) - 2 - 3(11) = -25 - 2 - 33 = -60 \] Step 3: Finding the adjugate matrix.
Next, we find the adjugate of \( A \), which is the transpose of the cofactor matrix. This involves calculating the minors for each element of the matrix and then taking the transpose of the cofactor matrix. After computing the adjugate matrix, we multiply it by \( \frac{1}{\text{det}(A)} = \frac{1}{-60} \) to find \( A^{-1} \). Step 4: Multiply \( A^{-1} \) by \( B \).
Finally, multiply the inverse of \( A \) by \( B \) to get the solution for \( X \). The resulting vector \( X \) will give the values of \( x \), \( y \), and \( z \).