Matrix method of 2x 3y 4z=29,x y 2z=13,3x 2y 1z=16

matrix method of 2x 3y 4z=29,x y 2z=13,3x 2y 1z=16

Solving a System of Linear Equations Using the Matrix Method

The system of linear equations given is:

  1. 2x + 3y + 4z = 29
  2. x + y + 2z = 13
  3. 3x + 2y + z = 16

We will solve this system using the matrix method, which involves converting the system into a matrix equation, finding the inverse of the coefficient matrix if applicable, and using it to solve for the variables. Let’s go step by step.


1. Represent the System as a Matrix Equation

The general system of linear equations can be represented in matrix form as:

A \cdot X = B

Where:

  • A is the coefficient matrix,
  • X is the column matrix of variables,
  • B is the constant matrix.

For this system:

Coefficient Matrix (A):

The coefficients of x, y, and z from the equations form the matrix:

A = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{bmatrix}

Variable Matrix (X):

This represents the variables x, y, and z:

X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}

Constant Matrix (B):

This represents the constants on the right-hand side of the equations:

B = \begin{bmatrix} 29 \\ 13 \\ 16 \end{bmatrix}

Now, substituting these matrices into A \cdot X = B, we get:

\begin{bmatrix} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 29 \\ 13 \\ 16 \end{bmatrix}

2. Solve for X Using the Matrix Method

From the matrix equation A \cdot X = B, we solve for X using:

X = A^{-1} \cdot B

Here, A^{-1} is the inverse of the coefficient matrix A (if it exists). To proceed, we’ll calculate A^{-1}.


a. Calculate the Determinant of A

To find A^{-1}, the determinant of A must be non-zero. The determinant of a 3 \times 3 matrix is calculated as:

\text{det}(A) = \begin{vmatrix} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{vmatrix}

Expanding along the first row:

\text{det}(A) = 2 \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} - 3 \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix}

Now compute the 2 \times 2 determinants:

  • \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (2)(2) = 1 - 4 = -3
  • \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = (1)(1) - (3)(2) = 1 - 6 = -5
  • \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = (1)(2) - (3)(1) = 2 - 3 = -1

Substitute these back into the determinant calculation:

\text{det}(A) = 2(-3) - 3(-5) + 4(-1)
\text{det}(A) = -6 + 15 - 4 = 5

Since \text{det}(A) \neq 0, the matrix is invertible, and we can proceed to calculate A^{-1}.


b. Find the Inverse of A

The inverse of a 3 \times 3 matrix is given by:

A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A)

Where \text{Adj}(A) is the adjugate (or adjoint) of A.

The adjugate is the transpose of the cofactor matrix of A. Compute the cofactor matrix first.

For A = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{bmatrix} , calculate the minors and cofactors for each element:

  1. For the element in position (1,1) (i.e., 2):
    Minor:
    \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = -3 , so the cofactor is +(-3) = -3.

  2. For (1,2) (3):
    Minor:
    \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = -5 , so the cofactor is -(-5) = +5.

  3. For (1,3) (4):
    Minor:
    \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = -1 , so the cofactor is +(-1) = -1.

  4. Repeat this process for all elements to find the cofactor matrix:
    After full computation, the cofactor matrix is:

    \text{Cofactor}(A) = \begin{bmatrix} -3 & 5 & -1 \\ -3 & -5 & 5 \\ 8 & -2 & -1 \end{bmatrix}
  5. Transpose the cofactor matrix to get the adjugate:

    \text{Adj}(A) = \begin{bmatrix} -3 & -3 & 8 \\ 5 & -5 & -2 \\ -1 & 5 & -1 \end{bmatrix}
  6. Now, compute A^{-1}:

    A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A)

    Since \text{det}(A) = 5:

    A^{-1} = \frac{1}{5} \begin{bmatrix} -3 & -3 & 8 \\ 5 & -5 & -2 \\ -1 & 5 & -1 \end{bmatrix}
    A^{-1} = \begin{bmatrix} -3/5 & -3/5 & 8/5 \\ 1 & -1 & -2/5 \\ -1,5