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:
- 2x + 3y + 4z = 29
- x + y + 2z = 13
- 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:
-
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. -
For (1,2) (3):
Minor:
\begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = -5 , so the cofactor is -(-5) = +5. -
For (1,3) (4):
Minor:
\begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = -1 , so the cofactor is +(-1) = -1. -
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} -
Transpose the cofactor matrix to get the adjugate:
\text{Adj}(A) = \begin{bmatrix} -3 & -3 & 8 \\ 5 & -5 & -2 \\ -1 & 5 & -1 \end{bmatrix} -
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