Solving a system of linear equations using matrices, specifically with the technique of matrix inverse, involves the following steps:
1. Write the system of linear equations in matrix form.
2. Check if the coefficient matrix has an inverse.
3. Calculate the inverse of the coefficient matrix.
4. Multiply the inverse of the coefficient matrix with the constant matrix to get the solution.
Let’s consider an example:
Suppose we have a system of linear equations:
$$\begin{aligned}
x - 2y &= 1\\
3x + 2y &= 11
\end{aligned}$$
Step 1: Write the system of linear equations in matrix form.
We can write the above system of linear equations as
AX = B
, where
$$A =
\begin{bmatrix}
1 & -2\\
3 & 2
\end{bmatrix}
, \quad
X =
\begin{bmatrix}
x\\
y
\end{bmatrix}
, \quad
B =
\begin{bmatrix}
1 \\
11
\end{bmatrix}$$
Step 2: Check if the coefficient matrix A has an inverse.
A square matrix has an inverse if and only if the determinant is nonzero, i.e.,
|det(A)| ≠ 0
.
The determinant of 2x2 matrix A is calculated as:
det(A) = (1)(2) − ( − 2)(3) = 2 + 6 = 8
Since |det(A)| = 8 ≠ 0, matrix A has an inverse.
Step 3: Calculate the inverse of the coefficient matrix A.
The inverse of a 2x2 matrix can be calculated using the formula:
$$A^{-1} = \frac{1}{\text{det}(A)}
\begin{bmatrix}
\phantom{-}d & -b\\
-c & \phantom{-}a
\end{bmatrix}$$
The inverse of matrix A is:
$$A^{-1} = \frac{1}{8}
\begin{bmatrix}
\phantom{-}2 & \phantom{-}2\\
-3 & \phantom{-}1
\end{bmatrix}$$
Step 4: Multiply the inverse of the coefficient matrix A with the constant matrix B to get the solution.
X = A − 1B
$$\begin{bmatrix}
x\\
y
\end{bmatrix}
=
\frac{1}{8}
\begin{bmatrix}
\phantom{-}2 & \phantom{-}2\\
-3 & \phantom{-}1
\end{bmatrix}
\begin{bmatrix}
1\\
11
\end{bmatrix}
=
\begin{bmatrix}
\phantom{-}\frac{5}{2} \\
\phantom{-}\frac{7}{4}
\end{bmatrix}$$
So, the solution to the system of linear equations is
$$x = \frac{5}{2}$$
, and
$$y = \frac{7}{4}$$
.