Solve the following system of linear equations using the matrix inverse method.
\begin{array}{ccc} 3x+2y&=&7 \\ 4x+3y&=&10\end{array}
To solve the system of linear equations using the matrix inverse method, we need to represent the system as a matrix equation and find the inverse of the coefficient matrix.
Step 1: Represent the system as a matrix equation
We can write the given system of equations as the following matrix equation: AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
The coefficient matrix A is:
A = \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix}
The variable matrix X is:
X = \begin{bmatrix} x \\ y \end{bmatrix}
The constant matrix B is:
B = \begin{bmatrix} 7 \\ 10 \end{bmatrix}
So, the matrix equation becomes:
\begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 10 \end{bmatrix}
Step 2: Find the inverse of the coefficient matrix
To find the inverse of a 2x2 matrix, we use the formula:
If A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, then A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
In our case, A = \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix}, so the determinant, denoted by |A|, is found using the formula:
|A| = ad - bc = (3)(3) - (2)(4) = 1
Therefore, the inverse of A is:
A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -4 & 3 \end{bmatrix}
Step 3: Solve for X
To solve for X, we multiply both sides of the matrix equation by the inverse of A:
A^{-1}AX = A^{-1}B
Since A^{-1}A is the identity matrix, we have:
IX = A^{-1}B
Multiplying the matrices, we get:
\begin{bmatrix} 3 & -2 \\ -4 & 3 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -4 & 3 \end{bmatrix} \begin{bmatrix} 7 \\ 10 \end{bmatrix}
Simplifying the equation, we have:
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 17 \\ -18 \end{bmatrix}
Which can be written as:
\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 17 \\ -18 \end{bmatrix}
Therefore, the solution to the system of equations is x = 17 and y = -18.