Solve the following equations using inverse of a matrix.
`{:(2x+5y=1),(3x+2y=7):}`

Answer

To solve the equations \(2x + 5y = 1\) and \(3x + 2y = 7\) using the inverse of a matrix, we can follow these steps:

### Step 1: Write the equations in matrix form
We can express the system of equations in the form \(Ax = b\), where:
- \(A\) is the coefficient matrix,
- \(x\) is the variable matrix,
- \(b\) is the constant matrix.

From the equations:
\[ A = \begin{pmatrix} 2 & 5 \\ 3 & 2 \end{pmatrix}, \quad x = \begin{pmatrix} x \\ y \end{pmatrix}, \quad b = \begin{pmatrix} 1 \\ 7 \end{pmatrix} \]

### Step 2: Find the inverse of matrix \(A\)
To find the inverse of matrix \(A\), we first need to calculate the determinant of \(A\):
\[ \text{det}(A) = (2)(2) - (5)(3) = 4 - 15 = -11 \]

Next, we find the adjoint of \(A\):
\[ \text{adj}(A) = \begin{pmatrix} 2 & -5 \\ -3 & 2 \end{pmatrix} \]

Now we can find \(A^{-1}\):
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{-11} \begin{pmatrix} 2 & -5 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{pmatrix} \]

### Step 3: Multiply \(A^{-1}\) by \(b\)
Now we can find \(x\) by multiplying \(A^{-1}\) with \(b\):
\[ x = A^{-1}b = \begin{pmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{pmatrix} \begin{pmatrix} 1 \\ 7 \end{pmatrix} \]

Calculating the multiplication:
\[ x = \begin{pmatrix} -\frac{2}{11} \cdot 1 + \frac{5}{11} \cdot 7 \\ \frac{3}{11} \cdot 1 - \frac{2}{11} \cdot 7 \end{pmatrix} = \begin{pmatrix} -\frac{2}{11} + \frac{35}{11} \\ \frac{3}{11} - \frac{14}{11} \end{pmatrix} = \begin{pmatrix} \frac{33}{11} \\ -\frac{11}{11} \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} \]

### Step 4: State the solution
Thus, the solution to the equations is:
\[ x = 3, \quad y = -1 \]