Let A =`[(2,3),(-1,5)]` . If `A^(-1)=xA+yI`, then the value of 2y+x , is ____

Answer

To solve the problem, we need to find the values of \(x\) and \(y\) such that \(A^{-1} = xA + yI\) for the matrix \(A = \begin{pmatrix} 2 & 3 \\ -1 & 5 \end{pmatrix}\). We will then calculate \(2y + x\).

### Step 1: Calculate the Inverse of Matrix A

The formula for the inverse of a \(2 \times 2\) matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is given by:

\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

For our matrix \(A\):

- \(a = 2\), \(b = 3\), \(c = -1\), \(d = 5\)
- Calculate the determinant \(ad - bc\):

\[ ad - bc = (2)(5) - (3)(-1) = 10 + 3 = 13 \]

Now, we can find \(A^{-1}\):

\[ A^{-1} = \frac{1}{13} \begin{pmatrix} 5 & -3 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{5}{13} & -\frac{3}{13} \\ \frac{1}{13} & \frac{2}{13} \end{pmatrix} \]

### Step 2: Set Up the Equation \(A^{-1} = xA + yI\)

We know that:

\[ A^{-1} = xA + yI \]

Where \(I\) is the identity matrix:

\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Thus,

\[ yI = y \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} y & 0 \\ 0 & y \end{pmatrix} \]

Now, substituting \(A\) and \(I\):

\[ xA = x \begin{pmatrix} 2 & 3 \\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 2x & 3x \\ -x & 5x \end{pmatrix} \]

So we have:

\[ A^{-1} = \begin{pmatrix} 2x + y & 3x \\ -x & 5x + y \end{pmatrix} \]

### Step 3: Equate the Matrices

Now, we equate \(A^{-1}\) with \(xA + yI\):

\[ \begin{pmatrix} \frac{5}{13} & -\frac{3}{13} \\ \frac{1}{13} & \frac{2}{13} \end{pmatrix} = \begin{pmatrix} 2x + y & 3x \\ -x & 5x + y \end{pmatrix} \]

This gives us the following equations:

1. \(2x + y = \frac{5}{13}\)
2. \(3x = -\frac{3}{13}\)
3. \(-x = \frac{1}{13}\)
4. \(5x + y = \frac{2}{13}\)

### Step 4: Solve for x

From equation 2:

\[ 3x = -\frac{3}{13} \implies x = -\frac{1}{13} \]

### Step 5: Solve for y

Substituting \(x = -\frac{1}{13}\) into equation 1:

\[ 2(-\frac{1}{13}) + y = \frac{5}{13} \implies -\frac{2}{13} + y = \frac{5}{13} \]

Solving for \(y\):

\[ y = \frac{5}{13} + \frac{2}{13} = \frac{7}{13} \]

### Step 6: Calculate \(2y + x\)

Now we can find \(2y + x\):

\[ 2y + x = 2 \left(\frac{7}{13}\right) + \left(-\frac{1}{13}\right) = \frac{14}{13} - \frac{1}{13} = \frac{13}{13} = 1 \]

### Final Answer

Thus, the value of \(2y + x\) is:

\[ \boxed{1} \]