If `A=[(1,x),(x^(2),4y)]` and `B=[(-3,1),(1,0)], adj(A)+B=[(1,0),(0,1)]` then the values of `x` and `y` are

Options

A. 1,1
B. `+-1,1`
C. 1,0
D. None of these

Explanation

To solve the problem, we need to find the values of \( x \) and \( y \) given the matrices \( A \) and \( B \) and the equation involving their adjoint. Let's break it down step by step.

### Step 1: Define the Matrices
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & x \\ x^2 & 4y \end{pmatrix} \]
The matrix \( B \) is given as:
\[ B = \begin{pmatrix} -3 & 1 \\ 1 & 0 \end{pmatrix} \]

### Step 2: Find the Adjoint of Matrix \( A \)
The adjoint of a matrix is the transpose of its cofactor matrix. We will first find the cofactors of matrix \( A \).

- The cofactor \( C_{11} \) is the determinant of the submatrix obtained by removing the first row and first column:
\[ C_{11} = 4y \]

- The cofactor \( C_{12} \) is:
\[ C_{12} = -x^2 \]

- The cofactor \( C_{21} \) is:
\[ C_{21} = -x \]

- The cofactor \( C_{22} \) is:
\[ C_{22} = 1 \]

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} 4y & -x^2 \\ -x & 1 \end{pmatrix} \]

Now, we take the transpose of the cofactor matrix to find the adjoint of \( A \):
\[ \text{adj}(A) = C^T = \begin{pmatrix} 4y & -x \\ -x^2 & 1 \end{pmatrix} \]

### Step 3: Set Up the Equation
We know from the problem that:
\[ \text{adj}(A) + B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Substituting the matrices, we have:
\[ \begin{pmatrix} 4y & -x \\ -x^2 & 1 \end{pmatrix} + \begin{pmatrix} -3 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 4: Combine the Matrices
Now we combine the matrices:
\[ \begin{pmatrix} 4y - 3 & -x + 1 \\ -x^2 + 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 5: Set Up the System of Equations
From the above matrix equation, we can set up the following equations:
1. \( 4y - 3 = 1 \)
2. \( -x + 1 = 0 \)
3. \( -x^2 + 1 = 0 \)
4. \( 1 = 1 \) (This is always true and does not provide any new information)

### Step 6: Solve the Equations
**From Equation 2:**
\[ -x + 1 = 0 \implies x = 1 \]

**From Equation 1:**
\[ 4y - 3 = 1 \implies 4y = 4 \implies y = 1 \]

### Final Values
Thus, the values of \( x \) and \( y \) are:
\[ x = 1, \quad y = 1 \]

### Summary
The final answer is:
\[ (x, y) = (1, 1) \]