`[{:(,x^(2)+x,-1),(,3,2):}]+[{:(,0,-1),(,-x+1,x):}]=[{:(,0,-2),(,5,1):}]` then x is equalto-

Options

A. `-1`
B. 2
C. `4`
D. No value of x

Answer

To solve the equation involving matrices, we start with the given equation:

\[ \begin{pmatrix} x^2 + x & -1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 0 & -1 \\ -x + 1 & x \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 5 & 1 \end{pmatrix} \]

### Step 1: Add the two matrices on the left side.

We add the corresponding elements of the two matrices:

\[ \begin{pmatrix} x^2 + x + 0 & -1 + (-1) \\ 3 + (-x + 1) & 2 + x \end{pmatrix} = \begin{pmatrix} x^2 + x & -2 \\ 4 - x & 2 + x \end{pmatrix} \]

### Step 2: Set the resulting matrix equal to the right side.

Now we equate the resulting matrix to the matrix on the right side:

\[ \begin{pmatrix} x^2 + x & -2 \\ 4 - x & 2 + x \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 5 & 1 \end{pmatrix} \]

### Step 3: Equate the corresponding elements.

From the equality of the matrices, we have the following equations:

1. \(x^2 + x = 0\)
2. \(-2 = -2\) (This is always true and doesn't provide information)
3. \(4 - x = 5\)
4. \(2 + x = 1\)

### Step 4: Solve the equations.

**From the first equation:**

\[ x^2 + x = 0 \implies x(x + 1) = 0 \]

This gives us two solutions:

\[ x = 0 \quad \text{or} \quad x = -1 \]

**From the third equation:**

\[ 4 - x = 5 \implies -x = 1 \implies x = -1 \]

**From the fourth equation:**

\[ 2 + x = 1 \implies x = 1 - 2 \implies x = -1 \]

### Conclusion

The common solution from all equations is:

\[ x = -1 \]

Thus, the value of \(x\) is \(-1\).

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