If the matrix `A=[{:(2-x," "1,1),(" "1,3-x,0),(-1,-3,-x):}]` is singular, then what is the solution set S ?

Options

A. S = {0, 2, 3}
B. S = {-1, 2, 3}
C. S = {1, 2, 3}
D. S = {2, 3}

Explanation

To determine the solution set \( S \) for the matrix \( A \) given by

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

we need to find the values of \( x \) for which the matrix is singular. A matrix is singular if its determinant is equal to zero.

### Step 1: Calculate the Determinant of Matrix \( A \)

The determinant of a \( 3 \times 3 \) matrix

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

is given by the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

- \( a = 2 - x \)
- \( b = 1 \)
- \( c = 1 \)
- \( d = 1 \)
- \( e = 3 - x \)
- \( f = 0 \)
- \( g = -1 \)
- \( h = -3 \)
- \( i = -x \)

Now, we can calculate the determinant:

\[ \text{det}(A) = (2 - x)((3 - x)(-x) - (0)(-3)) - (1)((1)(-x) - (0)(-1)) + (1)((1)(-3) - (3 - x)(-1)) \]

Calculating each term:

1. \( (3 - x)(-x) = -3x + x^2 \)
2. \( (1)(-x) = -x \)
3. \( (1)(-3) - (3 - x)(-1) = -3 + 3 - x = -x \)

Putting it all together:

\[ \text{det}(A) = (2 - x)(-3x + x^2) + x - (-x) = (2 - x)(-3x + x^2) + x + x \]

This simplifies to:

\[ \text{det}(A) = (2 - x)(x^2 - 3x) + 2x \]

### Step 2: Set the Determinant to Zero

To find when the matrix is singular, we set the determinant to zero:

\[ (2 - x)(x^2 - 3x) + 2x = 0 \]

Expanding this:

\[ (2 - x)(x^2 - 3x) = 2x^2 - 6x - x^3 + 3x^2 = -x^3 + 5x^2 - 6x \]

Thus, we have:

\[ -x^3 + 5x^2 - 6x + 2x = 0 \]

This simplifies to:

\[ -x^3 + 5x^2 - 4x = 0 \]

Factoring out \( -x \):

\[ -x(x^2 - 5x + 4) = 0 \]

### Step 3: Solve the Factored Equation

Setting each factor to zero gives us:

1. \( -x = 0 \) which implies \( x = 0 \)
2. \( x^2 - 5x + 4 = 0 \)

Now we can solve the quadratic equation \( x^2 - 5x + 4 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = -5, c = 4 \):

\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2} \]

Calculating the two possible values:

1. \( x = \frac{8}{2} = 4 \)
2. \( x = \frac{2}{2} = 1 \)

### Step 4: Conclusion

The values of \( x \) for which the matrix \( A \) is singular are:

\[ x = 0, 1, 4 \]

Thus, the solution set \( S \) is:

\[ S = \{0, 1, 4\} \]