For how many values of 'x' in the closed interval `[-4,-1]` is the matrix `[(3,-1+x,2),(3,-1,x+2),(x+3,-1,2)]` singular ? (A) `2` (B) `0` (C) `3` (D) `1`

Options

A. 0
B. 2
C. 1
D. 3

Explanation

To determine how many values of \( x \) in the closed interval \([-4, -1]\) make the matrix

\[ \begin{pmatrix} 3 & -1+x & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{pmatrix} \]

singular, we need to find when the determinant of this matrix is equal to zero.

### Step 1: Calculate the determinant of the matrix.

The determinant of a 3x3 matrix

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

is given by the formula:

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

For our matrix, we have:

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

Substituting these values into the determinant formula:

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

### Step 2: Simplify the determinant expression.

Calculating each part:

1. \( (-1)(2) - (-1)(x + 2) = -2 + x + 2 = x \)
2. \( 3(2) - (x + 2)(x + 3) = 6 - (x^2 + 5x + 6) = -x^2 - 5x \)
3. \( 3(-1) - (-1)(x + 3) = -3 + x + 3 = x \)

Now substituting back into the determinant:

\[ \text{det} = 3(x) - (-1 + x)(-x^2 - 5x) + 2(x) \]

Expanding this:

\[ \text{det} = 3x + (1 - x)(x^2 + 5x) + 2x \]

Distributing \( (1 - x) \):

\[ \text{det} = 3x + (x^2 + 5x - x^3 - 5x^2) + 2x \]

Combining like terms:

\[ \text{det} = -x^3 + 3x^2 + 10x \]

### Step 3: Set the determinant equal to zero.

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

\[ -x^3 + 3x^2 + 10x = 0 \]

Factoring out \( x \):

\[ x(-x^2 + 3x + 10) = 0 \]

This gives us one solution:

\[ x = 0 \]

Now we need to solve the quadratic equation:

\[ -x^2 + 3x + 10 = 0 \]

### Step 4: Use the quadratic formula.

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = -1, b = 3, c = 10 \):

\[ x = \frac{-3 \pm \sqrt{3^2 - 4(-1)(10)}}{2(-1)} = \frac{-3 \pm \sqrt{9 + 40}}{-2} = \frac{-3 \pm \sqrt{49}}{-2} = \frac{-3 \pm 7}{-2} \]

Calculating the two solutions:

1. \( x = \frac{4}{-2} = -2 \)
2. \( x = \frac{-10}{-2} = 5 \)

### Step 5: Identify the valid solutions in the interval \([-4, -1]\).

The solutions we have are:

1. \( x = 0 \) (not in the interval)
2. \( x = -2 \) (in the interval)
3. \( x = 5 \) (not in the interval)
4. \( x = -4 \) (boundary of the interval)

Thus, the only valid solution in the interval \([-4, -1]\) is \( x = -2 \) and \( x = -4 \).

### Conclusion

The number of values of \( x \) in the closed interval \([-4, -1]\) for which the matrix is singular is **2**.

### Final Answer

**(A) 2**