If the matrix `[{:(alpha,2,2),(-3,0,4),(1,-1,1):}]` is not invertible, then :

Options

A. `alpha = -5`
B. `alpha = 5`
C. `alpha = 0`
D. `alpha = 1`

Explanation

To determine the value of \( \alpha \) for which the matrix

\[ A = \begin{pmatrix} \alpha & 2 & 2 \\ -3 & 0 & 4 \\ 1 & -1 & 1 \end{pmatrix} \]

is not invertible, we need to find the determinant of the matrix and set it equal to zero. A matrix is not invertible if its determinant is zero.

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

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 \( A \):

- \( a = \alpha \)
- \( b = 2 \)
- \( c = 2 \)
- \( d = -3 \)
- \( e = 0 \)
- \( f = 4 \)
- \( g = 1 \)
- \( h = -1 \)
- \( i = 1 \)

Substituting these values into the determinant formula, we get:

\[ \text{det}(A) = \alpha(0 \cdot 1 - 4 \cdot (-1)) - 2(-3 \cdot 1 - 4 \cdot 1) + 2(-3 \cdot (-1) - 0 \cdot 1) \]

### Step 2: Simplify the determinant expression

Calculating each term:

1. The first term:
\[ \alpha(0 + 4) = 4\alpha \]

2. The second term:
\[ -2(-3 - 4) = -2(-7) = 14 \]

3. The third term:
\[ 2(3 - 0) = 6 \]

Combining these, we have:

\[ \text{det}(A) = 4\alpha + 14 + 6 = 4\alpha + 20 \]

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

To find the value of \( \alpha \) for which the matrix is not invertible, we set the determinant to zero:

\[ 4\alpha + 20 = 0 \]

### Step 4: Solve for \( \alpha \)

Rearranging the equation gives:

\[ 4\alpha = -20 \]

Dividing both sides by 4:

\[ \alpha = -5 \]

### Conclusion

Thus, the value of \( \alpha \) for which the matrix is not invertible is:

\[ \alpha = -5 \]

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