The inverse of the matrix $\begin{bmatrix}1&a&2\ 1&2&5\ 2&1&1\end{bmatrix}$ will not exist when a is equal to:

Q: The inverse of the matrix $\begin{bmatrix}1&a&2\\ 1&2&5\\ 2&1&1\end{bmatrix}$ will not exist when a is equal to:

The inverse of the matrix $\begin{bmatrix}1&a&2\\ 1&2&5\\ 2&1&1\end{bmatrix}$ will not exist when a is equal to:

Question

Accepted Answer

To determine when the inverse of the matrix $ \begin{bmatrix} 1 & a & 2 \ 1 & 2 & 5 \ 2 & 1 & 1 \end{bmatrix} $ does not exist, we need to find when the determinant of the matrix is zero. The determinant of a matrix is a scalar value that is a function of its entries and provides important information about the matrix, such as whether it is invertible. A matrix is invertible if and only if its determinant is non-zero.

Let's calculate the determinant of the given matrix:

$$
\text{det} = \begin{vmatrix} 1 & a & 2 \ 1 & 2 & 5 \ 2 & 1 & 1 \end{vmatrix}
$$

Using the formula for the determinant of a 3x3 matrix, we have:

$$
\text{det} = 1 \cdot \begin{vmatrix} 2 & 5 \ 1 & 1 \end{vmatrix} - a \cdot \begin{vmatrix} 1 & 5 \ 2 & 1 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & 2 \ 2 & 1 \end{vmatrix}
$$

Calculating each of these 2x2 determinants:

1. $\begin{vmatrix} 2 & 5 \ 1 & 1 \end{vmatrix} = (2)(1) - (5)(1) = 2 - 5 = -3$

2. $\begin{vmatrix} 1 & 5 \ 2 & 1 \end{vmatrix} = (1)(1) - (5)(2) = 1 - 10 = -9$

3. $\begin{vmatrix} 1 & 2 \ 2 & 1 \end{vmatrix} = (1)(1) - (2)(2) = 1 - 4 = -3$

Substituting these values back into the determinant expression:

$$
\text{det} = 1(-3) - a(-9) + 2(-3) = -3 + 9a - 6
$$

Simplifying:

$$
\text{det} = 9a - 9
$$

Setting the determinant to zero to find when the matrix is not invertible:

$$
9a - 9 = 0
$$

Solving for $a$:

$$
9a = 9 \
a = 1
$$

Thus, the determinant is zero when $a = 1$, meaning the matrix is not invertible when $a = 1$. Therefore, Option 2 is correct.

Now, let's explain why the other options are incorrect:

- **Option 1 (a = 2):** Substituting $a = 2$ into the determinant expression gives $9(2) - 9 = 18 - 9 = 9$, which is non-zero. Therefore, the matrix is invertible when $a = 2$.

- **Option 3 (a = 0):** Substituting $a = 0$ into the determinant expression gives $9(0) - 9 = -9$, which is non-zero. Therefore, the matrix is invertible when $a = 0$.

- **Option 4 (a = -1):** Substituting $a = -1$ into the determinant expression gives $9(-1) - 9 = -9 - 9 = -18$, which is non-zero. Therefore, the matrix is invertible when $a = -1$.

In conclusion, the matrix is not invertible only when $a = 1$, which supports Option 2 as the correct answer.