Consider the following system of linear equations \[\left[ {\begin{array}{*{20}{c}} 2&1&{ – 4} \\ 4&3&{ – 12} \\ 1&2&{ – 8} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \\ {\text{z}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} \alpha \\ 5 \\ 7 \end{array}} \right]\] Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of \[\alpha \], does this system of equations have infinitely many solutions? A. 0 B. 1 C. 2 D. infinitely many

0
1
2
infinitely many

The correct answer is $\boxed{\text{D}}$.

A system of linear equations has infinitely many solutions if the augmented matrix has a nontrivial row of all zeros. In this case, the augmented matrix is

$$\left[ {\begin{array}{*{20}{c}} 2&1&{ – 4} & \alpha \ 4&3&{ – 12} & 5 \ 1&2&{ – 8} & 7 \end{array}} \right]$$

The second and third columns of this matrix are linearly dependent, so the matrix has a nontrivial row of all zeros. Therefore, the system of equations has infinitely many solutions.

To see this, let $x$ and $y$ be any two real numbers. Then we can solve the system of equations for $z$ as follows:

$$z = \frac{4x + 12y – 5\alpha}{1 – 2x}$$

Since $x$ and $y$ can be any real numbers, $z$ can also be any real number. Therefore, the system of equations has infinitely many solutions.

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