[amp_mcq option1=”0″ option2=”either 0 or 1″ option3=”one of 0, 1 or -1″ option4=”any real number other than 5″ correct=”option4″]<\/p>\n
The correct answer is $\\boxed{\\text{B. either 0 or 1}}$.<\/p>\n
To solve the system of equations, we can use Gaussian elimination. First, we can subtract the first equation from the second equation to get $x_2 + x_3 = 1$. Then, we can subtract $4$ times the first equation from the third equation to get $-3x_2 – 3ax_3 = -3$.<\/p>\n
Now, we can express $x_2$ and $x_3$ in terms of $a$:<\/p>\n
$$x_2 = \\frac{1}{5} – a$$
\n$$x_3 = \\frac{1}{5} – \\frac{3a}{5}$$<\/p>\n
Substituting these expressions into the first equation, we get:<\/p>\n
$$x_1 + \\frac{1}{5} – a + 2 \\left( \\frac{1}{5} – \\frac{3a}{5} \\right) = 1$$<\/p>\n
Simplifying, we get:<\/p>\n
$$x_1 – 2a = -\\frac{4}{5}$$<\/p>\n
Therefore, $x_1 = 2a – \\frac{4}{5}$.<\/p>\n
We know that the system of equations has a unique solution if the determinant of the coefficient matrix is not equal to zero. The coefficient matrix for this system of equations is:<\/p>\n
$$\\begin{bmatrix} 1 & 1 & 2 \\\\ 1 & 2 & 3 \\\\ 1 & 4 & a \\end{bmatrix}$$<\/p>\n
The determinant of this matrix is:<\/p>\n
$$\\begin{vmatrix} 1 & 1 & 2 \\\\ 1 & 2 & 3 \\\\ 1 & 4 & a \\end{vmatrix} = -a^2 + 7a – 12$$<\/p>\n
For the system of equations to have a unique solution, this determinant must not be equal to zero. Therefore, $a^2 – 7a + 12 \\neq 0$.<\/p>\n
Solving this equation, we get $a = 0$ or $a = 1$. Therefore, the only possible values for $a$ are $\\boxed{\\text{0 or 1}}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
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