Matrices & Determinants
System of linear equations
Grade None
Question:
<p>If \(x = a\), \(y = b\), \(z = c\) is a solution of the system of linear equations<br>\(x + 8y + 7z = 0\)<br>\(9x + 2y + 3z = 0\)<br>\(x + y + z = 0\)<br>such that the point \((a, b, c)\) lies on the plane \(x + 2y + z = 6\), then \(2a + b + c\) equals</p>
<p>1</p>
<p>2</p>
<p>\(-1\)</p>
<p>0</p>
Step-by-Step Solution
Key Concept: For a non-trivial solution to exist in a homogeneous system, the coefficient matrix determinant must be zero. This gives a relationship between variables, which combined with the plane equation yields the specific solution.
<p><strong>Step 1:</strong> For the homogeneous system to have non-trivial solutions, the coefficient matrix determinant must equal zero:</p><p>$$\begin{vmatrix} 1 & 8 & 7 \\ 9 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix} = 0$$</p><p>Expanding: $1(2-3) - 8(9-3) + 7(9-2) = -1 - 48 + 49 = 0$ ✓</p><p><strong>Step 2:</strong> Since the determinant is zero, the solution space is non-trivial. Solve for the relationship between variables using any two equations:</p><p>From equations (1) and (3): $x + 8y + 7z = 0$ and $x + y + z = 0$</p><p>Subtracting: $7y + 6z = 0 \Rightarrow y = -\frac{6z}{7}$</p><p>From equation (3): $x = -y - z = \frac{6z}{7} - z = -\frac{z}{7}$</p><p>So the solution is proportional to $(a, b, c) = (-1, 6, 7)k$ for some scalar $k$.</p><p><strong>Step 3:</strong> Use the plane constraint $x + 2y + z = 6$:</p><p>$$-k + 12k + 7k = 6$$</p><p>$$18k = 6 \Rightarrow k = \frac{1}{3}$$</p><p>Therefore: $(a, b, c) = (-\frac{1}{3}, 2, \frac{7}{3})$</p><p><strong>Step 4:</strong> Calculate $2a + b + c$:</p><p>$$2a + b + c = 2(-\frac{1}{3}) + 2 + \frac{7}{3} = -\frac{2}{3} + 2 + \frac{7}{3} = \frac{-2 + 6 + 7}{3} = \frac{11}{3}$$</p><p>∴ Answer: B</p>
Correct Answer: B