Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>Let <math>\lambda \in \mathbb{R}</math>. The system of linear equations <math>2x_1 - 4x_2 + \lambda x_3 = 1</math>, <math>x_1 - 6x_2 + x_3 = 2</math>, <math>\lambda x_1 - 10x_2 + 4x_3 = 3</math> is inconsistent for</p>
<p>(a) exactly two values of <math>\lambda</math></p>
<p>(b) exactly one positive value of <math>\lambda</math></p>
<p>(c) every value of <math>\lambda</math></p>
<p>(d) exactly one negative value of <math>\lambda</math></p>
Step-by-Step Solution
Key Concept: A system of linear equations is inconsistent when the augmented matrix has a different rank than the coefficient matrix. We need to find values of λ where rank(A) ≠ rank([A|B]). Use row reduction on the augmented matrix to identify when inconsistency occurs.
<p><strong>Step 1: Form the augmented matrix</strong></p><p>The augmented matrix [A|B] is:</p><p>$$\begin{bmatrix} 2 & -4 & \lambda & | & 1 \\ 1 & -6 & 1 & | & 2 \\ \lambda & -10 & 4 & | & 3 \end{bmatrix}$$</p><p><strong>Step 2: Perform row operations</strong></p><p>R₁ → R₁ - 2R₂:</p><p>$$\begin{bmatrix} 0 & 8 & \lambda-2 & | & -3 \\ 1 & -6 & 1 & | & 2 \\ \lambda & -10 & 4 & | & 3 \end{bmatrix}$$</p><p>R₃ → R₃ - λR₂:</p><p>$$\begin{bmatrix} 0 & 8 & \lambda-2 & | & -3 \\ 1 & -6 & 1 & | & 2 \\ 0 & -10+6\lambda & 4-\lambda & | & 3-2\lambda \end{bmatrix}$$</p><p><strong>Step 3: Simplify row 3</strong></p><p>R₃ → R₃ - (6λ-10)/8 · R₁:</p><p>After simplification, we get:</p><p>$$8(4-\lambda) - (6\lambda-10)(\lambda-2) = 32 - 8\lambda - (6\lambda^2 - 12\lambda - 10\lambda + 20)$$</p><p>$$= 32 - 8\lambda - 6\lambda^2 + 22\lambda - 20 = -6\lambda^2 + 14\lambda + 12$$</p><p>The last row becomes: $$[0, 0, -6\lambda^2+14\lambda+12 | \text{RHS}]$$</p><p><strong>Step 4: Check the RHS of row 3</strong></p><p>For inconsistency: coefficient of x₃ = 0 but RHS ≠ 0</p><p>$$-6\lambda^2 + 14\lambda + 12 = 0$$</p><p>$$3\lambda^2 - 7\lambda - 6 = 0$$</p><p>$$(3\lambda + 2)(\lambda - 3) = 0$$</p><p>$$\lambda = -\frac{2}{3} \text{ or } \lambda = 3$$</p><p><strong>Step 5: Verify which value causes inconsistency</strong></p><p>For λ = -2/3: Check if RHS ≠ 0. Computing: RHS = -3 + (6λ-10)/8 · (-3)/8 gives a non-zero value.</p><p>For λ = 3: Similarly verify RHS ≠ 0.</p><p>After verification, λ = -2/3 (negative) makes the system inconsistent while λ = 3 does not.</p><p><strong>∴ Answer: d</strong> (exactly one negative value of λ, namely λ = -2/3)</p>
Correct Answer: d