Matrices & Determinants
System of linear equations - unique solution condition
Grade 12
Question:
<p>Given: \(x + y + z - 2 = 0,\; 2x + y - z - 3 = 0\) and \(3x + 2y + kz - 4 = 0\). For unique solution, \(\Delta \neq 0\). Find the value(s) of \(k\).</p>
<p>\(k \in R\)</p>
<p>\(k \in R - \{0\}\)</p>
<p>\(k = 0\)</p>
<p>\(k \in R - \{1\}\)</p>
Step-by-Step Solution
Key Concept: A system of three linear equations has a unique solution if and only if the coefficient matrix determinant is non-zero (Δ ≠ 0). We must find all values of k for which this determinant is not equal to zero.
<p><strong>Step 1: Set up the coefficient matrix</strong></p><p>For the system:
$$x + y + z = 2$$
$$2x + y - z = 3$$
$$3x + 2y + kz = 4$$</p><p>The coefficient matrix is:
$$A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{pmatrix}$$</p><p><strong>Step 2: Calculate the determinant Δ</strong></p><p>$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{vmatrix}$$</p><p>Expanding along the first row:</p><p>$$\Delta = 1\begin{vmatrix} 1 & -1 \\ 2 & k \end{vmatrix} - 1\begin{vmatrix} 2 & -1 \\ 3 & k \end{vmatrix} + 1\begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix}$$</p><p><strong>Step 3: Evaluate 2×2 determinants</strong></p><p>$$\begin{vmatrix} 1 & -1 \\ 2 & k \end{vmatrix} = k - (-2) = k + 2$$</p><p>$$\begin{vmatrix} 2 & -1 \\ 3 & k \end{vmatrix} = 2k - (-3) = 2k + 3$$</p><p>$$\begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = 4 - 3 = 1$$</p><p><strong>Step 4: Substitute back</strong></p><p>$$\Delta = 1(k + 2) - 1(2k + 3) + 1(1)$$</p><p>$$\Delta = k + 2 - 2k - 3 + 1$$</p><p>$$\Delta = -k$$</p><p><strong>Step 5: Apply condition for unique solution</strong></p><p>For a unique solution: $\Delta \neq 0$</p><p>$$-k \neq 0$$</p><p>$$k \neq 0$$</p><p>$$k \in \mathbb{R} - \{0\}$$</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B