Matrices & Determinants
System of Linear Equations and Progressions
Grade None
Question:
<p>If the system of equations \(x = 2a\), \(y = 3b\), \(z = c\) has non-zero solutions, i.e., \(\Delta = 0\) where \(\Delta = \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0\), then \(a\), \(b\), \(c\) are in:</p>
<p>(1) AP</p>
<p>(2) GP</p>
<p>(3) HP</p>
<p>(4) None of these</p>
Step-by-Step Solution
Key Concept: Apply column operations to simplify the determinant, then recognize that the resulting condition establishes a specific relationship between coefficients. The determinant equals zero forces the rows (or columns after transformation) into linear dependence, which reveals an arithmetic/geometric progression relationship.
<p><strong>Step 1:</strong> Start with the determinant condition:</p><p>$$\Delta = \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Perform column operations: $C_2 \to C_2 - 2C_3$ and $C_1 \to C_1 - C_3$:</p><p>$$\Delta = \begin{vmatrix} 1-a & 2a-2a & a \\ 1-b & 3b-2b & b \\ 1-c & 4c-2c & c \end{vmatrix} = \begin{vmatrix} 1-a & 0 & a \\ 1-b & b & b \\ 1-c & 2c & c \end{vmatrix}$$</p><p><strong>Step 3:</strong> Expand along column 2:</p><p>$$\Delta = -0 + b\begin{vmatrix} 1-a & a \\ 1-c & c \end{vmatrix} - 2c\begin{vmatrix} 1-a & a \\ 1-b & b \end{vmatrix}$$</p><p>$$= b[(1-a)c - a(1-c)] - 2c[(1-a)b - a(1-b)]$$</p><p>$$= b[c - ac - a + ac] - 2c[b - ab - a + ab]$$</p><p>$$= b(c-a) - 2c(b-a) = bc - ab - 2bc + 2ac = 2ac - ab - bc$$</p><p><strong>Step 4:</strong> Set $\Delta = 0$:</p><p>$$2ac - ab - bc = 0$$</p><p>Divide by $abc$ (non-zero):</p><p>$$\frac{2}{b} - \frac{1}{c} - \frac{1}{a} = 0$$</p><p>$$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$</p><p><strong>Step 5:</strong> This is the harmonic mean condition! Rearranging:</p><p>$$\frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in AP}$$</p><p><strong>Therefore:</strong> $a, b, c$ are in <strong>Harmonic Progression (HP)</strong></p><p>∴ Answer: C</p>
Correct Answer: C