First slide

Introduction to Determinants

Question

If x, y, z are in A.P., then the value of the determinant $|\begin{array}{lll} a + 2 a + 3 a + 2 x \\ a + 3 a + 4 a + 2 y \\ a + 4 a + 5 a + 2 z \end{array}|$ is

Moderate

A

1
B

0
C

2a
D

a

Solution

Since x, y, z are in A.P., therefore, x + z - 2y = 0. Now,
$|\begin{array}{lll} a + 2 a + 3 a + 2 x \\ a + 3 a + 4 a + 2 y \\ a + 4 a + 5 a + 2 z \end{array}| = |\begin{array}{ccc} 0 & 0 & 2 (x + z - 2 y) \\ a + 3 & a + 4 & a + 2 y \\ a + 4 & a + 5 & a + 2 z \end{array}| [Applying R_{1} \to R_{1} + R_{3} - 2 R_{2}]$
$\begin{array}{l} = |\begin{array}{ccc} 0 & 0 & 0 \\ a + 3 & a + 4 & a + 2 y \\ a + 4 & a + 5 & a + 2 z \end{array}| [∵ x + z - 2 y = 0] \\ = 0 \end{array}$

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