First slide

Introduction to Determinants

Question

If $|\begin{array}{lll} b + c c + a a + b \\ a + b b + c c + a \\ c + a a + b b + c \end{array}| = p |\begin{array}{lll} a b c \\ c a b \\ b c a \end{array}|$ , then the value of p is

Moderate

A

1
B

2
C

3
D

4

Solution

We have
$|\begin{array}{lll} b + c c + a a + b \\ a + b b + c c + a \\ c + a a + b b + c \end{array}| = p |\begin{array}{lll} a b c \\ c a b \\ b c a \end{array}|$
or $|\begin{array}{lll} 2 (a + b + c) c + a a + b \\ 2 (a + b + c) b + c c + a \\ 2 (a + b + c) a + b b + c \end{array}| = p |\begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array}| [Applying C_{1} \to C_{1} + (C_{2} + C_{3}) on L . H . S .]$
or $2 |\begin{array}{lll} a + b + c - b - c \\ a + b + c - a - b \\ a + b + c - c - a \end{array}| = p |\begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array}| [Applying C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1} on L . H . S]$
or $|\begin{array}{lll} a - b - c \\ c - a - b \\ b - c - a \end{array}| = p |\begin{array}{lll} a b c \\ c a b \\ b c a \end{array}| [Applying C_{1} \to C_{1} + C_{2} + C_{3} on L . H . S]$
or $2 |\begin{array}{lll} a b c \\ c a b \\ b c a \end{array}| = p |\begin{array}{lll} a b c \\ c a b \\ b c a \end{array}|$
$∴ p = 2$

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