First slide

Introduction to Determinants

Question

Let $a, b, c \in R$ not all are equal and $Δ_{1} = |\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}|$ , $Δ_{2} = |\begin{matrix} a + 2 b & b + 3 c & c + 4 a \\ b + 2 c & c + 3 a & a + 4 b \\ c + 2 a & a + 3 b & b + 4 c \end{matrix}|$ then $\frac{Δ_{2}}{Δ_{1}} =$ _________ .

Moderate

Solution

$Δ_{2} = \underset{∆'}{|\begin{matrix} a & b + 3 c & c + 4 a \\ b & c + 3 a & a + 4 b \\ c & a + 3 b & b + 4 c \end{matrix}|} + 2 \underset{∆''}{|\begin{array}{ccc} b & b + 3 c & c + 4 a \\ c & c + 3 a & a + 4 b \\ a & a + 3 b & b + 4 c \end{array}|}$
Applying $C_{3} \to C_{3} - 4 C_{1} in Δ' and C_{2} \to C_{2} - C_{1}$ in $∆''$ , we get
$Δ_{2} = |\begin{matrix} a & b + 3 c & c \\ b & c + 3 a & a \\ c & a + 3 b & b \end{matrix}| + 2 \times 3 |\begin{matrix} b & c & c + 4 a \\ c & a & a + 4 b \\ a & b & b + 4 c \end{matrix}|$
Applying $C_{2} \to C_{2} - 3 C_{3} in Δ' and C_{3} \to C_{3} - C_{2} in Δ''$ , we get
$\begin{aligned} = |\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}| + 6 |\begin{array}{ccc} b & c & 4 a \\ c & a & 4 b \\ a & b & 4 c \end{array}| \\ = |\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}| + 24 |\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}| \end{aligned} = 25 |\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}| = 25 Δ_{1}$
$\Rightarrow \frac{Δ_{2}}{Δ_{1}} = 25$

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