By using properties of determinants- show that-i - a-b-c 2 a 2 a- 2 b b-c-a 2 b- 2 c 2 c c-a-b -a-b-c3ii - x-y-2 z x y- z y-z-2 x y- z x z-x-2 y -2x-y-z3

(i) Applying R1 → R1 + R2 + R3, we have: Applying C2 → C2 − C1, C3 → C3 − C1, we have: Expanding along C3, we have: Hence, the given result is proved. ...

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Byju's Answer

Standard XII

Mathematics

Property 4

By using prop...

Question

By usingproperties of determinants, show that:
(i) $∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣$ $= (a + b + c)^{3}$
(ii) $∣ ∣ ∣ ∣ \begin{matrix} x + y + 2 z & x & y z & y + z + 2 x & y z & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣$ $= 2 (x + y + c)^{3}$

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Solution

(i) LHS= $∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣$
Applying R₁ → R₁+ R₂+ R₃, we have:
= $∣ ∣ ∣ ∣ \begin{matrix} a + b + c & a + b + c & a + b + c 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣$
= $(a + b + c)$ $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣$
Applying C₂ → C₂− C₁, C₃ →C₃ − C₁, we have:
= $(a + b + c)$ $∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 2 b & - b - c - a & 0 2 c & 0 & - c - a - b \end{matrix} ∣ ∣ ∣ ∣$
= $(a + b + c)^{3}$ $∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 2 b & - 1 & 0 2 c & 0 & - 1 \end{matrix} ∣ ∣ ∣ ∣$
Expanding along first row, we have:
= $(a + b + c)^{3} [1 ((- 1) (- 1) - 0)]$
= $(a + b + c)^{3}$
=RHS.
Hence, the given result is proved.
(ii) Given $∣ ∣ ∣ ∣ \begin{matrix} x + y + 2 z & x & y z & y + z + 2 x & y z & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣$
Applying C₁ → C₁+ C₂+ C₃, we have:
= $∣ ∣ ∣ ∣ \begin{matrix} 2 x + 2 y + 2 z & x & y 2 x + 2 y + 2 z & y + z + 2 x & y 2 x + 2 y + 2 z & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣$
= $(2 x + 2 y + 2 z) ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & y 1 & y + z + 2 x & y 1 & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣$
Applying R₂ → R₂− R₁and R₃ →R₃ − R₁, we have:
= $2 (x + y + z) ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & y 0 & y + z + x & 0 0 & 0 & z + x + y \end{matrix} ∣ ∣ ∣ ∣$
$= 2 (x + y + z)^{3} ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & y 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣$
Expanding along R₃, we have:
$= 2 (x + y + z)^{3} (1) (1 - 0)$
$= 2 (x + y + z)^{3}$
Hence, the given result is proved.

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Similar questions

Using the properties of determinants, show that:

(i)

∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣ = (a + b + {c)}^{2}

(ii)

∣ ∣ ∣ ∣ \begin{matrix} x + y + 2 z & x & y z & y + z + 2 x & y z & x & z + x + 2 y \end{matrix} ∣ ∣ ∣ ∣ = 2 (x + y + {z)}^{3}

Q. Using properties of determinants, prove that

∣ ∣ ∣ ∣ ∣ \begin{matrix} (x + y)^{2} & z x & z y z x & (z + y)^{2} & x y z y & x y & (z + x)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2 x y z (x + y + z)^{3}

Q. a -b-c2a2a1. ) 2b b-c-a 2b -(a+b+c)2c2c c-a-b(ii) z2y+z+2x2-2(x y+ z)z+x+2y

By using properties of determinants, show that:

(i)

(ii)

Q. Using properties of determinants,prove that

∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b q + r & r + p & p + q y + z & z + x & x + y \end{matrix} ∣ ∣ ∣ ∣ = 2 ∣ ∣ ∣ ∣ \begin{matrix} a & b & c p & q & r x & y & z \end{matrix} ∣ ∣ ∣ ∣

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By usingproperties of determinants, show that:(i)∣∣ ∣∣a−b−c2a2a2bb−c−a2b2c2cc−a−b∣∣ ∣∣=(a+b+c)3(ii)∣∣ ∣∣x+y+2zxyzy+z+2xyzxz+x+2y∣∣ ∣∣=2(x+y+c)3