A-b-c-c-b-ca-b-c-a-b-aa-b-c-2a-b -160-b-c -160-c-a

Let #160;LHS #160;= #8710;=a+b+c #160; #160; #160; #160; #160; #160; #160; #160; #160; c #160; #160; #160; #160; #160; #160; ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Identity in Binary Operation

a+b+c-c-b-ca+...

Question

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

Open in App

Solution

$Let LHS = ∆ = |a + b + c - c - b - c a + b + c - a - b - a a + b + c| = |a - c - b b a + b + c - a c - a a + b + c| [Applying C_{1} \to C_{1} + C_{2} + C_{3}] = |a + b a + b - (a + b) b + c b + c b + c c - a a + b + c| [Applying R_{1} \to R_{1} + R_{2} and R_{2} \to R_{2} + R_{3}] = (a + b) (b + c) |1 1 - 1 1 1 1 c - a a + b + c| [Taking out common factor from R_{1} and R_{2}] = (a + b) (b + c) |0 0 - 2 1 1 1 c - a a + b + c| [Applying R_{1} \to R_{1} - R_{2}] = (a + b) (b + c) \{(- 2) (- a - c)\} [Expanding along R_{1}] = 2 (a + b) (b + c) (c + a) = RHS$

Hence proved.

Suggest Corrections

Similar questions

Q. If a, b, c are real numbers such that

|\begin{matrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix}| = 0

, then show that either

a + b + c = 0 or, a = b = c

Q. If

a, b

and

c

are real numbers and

Δ = ∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b c + a & a + b & b + c a + b & b + c & c + a \end{matrix} ∣ ∣ ∣ ∣ = 0

,
show that either

a + b + c = 0

a = b = c

Q. Prove the following identity:

\frac{a {(b - c)}^{2}}{(c - a) (a - b)} + \frac{b {(c - a)}^{2}}{(a - b) (b - c)} + \frac{c {(a - b)}^{2}}{(b - c) (c - a)} = a + b + c

Q. Prove that

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

If a, b and c are real numbers, and,

Show that either a + b + c = 0 or a = b = c.

Join BYJU'S Learning Program

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

COURSES

EXAMS

CLASSES

EXAM PREPARATION

RESOURCES

COMPANY

FOLLOW US

FREE TEXTBOOK SOLUTIONS

STATE BOARDS

FOLLOW US

a+b+c-c-b-ca+b+c-a-b-aa+b+c=2a+bb+cc+a

LetLHS=∆=a+b+c-c-b-ca+b+c-a-b-aa+b+c=a-c-bba+b+c-ac-aa+b+cApplyingC1→C1+C2+C3=a+ba+b-a+bb+cb+cb+cc-aa+b+cApplyingR1→R1+R2andR2→R2+R3=a+bb+c11-1111c-aa+b+cTakingoutcommonfactorfromR1andR2=a+bb+c00-2111c-aa+b+cApplyingR1→R1-R2=a+bb+c-2-a-cExpandingalongR1=2a+bb+cc+a=RHS Hence proved.

COURSES

EXAMS

CLASSES

EXAM PREPARATION

RESOURCES

COMPANY

FOLLOW US

FREE TEXTBOOK SOLUTIONS

STATE BOARDS

FOLLOW US

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