If a-x-b-y-c-z-1 where a-b-c-x-y-zare non-zero distinct real numbers- then - x a-y x-a- y b-y y-b- z c-y z-c - is equals to

Step 1. Find The value of |[ x a+y x+a; y b+y y+b; z c+y z+c ]|: Given that a,b,c,x,y,z are non zero distinct real numbers and a+x=b+y=c+z+1Now, |[ ...

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Standard XII

Mathematics

Equivalence Relation

If a+x=b+y=c+...

Question

If $a + x = b + y = c + z + 1$ where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $|\begin{matrix} x & a + y & x + a \\ y & b + y & y + b \\ z & c + y & z + c \end{matrix}|$ is equals to

$y (a - b)$

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$0$

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$y (b - a)$

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$y (a - c)$

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Solution

The correct option is A
$y (a - b)$

Step 1. Find The value of $|\begin{matrix} x & a + y & x + a \\ y & b + y & y + b \\ z & c + y & z + c \end{matrix}|$ :
Given that $a, b, c, x, y, z$ are non zero-distinct real numbers
and $a + x = b + y = c + z + 1$
Now,
$|\begin{matrix} x & a + y & x + a \\ y & b + y & y + b \\ z & c + y & z + c \end{matrix}|$
Step 2.Operating $(C_{3} \to C_{3} - C_{1})$
$\Rightarrow |\begin{matrix} x & a + y & a \\ y & b + y & b \\ z & c + y & c \end{matrix}|$
Step 3. Operating $(C_{2} \to C_{2} - C_{3})$
$\Rightarrow |\begin{matrix} x & y & a \\ y & y & b \\ z & y & c \end{matrix}|$
$\Rightarrow y |\begin{matrix} x & 1 & a \\ y & 1 & b \\ z & 1 & c \end{matrix}|$
Step 4. Operating $R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}$
$\Rightarrow y |\begin{matrix} x & 1 & a \\ y - x & 0 & b - a \\ z - x & 0 & c - a \end{matrix}|$
$= y [x \times 0 - 1 (y - x) (c - a) - (b - a) (z - x) + a \times 0] = y [b z - b x - a z + a x - (c y - a y - c x + a x)] = y [b z - b x - a z - c y + a y + c x] = y [b (z - x) + a (y - z) + c (x - y)] = y [b a - c - 1 + a (c - b + 1) + c (b - a)] = y [a b - b c - b + a c - a b + a + b c - a c] = y (a - b)$
Hence, Option ‘A’ is Correct.

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Similar questions

Q. If a, b, c are non-zero real numbers and if the system of equations
(a − 1) x = y + z
(b − 1) y = z + x
(c − 1) z = x + y
has a non-trivial solution, then prove that ab + bc + ca = abc.

(i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then show that,

\frac{y - z}{a (b - c)} = \frac{z - x}{b (c - a)} = \frac{x - y}{c (a - b)}

(ii) If

\frac{x}{3 x - y - z} = \frac{y}{3 y - z - x} = \frac{z}{3 z - x - y}

and

x + y + z \neq 0

then show that the value of each ratio is equal to 1.

(iii) If

\frac{a x + b y}{x + y} = \frac{b x + a z}{x + z} = \frac{a y + b z}{y + z}

and

x + y + z \neq 0

then show that

\frac{a + b}{2}

.

(iv) If

\frac{y + z}{a} = \frac{z + x}{b} = \frac{x + y}{c}

then show that

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}

.

(v) If

\frac{3 x - 5 y}{5 z + 3 y} = \frac{x + 5 z}{y - 5 x} = \frac{y - z}{x - z}

then show that every ratio =

\frac{x}{y}

Q. If

a + x = b + y = c + z + 1

, where

a, b, c, x, y, z

are non-zero distinct real numbers, then

∣ ∣ ∣ ∣ \begin{matrix} x & a + y & x + a y & b + y & y + b z & c + y & z + c \end{matrix} ∣ ∣ ∣ ∣

is equal to:

Q. If

\frac{x y}{x + y} = a, \frac{x z}{x + z} = b

and

\frac{y z}{y + z} = c

where a, b, c are all non - zero numbers, then x equals to :

Q. If a, b, c are non zero real numbers and if the equations

(a - 1) x = y + z, (b - 1) y = z + x, (c - 1) z = (x + y)

has a non trivial solution then ab + bc + ca equals

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If a+x=b+y=c+z+1 where a,b,c,x,y,zare non-zero distinct real numbers, then xa+yx+ayb+yy+bzc+yz+c is equals to

If $a + x = b + y = c + z + 1$ where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $|\begin{matrix} x & a + y & x + a \\ y & b + y & y + b \\ z & c + y & z + c \end{matrix}|$ is equals to