If a + x = b + y = c + z + 1, where a, b, c, x, y, z
are non-zero distinct real numbers, then
$$\left| {\matrix{
x & {a + y} & {x + a} \cr
y & {b + y} & {y + b} \cr
z & {c + y} & {z + c} \cr
} } \right|$$ is equal to :
Solution
$$\left| {\matrix{
x & {a + y} & {x + a} \cr
y & {b + y} & {y + b} \cr
z & {c + y} & {z + c} \cr
} } \right|$$
<br><br>C<sub>3</sub> $\to$ C<sub>3</sub> – C<sub>1</sub>
<br><br>= $$\left| {\matrix{
x & {a + y} & a \cr
y & {b + y} & b \cr
z & {c + y} & c \cr
} } \right|$$
<br><br>C<sub>2</sub> $\to$ C<sub>2</sub> – C<sub>3</sub>
<br><br>= $$\left| {\matrix{
x & y & a \cr
y & y & b \cr
z & y & c \cr
} } \right|$$
<br><br>R<sub>3</sub> $\to$ R<sub>3</sub> – R<sub>1</sub>, R<sub>2</sub> $\to$ R<sub>2</sub> – R<sub>1</sub>
<br><br>= $$\left| {\matrix{
x & y & a \cr
{y - x} & 0 & {b - a} \cr
{z - x} & 0 & {c - a} \cr
} } \right|$$
<br><br>= (–y)[(y – x) (c – a) – (b – a) (z – x)]
<br><br>Given, a + x = b + y = c + z + 1
<br><br>= (–y)[(a – b) (c – a) + (a – b) (a – c – 1)]
<br><br>= (–y)[(a – b) (c – a) + (a – b) (a – c) + b – a)
<br><br>= –y(b – a) = y(a – b)
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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