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  • The determinant equals A. abc (b–c) (c – a) (a – b)

The determinant \left|\begin{array}{ccc} b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2} \end{array}\right|  equals
 

A. abc (b–c) (c – a) (a – b)

B. (b–c) (c – a) (a – b)
C. (a + b + c) (b – c) (c – a) (a – b)
D. None of these

Answers (1)

D)

Given:

\left|\begin{array}{ccc} b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2} \end{array}\right|

\\=\left|\begin{array}{lll}b(b-a) & b-c & c(b-a) \\ a(b-a) & a-b & b(b-a) \\ c(b-a) & c-a & a(b-a)\end{array}\right|$ \\Take (b-a) common from both Column 1 and 3\\ $=(\mathrm{b}-\mathrm{a})(\mathrm{b}-\mathrm{a})\left|\begin{array}{lll}\mathrm{b} & \mathrm{b}-\mathrm{c} & \mathrm{c} \\ \mathrm{a} & \mathrm{a}-\mathrm{b} & \mathrm{b} \\ \mathrm{c} & \mathrm{c}-\mathrm{a} & \mathrm{a}\end{array}\right|$ \\Apply $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3}\\$ $=(\mathrm{b}-\mathrm{a})(\mathrm{b}-\mathrm{a})\left|\begin{array}{lll}\mathrm{b}-\mathrm{c} & \mathrm{b}-\mathrm{c} & \mathrm{c} \\ \mathrm{a}-\mathrm{b} & \mathrm{a}-\mathrm{b} & \mathrm{b} \\ \mathrm{c}-\mathrm{a} & \mathrm{c}-\mathrm{a} & \mathrm{a}\end{array}\right|

Whenever any two columns or rows in any determinant are equal, its value becomes = 0

Here Column 1 and 2 are identical 

\\\therefore\left|\begin{array}{lll}b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2}\end{array}\right|=0$

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