English

|0xyzx-zy-x0y zz-xz-y0| = ______.

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Question

`|(0, xyz, x - z),(y - x, 0, y  z),(z - x, z - y, 0)|` = ______.

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Solution

`|(0, xyz, x - z),(y - x, 0, y  z),(z - x, z - y, 0)|` = (y – z)(z – x)(y – x + xyz).

Explanation:

Let Δ = `|(0, xyz, x - z),(y - x, 0, y  z),(z - x, z - y, 0)|`

C1 → C1 – C3

= `|(z - x, xyz, x - z),(z - x, 0, y - z),(z - x, z - y, 0)|`

Taking (z – x) common from C1

= `(z - x) |(1, xyz, x - z),(1, 0, y - z),(1, z - y, 0)|`

R1 → R1 – R2, R2 → R2 – R3

= `(z - x) |(0, xyz,y),(0, y - x, y - z),(1, z - y, 0)|`

Taking (y – z) common from R2

= `(z -  x)(y - z) |(0, xyz, x - y),(0, 1, 1),(1,  z - y, 0)|`

Expanding along C1

= `(z - x)(y - z) [1|(xyz, x - y),(1, 1)|]`

= (z – x)(y – z)(xyz – x + y)

= (y – z)(z – x)(y – x + xyz)

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Chapter 4: Determinants - Exercise [Page 83]

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NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 4 Determinants
Exercise | Q 46 | Page 83
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