English

Evaluate: |3x-x+y-x+zx-y3yz-yx-zy-z3z|

Advertisements
Advertisements

Question

Evaluate: `|(3x, -x + y, -x + z),(x - y, 3y, z - y),(x - z, y - z, 3z)|`

Sum
Advertisements

Solution

We have, `|(3x, -x + y, -x + z),(x - y, 3y, z - y),(x - z, y - z, 3z)|`

[Applying C1 → C1 + C2 + C3]

= `|(x + y + z, -x + y, -x + z),(x + y + z, 3y, z - y),(x + y + z, y - z, 3z)|`

[Taking (x + y + z) common from colmn C1]

= `(x + y + z)|(1, -x + y, -x + z),(1, 3y, z - y),(1, y - z, 3z)|`

[Applying R1 → R2 – R1 and R3 → R3 – R1]

= `(x + y + z)|(1, -x + y, -x + z),(0, 2y + x, x - y),(0, x - z, 2z + x)|`

[Applying C2 → C2 – C3]

= `(x + y + z)|(1, -x + y, -x + z),(0, 3y, x - y),(0, -3z, 2z + x)|`

[Expanding along first column]

= `(x + y + z) * 1[3y(2z + x) + (3z)(x - y)]`

= (x + y + z)(3yz + 3yx + 3xz)

= 3(x + y + z)(xy + yz + zx)

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Determinants - Exercise [Page 77]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 4 Determinants
Exercise | Q 4 | Page 77
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×