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Example 10 - Chapter 4 Class 12 Determinants (Term 1)
Last updated at March 16, 2023 by Teachoo
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Chapter 4 Class 12 Determinants
Serial order wise
Transcript
Example 10 Show that |โ 8(a&b&[email protected]+2x&b+2y&[email protected]&y&z)| = 0 Solving L.H.S |โ 8(a&b&[email protected]+2x&b+2y&[email protected]&y&z)| Expressing elements of 2nd row as sum of two elements = |โ 8(๐&๐&๐@๐&๐&๐@x&y&z)| + |โ 8(a&b&[email protected]&2y&[email protected]&y&z)| By Property 5: If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms ,then the determinant is expressed as a sum of two (or more) determinants. R1 & R2 are identical Using property: If any two rows or columns are identical, then value of determinant is 0 = 0 + |โ 8(a&b&[email protected]&2y&[email protected]&y&z)| = |โ 8(a&b&[email protected]๐x&๐y&๐[email protected]&y&z)| Since, all elements of 2nd row are multiplied by 2, we can take 2 outside the determinant, = 2|โ 8(a&b&[email protected]&y&[email protected]&y&z)| R2 & R3 are identical = 2 ร 0 Using property: if any two rows or columns are identical, then value of determinant is 0 = 0 = R.H.S Thus L.H.S = R.H.S Hence proved
