# Example 10 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 23, 2020 by Teachoo

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Chapter 4 Class 12 Determinants (Term 1)

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Example 10 Show that |β 8(a&b&c@a+2x&b+2y&c+2z@x&y&z)| = 0 Solving L.H.S |β 8(a&b&c@a+2x&b+2y&c+2z@x&y&z)| Expressing elements of 2nd row as sum of two elements = |β 8(π&π&π@π&π&π@x&y&z)| + |β 8(a&b&c@2x&2y&2z@x&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 + |β 8(a&b&c@2x&2y&2z@x&y&z)| = |β 8(a&b&c@πx&πy&πz@x&y&z)| Since, all elements of 2nd row are multiplied by 2, we can take 2 outside the determinant, = 2|β 8(a&b&c@x&y&z@x&y&z)| R2 & R3 are identical = 2 Γ 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