# Example 11 - 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)

Serial order wise

Example 11 Prove that |β 8(π&π+π&π+π+π@2π&3π+2π&4π+3π+2π@3π&6π+3π&10π+6π+3π)| = a3 Let Ξ = |β 8(π&π+π&π+π+π@2π&3π+2π&4π+3π+2π@3π&6π+3π&10π+6π+3π)| Applying R2 β R2 β 2R1 = |β 8(π&π+π&π+π+π@ππβπ(π)&3π+3πβ2(π+π)&4π+3π+2πβ2(π+π+π)@3π&6π+3π&10π+6π+3π)| = |β 8(π&π+π&π+π+π@π&3π+3πβ2πβ2π&4π+3π+2πβ2πβ2πβ2π@3π&6π+3π&10π+6π+3π)| = |β 8(π&π+π&π+π+π@π&π&2π+π@3π&6π+3π&10π+6π+3π)| Applying R3 β R3 β 3R1 = |β 8(π&π+π&π+π+π@0&π&2π+π@ππβπ(π)&6π+3πβ3(π+π)&10π+6π+3πβ3π+π+π)| = |β 8(π&π+π&π+π+π@0&π&2π+π@π&6π+3πβ3πβ3π&10π+6π+3πβ3π+π+π)| = |β 8(π&π+π&π+π+π@0&π&2π+π@0&3π&7π+3π)| Expanding along C1 = = a |β 8(π&2π+π@3π&7π+3π)| β 0 + 0 = a (a(7a + 3b) β 3a (2a + b) = a (7a2 + 3ab β 6a2 β 3ab) = a(a2) = a3 = R.H.S Hence proved |β 8(π&2π+π@3π&7π+3π)| |β 8(π+π&π+π+π@3π&7π+3π)|