Using properties of determinants- prove that begin vmatrix 1 - 1-p - 1-p-q3 - 4-3 p - 2-4 p-3 q4 - 7-4 p - 2-7 p-4 q vmatrix -1-1

LHS : 1 amp; 1+p amp; 1+p+q 3 amp; 4+3p amp; 2+4p+3q 4 amp; 7+4p amp; 2+7p+4q =1. By R_2 → R_2 3R_1, R_3 → R 3 4R_1 = 1 amp; 1+p ...

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Using properties of determinants, prove that $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + p & 1 + p + q 3 & 4 + 3 p & 2 + 4 p + 3 q 4 & 7 + 4 p & 2 + 7 p + 4 q \end{matrix} ∣ ∣ ∣ ∣ = 1.$

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Using properties of determinants prove the following questions.

$∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + p & 1 + p + q 2 & 3 + 2 p & 4 + 3 p + 2 q 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} ∣ ∣ ∣ ∣ = 1$

∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + p & 1 + p + q 2 & 3 + 2 p & 4 + 3 p + 2 q 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} ∣ ∣ ∣ ∣

= 1

∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 2 a + 1 & a + 2 & 1 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ = (a - 1)^{3}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x^{2} & 1 & x x & x^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 - x^{3})^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c c a & c b & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 1 + a^{2} + b^{2} + c^{2}

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