Evaluate the following determinant-i 1- a 1111- a 1111- a - a 3-3 a 2ii a 2-2 a 2 a -112 a -1 a -21331- a -13

(ii) To Prove: nbsp;a2+2a2a+112a+1a+21331=a 13 LHS=a2+2a2a+112a+1a+21331Applying #160;R1 #8594;R1 R2 #160; #160; #160; #160; #160; #160;=a2+ ...

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Laws of Exponents

Evaluate the ...

Question

Evaluate the following determinant:

(i) $|\begin{matrix} 1 + a & 1 & 1 \\ 1 & 1 + a & 1 \\ 1 & 1 & 1 + a \end{matrix}| = a^{3} + 3 a^{2}$

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

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

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

C =

|\begin{matrix} 1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a \end{matrix}| = a^{3} + 3 a^{2}

∣ ∣ ∣ ∣ \begin{matrix} 1 + a 11 \end{matrix} \begin{matrix} 1 1 + a 1 \end{matrix} \begin{matrix} 11 1 + a \end{matrix} ∣ ∣ ∣ ∣

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