Prove that beginvmatrix -a- 2 - a b - a c l lba -b- 2 - b cca -cb - c - 2end vmatrix -4 a - 2 b - 2 c - 2

LHS= a^2 amp; ab amp; ac ba amp; b^2 amp; bc ca amp;cb amp; c^2 =abc a amp;b amp;c a amp; b amp; c a amp; b ...

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Byju's Answer

Standard XII

Mathematics

Adjoint of a Matrix

Prove that be...

Question

Prove that

$∣ ∣ ∣ ∣ ∣ \begin{matrix} - a^{2} & a b & a c b a & - b^{2} & b c c a & c b & - c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 4 a^{2} b^{2} c^{2}$

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Solution

$L H S = ∣ ∣ ∣ ∣ ∣ \begin{matrix} - a^{2} & a b & a c b a & - b^{2} & b c c a & c b & - c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = a b c ∣ ∣ ∣ ∣ \begin{matrix} - a & b & c a & - b & c a & b & - c \end{matrix} ∣ ∣ ∣ ∣$
[taking out factors a from $R_{1}$ , b from $R_{2}$ and c from $R_{3}$ ]

$= (a b c) (a b c) ∣ ∣ ∣ ∣ \begin{matrix} - 1 & 1 & 1 1 & - 1 & 1 1 & 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣$ (taking out factors a from $C_{1}$ b from $C_{2}$ and c form $C_{3}$ )

Expanding corresponding to first row $R_{1}$ , we get
$= a^{2} b^{2} c^{2} [0 ∣ ∣ ∣ \begin{matrix} - 2 & 1 1 & - 1 \end{matrix} ∣ ∣ ∣ - 0 ∣ ∣ ∣ \begin{matrix} 0 & 2 1 & - 1 \end{matrix} ∣ ∣ ∣ + 2 ∣ ∣ ∣ \begin{matrix} 0 & - 2 1 & 2 \end{matrix} ∣ ∣ ∣]$
$= a^{2} b^{2} c^{2} [0 - 0 + 2 (0 + 2)] = 4 a^{2} b^{2} c^{2} = R H S$ Hence proved.

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Similar questions

Prove that $∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b c & a c + c^{2} a^{2} + a b & b^{2} & a c a b & b^{2} + b c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 4 a^{2} b^{2} c^{2}$

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

Q. Prove that

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

= 4

a^{2} b^{2} c^{2}

Q. Using properties of determinants, prove the following:

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b c & a c + c^{2} a^{2} + a b & b^{2} & a c a b & b^{2} + b c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 4 a^{2} b^{2} c^{2}

Q. If

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

then the value of

k

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Prove that ∣∣ ∣ ∣∣−a2abacba−b2bccacb−c2∣∣ ∣ ∣∣=4a2b2c2

Prove that

$∣ ∣ ∣ ∣ ∣ \begin{matrix} - a^{2} & a b & a c b a & - b^{2} & b c c a & c b & - c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 4 a^{2} b^{2} c^{2}$