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Properties of Determinants

Using propert...

Question

Using properties of determinants, prove that $⎡ ⎢ ⎣ \begin{matrix} x + y & x & x 5 x + 4 y & 4 x & 2 x 10 x + 8 y & 8 x & 3 x \end{matrix} ⎤ ⎥ ⎦ = x^{3}$

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Solution

To prove, $∣ ∣ ∣ ∣ \begin{matrix} x + y & x & x 5 x + 4 y & 4 x & 2 x 10 x + 8 y & 8 x & 3 x \end{matrix} ∣ ∣ ∣ ∣ = x^{3}$
$L H S = ∣ ∣ ∣ ∣ \begin{matrix} x & x & x 5 x & 4 x & 2 x 10 x & 8 x & 3 x \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} y & x & x 4 y & 4 x & 2 x 8 y & 8 x & 3 x \end{matrix} ∣ ∣ ∣ ∣$
$= x^{3} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 5 & 4 & 2 10 & 8 & 3 \end{matrix} ∣ ∣ ∣ ∣ + y x^{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 4 & 4 & 2 8 & 8 & 3 \end{matrix} ∣ ∣ ∣ ∣$
Applying $C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3}$ in the first determinant
$= x^{3} ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 1 & 2 & 2 2 & 5 & 3 \end{matrix} ∣ ∣ ∣ ∣ + y x^{2} \times 0$
As the first two columns of the $2^{n d}$ determinant are same.
Expanding the first determinant through $R_{1}$
$= x^{3} .1 . ∣ ∣ ∣ \begin{matrix} 1 & 2 2 & 5 \end{matrix} ∣ ∣ ∣ = x^{3} (5 - 4)$
$= x^{3}$ = RHS
Hence proved.

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