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Applications of Cross Product

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Question

Using the properties of determinants, show that:

(i) $∣ ∣ ∣ ∣ \begin{matrix} x + 4 & 2 x & 2 x 2 x & x + 4 & 2 x 2 x & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣ = (5 x + 4) (4 - {x)}^{2}$
(ii) $∣ ∣ ∣ ∣ \begin{matrix} y + k & y & y y & y + k & y y & y & y + k \end{matrix} ∣ ∣ ∣ ∣ = k^{2} (3 y + k)$

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Solution

(i) Consider, $L H S = ∣ ∣ ∣ ∣ \begin{matrix} x + 4 & 2 x & 2 x 2 x & x + 4 & 2 x 2 x & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣$

$C_{1} \to C_{1} + C_{2} + C_{3}$

$= ∣ ∣ ∣ ∣ \begin{matrix} 5 x + 4 & 2 x & 2 x 5 x + 4 & x + 4 & 2 x 5 x + 4 & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣$

Taking $(5 x + 4)$ as a common factor from $C_{1}$
$= (5 x + 4) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 x & 2 x 1 & x + 4 & 2 x 1 & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣$

$R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$

$= (5 x + 4) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 x & 2 x 0 & 4 - x & 0 0 & 0 & 4 - x \end{matrix} ∣ ∣ ∣ ∣$

Taking $(4 - x)$ as common factor from $R_{2}$ and $R_{3}$

$= (5 x + 4) {(4 - x)}^{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 x & 2 x 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣$

Expanding along first column, we get
$= (5 x + 4) {(4 - x)}^{2} ∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣$
$= (5 x + 4) {(4 - x)}^{2}$
$= R H S$

(ii) $L H S = ∣ ∣ ∣ ∣ \begin{matrix} y + k & y & y y & y + k & y y & y & y + k \end{matrix} ∣ ∣ ∣ ∣$

$C_{1} \to C_{1} + C_{2} + C_{3}$

$L H S = ∣ ∣ ∣ ∣ \begin{matrix} 3 y + k & y & y 3 y + k & y + k & y 3 y + k & y & y + k \end{matrix} ∣ ∣ ∣ ∣$

Taking $(3 y + k)$ as a common factor from $C_{1}$
$= (3 y + k) ∣ ∣ ∣ ∣ \begin{matrix} 1 & y & y 1 & y + k & y 1 & y & y + k \end{matrix} ∣ ∣ ∣ ∣$

$R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$

$= (3 y + k) ∣ ∣ ∣ ∣ \begin{matrix} 1 & y & y 0 & k & 0 0 & 0 & k \end{matrix} ∣ ∣ ∣ ∣$

Taking $k$ as common factor from $R_{2}$ and $R_{3}$

$= (3 y + k) k^{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 x & 2 x 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣$

Expanding along first column, we get
$= (3 y + k) k^{2} ∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣$
$= k^{2} (3 y + k)$
$= R H S$

$∴ ∣ ∣ ∣ ∣ \begin{matrix} y + k & y & y y & y + k & y y & y & y + k \end{matrix} ∣ ∣ ∣ ∣ = k^{2} (3 y + k)$

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∣ ∣ ∣ ∣ \begin{matrix} y + k & y & y y & y + k & y y & y & y + k \end{matrix} ∣ ∣ ∣ ∣ = k^{2} (3 y + k)

By using properties of determinants, show that:

(i)

(ii)

∣ ∣ ∣ ∣ \begin{matrix} x + 4 & 2 x & 2 x 2 x & x + 4 & 2 x 2 x & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣ = (5 x + 4) (4 - x)^{2}

det ∣ ∣ ∣ ∣ \begin{matrix} x + 4 & 2 x & 2 x 2 x & x + 4 & 2 x 2 x & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣

= (5 x + 4) (4 - x)^{2}

∣ ∣ ∣ ∣ \begin{matrix} x + 4 & 2 x & 2 x 2 x & x + 4 & 2 x 2 x & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣ = (5 x + 4) (4 - x)^{2}

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