SATHEE: Matrices and Determinants 2 Question 40

Matrices and Determinants 2 Question 40

44. Show that

$| \begin{aligned} ^{x} C_{r}^{x} C_{r + 1}^{x} C_{r + 2} \\ ^{y} C_{r}^{y} C_{r + 1}^{y} C_{r + 2} \\ ^{z} C_{r}^{z} C_{r + 1}^{z} C_{r + 2} \end{aligned} |$ $| \begin{aligned} ^{x} C_{r}^{x} C_{r + 1}^{x + 2} C_{r + 2} \\ ^{y} C_{r}^{y} C_{r + 1}^{y + 2} C_{r + 2} \\ ^{z} C_{r}^{z} C_{r + 1}^{z + 2} C_{r + 2} \end{aligned} |$

$(1985, 3 M)$

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Answer:

Solution:

Let $Δ =$ $| \begin{aligned} ^{x} C_{r}^{x} C_{r + 1}^{x} C_{r + 2} \\ ^{y} C_{r}^{y} C_{r + 1}^{y} C_{r + 2} \\ ^{z} C_{r}^{z} C_{r + 1}^{z} C_{r + 2} \end{aligned} |$

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

$Δ =$ $| \begin{aligned} ^{x} C_{r}^{x} C_{r + 1}^{x + 1} C_{r + 2} \\ ^{y} C_{r}^{y} C_{r + 1}^{y + 1} C_{r + 2} \\ ^{z} C_{r}^{z} C_{r + 1}^{z + 1} C_{r + 2} \end{aligned} |$

$[∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}]$

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

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

Hence proved.

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Matrices and Determinants 2 Question 40

44. Show that

Answer:

Solution:

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