SATHEE: Matrices and Determinants 2 Question 39

Matrices and Determinants 2 Question 39

43. Let $Δ_{a} = | \begin{array}{ccc} a - 1 & n & 6 \\ (a - 1)^{2} & 2 n^{2} & 4 n - 2 \\ (a - 1)^{3} & 3 n^{3} & 3 n^{2} - 3 n \end{array} |$

Show that $\sum_{a = 1}^{n} Δ_{a} = c \in$ constant.

$(1989, 5 M)$

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Solution:

Given, $Δ_{a} = | \begin{array}{ccc} a - 1 & n & 6 \\ (a - 1)^{2} & 2 n^{2} & 4 n - 2 \\ (a - 1)^{3} & 3 n^{3} & 3 n^{2} - 3 n \end{array} |$

$∴ \sum_{a = 1}^{n} Δ_{a} =$ $| \begin{matrix} \sum_{a = 1}^{n} (a - 1) & n & 6 \\ \sum_{a = 1}^{n} (a - 1)^{2} & 2 n^{2} & 4 n - 2 \\ \sum_{a = 1}^{n} (a - 1)^{3} & 3 n^{3} & 3 n^{2} - 3 n \end{matrix} |$

$\begin{aligned} = | \begin{array}{ccc} \frac{n (n - 1)}{2} & n & 6 \\ \frac{n (n - 1) (2 n - 1)}{6} & 2 n^{2} & 4 n - 2 \\ \frac{n^{2} (n - 1)^{2}}{4} & 3 n^{3} & 3 n^{2} - 3 n \end{array} | \\ = \frac{n^{2} (n - 1)}{2} | \begin{array}{ccc} 1 & 1 & 6 \\ \frac{(2 n^{3} - 1)}{3} & 2 n & 4 n - 2 \\ \frac{n (n^{- 1)}}{2} & 3 n^{2} & 3 n^{2} - 3 n \end{array} | \\ = \frac{n^{3} (n - 1)}{12} | \begin{array}{ccc} 1 & 1 & 6 \\ 2 n - 1 & 6 n & 12 n - 6 \\ n - 1 & 6 n & 6 n - 6 \end{array} | \end{aligned}$

Applying $C_{3} \to C_{3} - 6 C_{1}$

$= \frac{n^{3} (n - 1)}{12} | \begin{array}{ccc} 1 & 1 & 0 \\ 2 n - 1 & 6 n & 0 \\ n - 1 & 6 n & 0 \end{array} | = 0$

$\Rightarrow \sum_{a = 1}^{n} Δ_{a} = c$

$[c = 0$ , i.e. constant $]$

Matrices and Determinants 2 Question 39

43. Let $Δ_{a} = | \begin{array}{ccc} a - 1 & n & 6 \\ (a - 1)^{2} & 2 n^{2} & 4 n - 2 \\ (a - 1)^{3} & 3 n^{3} & 3 n^{2} - 3 n \end{array} |$

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Matrices and Determinants 2 Question 39

43. Let Δa=|a−1n6(a−1)22n24n−2(a−1)33n33n2−3n|

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43. Let $Δ_{a} = | \begin{array}{ccc} a - 1 & n & 6 \\ (a - 1)^{2} & 2 n^{2} & 4 n - 2 \\ (a - 1)^{3} & 3 n^{3} & 3 n^{2} - 3 n \end{array} |$