SATHEE: Matrices and Determinants 2 Question 37

Matrices and Determinants 2 Question 37

41. If $a \neq p, b \neq q, c \neq r$ and $| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} |$

Then, find the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ .

$(1991, 4 M)$

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Answer:

Correct Answer: 41. (2)

Solution:

Let $Δ = | \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} |$

Applying $R_{1} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , we get

$\begin{aligned} Δ & = | \begin{array}{c} p & b & c \\ a - p & q - b & 0 \\ a - p & 0 & r - c \end{array} | \\ = c | \begin{array}{c} a - p & q - b \\ a - p & 0 \end{array} | + (r - c) | \begin{array}{c} p & b \\ a - p & q - b \end{array} | \\ = - c (a - p) (q - b) + (r - c) [p (q - b) - b (a - p)] \\ = - c (a - p) (q - b) + p (r - c) (q - b) - b (r - c) (a - p) \end{aligned}$

Since, $Δ = 0$

$\begin{aligned} \Rightarrow - c (a - p) (q - b) + p (r - c) (q - b) - b (r - c) (a - p) = 0 \\ \Rightarrow \frac{c}{r - c} + \frac{p}{p - a} + \frac{b}{q - b} = 0 \end{aligned}$

[on dividing both sides by $(a - p) (q - b) (r - c)$ ]

$\begin{aligned} \Rightarrow & \frac{p}{p - a} + \frac{b}{q - b} + 1 + \frac{c}{r - c} + 1 & = 2 \\ \Rightarrow & \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2 \end{aligned}$

Matrices and Determinants 2 Question 37

41. If $a \neq p, b \neq q, c \neq r$ and $| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} |$

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Matrices and Determinants 2 Question 37

41. If a≠p,b≠q,c≠r and |pbcaqcabr|

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41. If $a \neq p, b \neq q, c \neq r$ and $| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} |$