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Chapter 4: Problem 306

If \(\mathrm{P}, \mathrm{Q}, \mathrm{R}\) represent the angles of an acute angled triangle, no two of them being equal them the value of $$ \left|\begin{array}{ccc} 1 & 1+\cos P & \cos P(1+\cos P) \\ 1 & 1+\cos Q & \cos Q(1+\cos Q) \\ 1 & 1+\cos R & \cos R(1+\cos R) \end{array}\right| \text { is } \ldots $$ (a) positive (b) 0 (c) negative (d) cannot be determined

Short Answer

Expert verified
The value of the determinant is (a) positive.

Step by step solution

01

Write down the formula for a 3x3 determinant

The determinant of a 3x3 matrix is given by the formula: \( \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \) Now, we'll apply this formula to the given matrix.
02

Plug in the values from the given matrix

Plugging the values from the given matrix, we have: \( \begin{vmatrix} 1 & 1 + \cos P & \cos P (1 + \cos P) \\ 1 & 1 + \cos Q & \cos Q (1 + \cos Q) \\ 1 & 1 + \cos R & \cos R (1 + \cos R) \end{vmatrix} \)
03

Evaluate the determinant

Let's evaluate the determinant using the formula from Step 1: \( 1[(1 + \cos Q)(1 + \cos R) - \cos Q(1 + \cos Q)\cos R(1 + \cos R)] \\ - (1 + \cos P)([\cos Q(1 + \cos R) - \cos R(1 + \cos Q)] \\ + (\cos P (1 + \cos P)[\cos Q + \cos Q\cos R - \cos R - \cos R\cos Q] \) Simplifying the expression, we get: \(=(1 + \cos Q + \cos R + \cos Q \cos R) - (\cos Q + \cos Q \cos R + \cos R + \cos R \cos Q) = (1 - \cos Q - \cos R)\) Now, we know that for an acute angled triangle, all angles are less than 90°, so all cosines are positive. Thus, \( 0 < \cos P, \cos Q, \cos R < 1\).
04

Determine the value of the determinant

Given that \(0 < \cos P, \cos Q, \cos R < 1\), we have: \(0 < 1 - \cos Q - \cos R < 2\) So the value of the determinant is positive, between 0 and 2. Therefore, the correct answer is: (a) positive.

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Most popular questions from this chapter

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