Question

If matrix A and its inverse \(A^{-1}\) are given by \(A = \left[ \begin{array}{ccc}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1\end{array} \right]\) and \(A^{-1} = \left[ \begin{array}{ccc}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & y \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array} \right]\), then the polar co-ordinates of the points whose Cartesian co-ordinates are \((x,y)\) are \(\ldots\)

• A. \((2,\frac{7π}{4})\)
• B. \((\sqrt{2},\frac{π}{4})\)
• C. \((\sqrt{2},\frac{7π}{4})\)
• D. \((2,\frac{π}{4})\)

Hint:

Use A times A inverse equals I to find x and y, then convert (x,y) into polar form.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Since \(A^{-1}\) is the inverse of \(A\), their product is the identity matrix. Comparing particular entries of \(AA^{-1} = I\) gives \(x\) and \(y\).

Step 2: Key Formula or Approach:
Row \(i\) of \(A\) times column \(j\) of \(A^{-1}\) equals 1 if \(i = j\) and 0 otherwise.

Step 3: Detailed Explanation:
Row 3 of \(A\) is \((3, x, 1)\) and column 1 of \(A^{-1}\) is \(\left(\tfrac12, -4, \tfrac52\right)\). Their product must be 0:
\[ \frac{3}{2} - 4x + \frac{5}{2} = 0 \Rightarrow 4 - 4x = 0 \Rightarrow x = 1 \]
Row 2 of \(A\) is \((1, 2, 3)\) and column 3 of \(A^{-1}\) is \(\left(\tfrac12, y, \tfrac12\right)\). Their product must be 0:
\[ \frac{1}{2} + 2y + \frac{3}{2} = 0 \Rightarrow 2 + 2y = 0 \Rightarrow y = -1 \]
The Cartesian point is \((1, -1)\).
Polar form: \(r = \sqrt{1^2 + (-1)^2} = \sqrt{2}\). The point lies in the fourth quadrant and \(\tan\theta = -1\), so \(\theta = \dfrac{7\pi}{4}\).
Option (B) has the same radius but \(\theta = \frac{\pi}{4}\), which is in the first quadrant and would be the point \((1, 1)\).

Final Answer:
The polar coordinates are \(\left(\sqrt{2}, \dfrac{7\pi}{4}\right)\), option (C). \[ \boxed{\left(\sqrt{2},\tfrac{7\pi}{4}\right) \text{ (C)}} \]