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: Use the determinant of A:
From the given inverse, $\det(A)\cdot\det(A^{-1}) = 1$. Instead, use a direct check on one diagonal entry.
Row 3 of $A$ times column 3 of $A^{-1}$ must be 1: $\frac32 + xy + \frac12 = 1$, so $xy = -1$.

Step 2: Use a zero entry:
Row 3 times column 1 gives $0$: $\frac32 - 4x + \frac52 = 0$, so $x = 1$. Then $y = -1$ from $xy = -1$.

Step 3: Convert to polar:
For $(1,-1)$: $r = \sqrt2$. The point is in quadrant IV, so $\theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$.
Options (A) and (D) have $r=2$, which does not match $\sqrt2$, and option (B) has the wrong angle.

Final Answer:
$(\sqrt2, 7\pi/4)$, option (C). \[ \boxed{\left(\sqrt{2},\tfrac{7\pi}{4}\right) \text{ (C)}} \]