First Group
(1) If \( A = \begin{pmatrix} x+y & 3 \\ 4 & x-y \end{pmatrix} \), \( B = \begin{pmatrix} 9 & 3 \\ 4 & 5 \end{pmatrix} \) and \( A = B \), find \( x, y \).
Answer: \( x = 7, y = 2 \)(2) If \( A = \begin{pmatrix} 3 & -6 \\ 2d & 2 \end{pmatrix} \), \( B = \begin{pmatrix} 2d & 5 \\ 3e & 2 \end{pmatrix} \) and \( A = B^t \), find \( d, e \).
Answer: \( d = 1.5, e = -2 \)(3) Simplest form: \( \frac{1}{\cos^2 \theta} - \frac{1}{\cot^2 \theta} \).
Answer: \( 1 \)(4) If \( \vec{OA} = (6, -8) \), find \( ||\vec{OA}|| \).
Answer: \( 10 \) units(5) Polar form of \( \vec{OA} = (6, 6\sqrt{3}) \).
Answer: \( (12, 60^\circ) \)Second Group
(1) In \( A = \begin{pmatrix} -8 & -5 & 0 \\ 3 & 6 & -6 \end{pmatrix} \), find \( a_{21}, a_{23} \).
Answer: \( a_{21} = 3, a_{23} = -6 \)(2) If \( A = \begin{pmatrix} 6 & 12 \\ 7 & -2 \end{pmatrix} \), \( B = \begin{pmatrix} -3d & 7 \\ 4e & -2 \end{pmatrix} \) and \( A = B^t \), find \( d, e \).
Answer: \( d = -2, e = 3 \)(3) Simplest form: \( \frac{1}{\sin^2 \theta} - \frac{1}{\tan^2 \theta} \).
Answer: \( 1 \)(4) If \( \vec{OA} = (5, 12) \), find \( ||\vec{OA}|| \).
Answer: \( 13 \) units(5) Polar form of \( \vec{OA} = (-4, 4) \).
Answer: \( (4\sqrt{2}, 135^\circ) \)Third Group
(1) Find \( x, y \) if \( \begin{pmatrix} 2x-5 \\ 3y+12 \end{pmatrix} = \begin{pmatrix} 25 \\ y \end{pmatrix} \).
Answer: \( x = 15, y = -6 \)(2) If \( A = \begin{pmatrix} 2 & 5 \\ -1 & -3 \end{pmatrix} \), \( B = \begin{pmatrix} a & 5 \\ b+1 & -3 \end{pmatrix} \) and \( A = B \), find \( a, b \).
Answer: \( a = 2, b = -2 \)(3) Simplify: \( (\sin \theta + \cos \theta)^2 - 2 \sin \theta \cos \theta \).
Answer: \( 1 \)(4) If \( \vec{OA} = (3, -4) \), find \( ||\vec{OA}|| \).
Answer: \( 5 \) units(5) Polar form of \( \vec{OA} = (0, 5) \).
Answer: \( (5, 90^\circ) \)Mastery & Detailed Explanations
01 First Group: In-Depth Analysis
Q1: Matrix Equality Logic
To solve \( A=B \), we equate elements in the same position.
1. \( x+y = 9 \)
2. \( x-y = 5 \)
By adding the two equations: \( (x+y) + (x-y) = 9+5 \Rightarrow 2x = 14 \Rightarrow \mathbf{x=7} \).
Substitute \( x \) in eq.1: \( 7+y=9 \Rightarrow \mathbf{y=2} \).
Q3: Trigonometric Identities
\( \frac{1}{\cos^2 \theta} \) is \( \sec^2 \theta \), and \( \frac{1}{\cot^2 \theta} \) is \( \tan^2 \theta \).
The fundamental identity is \( 1 + \tan^2 \theta = \sec^2 \theta \).
Rearranging it: \( \sec^2 \theta - \tan^2 \theta = \mathbf{1} \).
"Try this: If \( A = \begin{pmatrix} 2x+y & 10 \\ 5 & 8 \end{pmatrix} \) and \( A = \begin{pmatrix} 7 & 10 \\ 5 & x-y \end{pmatrix} \), find \( x, y \)."
Hint: \( 2x+y=7 \) and \( x-y=8 \). Solve them to get \( x=5, y=-3 \).
02 Second Group: Key Concepts
Q2: Transpose Strategy
\( B = \begin{pmatrix} -3d & 7 \\ 4e & -2 \end{pmatrix} \Rightarrow B^t = \begin{pmatrix} -3d & 4e \\ 7 & -2 \end{pmatrix} \).
Comparing with \( A = \begin{pmatrix} 6 & 12 \\ 7 & -2 \end{pmatrix} \):
\( -3d = 6 \Rightarrow \mathbf{d = -2} \).
\( 4e = 12 \Rightarrow \mathbf{e = 3} \).
Q5: Polar Form & Quadrants
Vector \( (-4, 4) \) is in the 2nd Quadrant (x is negative, y is positive).
Magnitude \( r = \sqrt{(-4)^2 + 4^2} = \sqrt{32} = \mathbf{4\sqrt{2}} \).
Angle \( \alpha = \tan^{-1}|\frac{4}{-4}| = 45^\circ \).
Since it's 2nd Quad: \( \theta = 180^\circ - 45^\circ = \mathbf{135^\circ} \).
"Try this: Find the polar form of \( \vec{V} = (-3, -3\sqrt{3}) \)."
Hint: 3rd Quadrant! \( r = 6 \), \( \theta = 180+60 = 240^\circ \).
03 Third Group: Advanced Steps
Q3: Perfect Square Expansion
Recall \( (a+b)^2 = a^2 + b^2 + 2ab \).
So, \( (\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta \).
Substitute this into the expression:
\( (\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta) - 2\sin\theta\cos\theta \)
\( = \sin^2\theta + \cos^2\theta = \mathbf{1} \).
Q5: Quadrantal Vectors
Vector \( (0, 5) \) lies exactly on the positive y-axis.
Magnitude \( r = \sqrt{0^2 + 5^2} = \mathbf{5} \).
Any point on the positive y-axis has an angle of \( \mathbf{90^\circ} \).
"Try this: Simplify \( (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) \)."
Hint: Difference of squares! It becomes \( \sec^2\theta - \tan^2\theta = 1 \).