1. If A = [[0.8, 0.6], [-0.6, 0.8]], find A³. 2. Prove that the product of two matrices [[cos²(θ), cos(θ)sin(θ)], [cos(θ)sin(θ), sin²(θ)]] and [[cos²(φ), cos(φ)sin(φ)], [cos(φ)sin(... 1. If A = [[0.8, 0.6], [-0.6, 0.8]], find A³. 2. Prove that the product of two matrices [[cos²(θ), cos(θ)sin(θ)], [cos(θ)sin(θ), sin²(θ)]] and [[cos²(φ), cos(φ)sin(φ)], [cos(φ)sin(φ), sin²(φ)]] is a zero matrix when θ and φ differ by an odd multiple of π/2. 3. Find the product of the following two matrices [[0, c, -b], [-c, 0, a]] and [[a², ab, ac], [ab, b², bc]] Read more

Steps to Solve

1. Calculate $A^2$

To find $A^2$, multiply matrix $A$ by itself.

$A^2 = A \cdot A = \begin{bmatrix} 0.8 & 0.6 \ -0.6 & 0.8 \end{bmatrix} \cdot \begin{bmatrix} 0.8 & 0.6 \ -0.6 & 0.8 \end{bmatrix} = \begin{bmatrix} (0.8)(0.8) + (0.6)(-0.6) & (0.8)(0.6) + (0.6)(0.8) \ (-0.6)(0.8) + (0.8)(-0.6) & (-0.6)(0.6) + (0.8)(0.8) \end{bmatrix} = \begin{bmatrix} 0.64 - 0.36 & 0.48 + 0.48 \ -0.48 - 0.48 & -0.36 + 0.64 \end{bmatrix} = \begin{bmatrix} 0.28 & 0.96 \ -0.96 & 0.28 \end{bmatrix}$

2. Calculate $A^3$

To find $A^3$, multiply $A^2$ by $A$.

$A^3 = A^2 \cdot A = \begin{bmatrix} 0.28 & 0.96 \ -0.96 & 0.28 \end{bmatrix} \cdot \begin{bmatrix} 0.8 & 0.6 \ -0.6 & 0.8 \end{bmatrix} = \begin{bmatrix} (0.28)(0.8) + (0.96)(-0.6) & (0.28)(0.6) + (0.96)(0.8) \ (-0.96)(0.8) + (0.28)(-0.6) & (-0.96)(0.6) + (0.28)(0.8) \end{bmatrix} = \begin{bmatrix} 0.224 - 0.576 & 0.168 + 0.768 \ -0.768 - 0.168 & -0.576 + 0.224 \end{bmatrix} = \begin{bmatrix} -0.352 & 0.936 \ -0.936 & -0.352 \end{bmatrix}$

3. Prove that the product of the two matrices is a zero matrix

Let $B = \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \ \cos \theta \sin \theta & \sin^2 \theta \end{bmatrix}$ and $C = \begin{bmatrix} \cos^2 \phi & \cos \phi \sin \phi \ \cos \phi \sin \phi & \sin^2 \phi \end{bmatrix}$. Then, $BC = \begin{bmatrix} \cos^2 \theta \cos^2 \phi + \cos \theta \sin \theta \cos \phi \sin \phi & \cos^2 \theta \cos \phi \sin \phi + \cos \theta \sin \theta \sin^2 \phi \ \cos \theta \sin \theta \cos^2 \phi + \sin^2 \theta \cos \phi \sin \phi & \cos \theta \sin \theta \cos \phi \sin \phi + \sin^2 \theta \sin^2 \phi \end{bmatrix}$ $BC = \begin{bmatrix} \cos \theta \cos \phi (\cos \theta \cos \phi + \sin \theta \sin \phi) & \cos \theta \sin \phi (\cos \theta \cos \phi + \sin \theta \sin \phi) \ \sin \theta \cos \phi (\cos \theta \cos \phi + \sin \theta \sin \phi) & \sin \theta \sin \phi (\cos \theta \cos \phi + \sin \theta \sin \phi) \end{bmatrix}$ $BC = \begin{bmatrix} \cos \theta \cos \phi \cos(\theta - \phi) & \cos \theta \sin \phi \cos(\theta - \phi) \ \sin \theta \cos \phi \cos(\theta - \phi) & \sin \theta \sin \phi \cos(\theta - \phi) \end{bmatrix}$

Given that $\theta$ and $\phi$ differ by an odd multiple of $\frac{\pi}{2}$, then $\theta - \phi = (2n+1)\frac{\pi}{2}$, where n is an integer. Thus $\cos(\theta - \phi) = \cos((2n+1)\frac{\pi}{2}) = 0$. Therefore, $BC = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.

4. Find the product of the matrices

Let $D = \begin{bmatrix} 0 & c & -b \ -c & 0 & a \end{bmatrix}$ and $E = \begin{bmatrix} a^2 & ab & ac \ ab & b^2 & bc \ ac & bc & c^2 \end{bmatrix}$. Then, $DE = \begin{bmatrix} (0)(a^2) + (c)(ab) + (-b)(ac) & (0)(ab) + (c)(b^2) + (-b)(bc) & (0)(ac) + (c)(bc) + (-b)(c^2) \ (-c)(a^2) + (0)(ab) + (a)(ac) & (-c)(ab) + (0)(b^2) + (a)(bc) & (-c)(ac) + (0)(bc) + (a)(c^2) \end{bmatrix}$ $DE = \begin{bmatrix} 0 + abc - abc & 0 + cb^2 - b^2c & 0 + bc^2 - bc^2 \ -ca^2 + 0 + a^2c & -abc + 0 + abc & -ac^2 + 0 + ac^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$