Question

Let $A = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$, $B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $C = ABA^T$, then $A^T C^3 A$ is equal to

• A. $\begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ 1 & 0 \end{bmatrix}$
• B. $\begin{bmatrix} 1 & 0 \\ \frac{\sqrt{3}}{2} & 1 \end{bmatrix}$
• C. $\begin{bmatrix} 1 & \frac{\sqrt{3}}{2} \\ 0 & 3 \end{bmatrix}$
• D. $\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$

Answer 1

Answer: (D)

$\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$

Answer 2

Let the given matrices be $A = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. We are given $C = ABA^T$. We need to find $A^T C^3 A$.

First, find the transpose of $A$, $A^T$:

$A^T = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$.

Calculate $A A^T$:

$A A^T = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} (\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2 & \frac{\sqrt{3}}{2}(\frac{-1}{2}) + \frac{1}{2}(\frac{\sqrt{3}}{2}) \\ (\frac{-1}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{3}}{2})(\frac{1}{2}) & (\frac{-1}{2})^2 + (\frac{\sqrt{3}}{2})^2 \end{bmatrix} = \begin{bmatrix} \frac{3}{4} + \frac{1}{4} & \frac{-\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4} + \frac{\sqrt{3}}{4} & \frac{1}{4} + \frac{3}{4} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.

Similarly, $A^T A = I$. This shows that $A$ is an orthogonal matrix, and $A^T = A^{-1}$.

Now consider the expression $A^T C^3 A$. Substitute $C = ABA^T$:

$A^T C^3 A = A^T (ABA^T)^3 A$.

Using the property of matrix multiplication, $(PQR)^n = P Q^n R$ if $Q$ commutes with $P$ and $R$, or if $P$ and $R$ are inverses and $Q$ is in the middle. In this case, we can expand the power:

$A^T C^3 A = A^T (ABA^T)(ABA^T)(ABA^T) A$.

Using the associative property of matrix multiplication and the fact that $A^T A = I$:

$A^T (ABA^T)(ABA^T)(ABA^T) A = A^T A B A^T A B A^T A B A^T A = (A^T A) B (A^T A) B (A^T A) B (A^T A) = I B I B I B I = B B B = B^3$.

So, the problem reduces to calculating $B^3$.

$B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.

Calculate the powers of $B$:

$B^2 = B \cdot B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + 1 \cdot 0 & 1 \cdot 1 + 1 \cdot 1 \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot 1 + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.

$B^3 = B^2 \cdot B = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + 2 \cdot 0 & 1 \cdot 1 + 2 \cdot 1 \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot 1 + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 1 & 1+2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.

Alternatively, for a matrix of the form $\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$, the $n$-th power is given by $\begin{bmatrix} 1 & nk \\ 0 & 1 \end{bmatrix}$.

For $B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, $k=1$. So $B^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.

For $n=3$, $B^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.

Thus, $A^T C^3 A = B^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.