Question

Let $A = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix}$ and $C = (A(ABA^T)^{25} (AB^T A)^{10} A^T )^T$, If trace of matrix C is $\lambda$ and $|C| = \mu$ then $\frac{\lambda}{\mu}$ is

• A. 2
• B. 6
• C. 4
• D. 8

Answer 1

Answer: (B)

6

Answer 2

Given matrices $A = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix}$. We are given $C = (A(ABA^T)^{25} (AB^T A)^{10} A^T )^T$. We need to find $\frac{\lambda}{\mu}$ where $\lambda = \text{trace}(C)$ and $\mu = |C|$.

First, let's compute $A^T$ and $B^T$: $A^T = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}$ $B^T = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} = B$

Now, let's evaluate the terms inside the expression for $C$:

1. $ABA^T$: $AB = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$ $ABA^T = (AB)A^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$ So, $(ABA^T)^{25} = I^{25} = I$.

2. $AB^T A$: Since $B^T = B$, this is $ABA$. $ABA = (AB)A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix}$. Let $M = ABA = \begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix}$.

Now, the expression inside the transpose for $C$ is $P = A(I)^{25} M^{10} A^T = A M^{10} A^T$. So, $C = P^T = (A M^{10} A^T)^T$. Using the property $(XYZ)^T = Z^T Y^T X^T$, we get: $C = (A^T)^T (M^{10})^T A^T = A (M^{10})^T A^T$.

The determinant of $C$ is: $|C| = |(A(ABA^T)^{25} (AB^T A)^{10} A^T )^T| = |A(ABA^T)^{25} (AB^T A)^{10} A^T|$ $|C| = |A| |(ABA^T)^{25}| |(AB^T A)^{10}| |A^T|$. $|A|=1$, $|A^T|=1$. $|ABA^T| = |I| = 1$. So $|(ABA^T)^{25}| = 1^{25}=1$. $|AB^T A| = |M| = (1)(-3) - (-2)(2) = -3 + 4 = 1$. So $|(AB^T A)^{10}| = 1^{10}=1$. Thus, $|C| = 1 \times 1 \times 1 \times 1 = 1$. So $\mu = 1$.

To find the trace of $C$, we use the property trace($XY$) = trace($YX$). $C = A (M^{10})^T A^T$. trace($C$) = trace($A (M^{10})^T A^T$) = trace($A^T A (M^{10})^T$). $A^T A = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix}$.

The characteristic equation of $M$ is $|M - \lambda I| = 0$: $\begin{vmatrix} 1-\lambda & -2 \\ 2 & -3-\lambda \end{vmatrix} = (1-\lambda)(-3-\lambda) - (-2)(2) = -3 - \lambda + 3\lambda + \lambda^2 + 4 = \lambda^2 + 2\lambda + 1 = (\lambda+1)^2 = 0$. The only eigenvalue is $\lambda = -1$. Let $M = -I + N$, where $N = M+I = \begin{bmatrix} 2 & -2 \\ 2 & -2 \end{bmatrix}$. $N^2 = \begin{bmatrix} 2 & -2 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$. $M^{10} = (-I+N)^{10} = \binom{10}{0}(-I)^{10}N^0 + \binom{10}{1}(-I)^9N^1 = I - 10N$. $M^{10} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - 10\begin{bmatrix} 2 & -2 \\ 2 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 20 & -20 \\ 20 & -20 \end{bmatrix} = \begin{bmatrix} -19 & 20 \\ -20 & 21 \end{bmatrix}$. $(M^{10})^T = \begin{bmatrix} -19 & -20 \\ 20 & 21 \end{bmatrix}$.

trace($C$) = trace($\begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} \begin{bmatrix} -19 & -20 \\ 20 & 21 \end{bmatrix}$) = trace($\begin{bmatrix} 1(-19)+(-2)(20) & \dots \\ (-2)(-19)+5(20) & \dots \end{bmatrix}$) = trace($\begin{bmatrix} -19-40 & \dots \\ \dots & \dots \end{bmatrix}$) = trace($\begin{bmatrix} -59 & \dots \\ \dots & \dots \end{bmatrix}$) The diagonal elements of $(A^T A)(M^{10})^T$ are: $(A^T A)_{11} (M^{10})^T_{11} + (A^T A)_{12} (M^{10})^T_{21} = 1(-19) + (-2)(20) = -19 - 40 = -59$. $(A^T A)_{21} (M^{10})^T_{12} + (A^T A)_{22} (M^{10})^T_{22} = (-2)(-20) + 5(21) = 40 + 105 = 145$. So trace($C$) = $-59 + 145 = 86$.

The ratio $\frac{\lambda}{\mu} = \frac{86}{1} = 86$. Since 86 is not an option, there might be an error in the question or options. However, if we consider the possibility that the expression simplifies such that $C=AA^T$, then: $AA^T = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix}$. Trace($AA^T$) = $5+1 = 6$. Determinant($AA^T$) = $5 \times 1 - (-2) \times (-2) = 5 - 4 = 1$. In this hypothetical case, $\frac{\lambda}{\mu} = \frac{6}{1} = 6$. This matches option (2). This would occur if $(AB^T A)^{10} = I$. $M^{10} = I$. $M^{10} = \begin{bmatrix} -19 & 20 \\ -20 & 21 \end{bmatrix} \neq I$. Given the options, it is highly probable that the intended answer is 6, which arises from a simplification that leads to $C=AA^T$.