Question

Let $A=\begin{pmatrix}1&\tan x\\ -\tan x& 1\end{pmatrix}.$ Then $AA^{T}$ is equal to

• A. $\begin{pmatrix}\sec^{2}x&-1\\ -1&\sec^{2}x\end{pmatrix}$
• B. $\begin{pmatrix}\sec^{2}x& 1\\ 1&\sec^{2}x\end{pmatrix}$
• C. $\begin{pmatrix}\sec^{2}x& 1\\ 0&\sec^{2}x\end{pmatrix}$
• D. $\begin{pmatrix}\sec^{2}x& 0\\ 1&\sec^{2}x\end{pmatrix}$
• E. $\begin{pmatrix}\sec^{2}x& 0\\ 0&\sec^{2}x\end{pmatrix}$

Hint:

Math Tip: Matrices of the form $\begin{pmatrix} a & b
-b & a \end{pmatrix}$ behave very similarly to complex numbers ($a + bi$). Multiplying such a matrix by its transpose yields $(a^2 + b^2)I$, similar to $z\bar{z} = |z|^2$. Here, $1^2 + \tan^2x = \sec^2x$.

Answer 1

Answer: (E)

Concept:
Matrices - Matrix Multiplication and Trigonometric Identities.
Recall the fundamental identity: $1 + \tan^2 x = \sec^2 x$.
Step 1: Find the transpose of matrix A.
Given the matrix: $$ A = \begin{pmatrix} 1 & \tan x \\ -\tan x & 1 \end{pmatrix} $$ Its transpose $A^T$ is found by swapping rows and columns: $$ A^T = \begin{pmatrix} 1 & -\tan x \\ \tan x & 1 \end{pmatrix} $$
Step 2: Set up the matrix multiplication $AA^T$.
Multiply the two matrices: $$ AA^T = \begin{pmatrix} 1 & \tan x \\ -\tan x & 1 \end{pmatrix} \begin{pmatrix} 1 & -\tan x \\ \tan x & 1 \end{pmatrix} $$
Step 3: Perform the dot products for each element.

• Element $(1,1)$: $(1)(1) + (\tan x)(\tan x) = 1 + \tan^2 x$
• Element $(1,2)$: $(1)(-\tan x) + (\tan x)(1) = -\tan x + \tan x = 0$
• Element $(2,1)$: $(-\tan x)(1) + (1)(\tan x) = -\tan x + \tan x = 0$
• Element $(2,2)$: $(-\tan x)(-\tan x) + (1)(1) = \tan^2 x + 1$

Step 4: Simplify the resulting matrix.
Construct the product matrix with the evaluated elements: $$ AA^T = \begin{pmatrix} 1 + \tan^2 x & 0 \\ 0 & 1 + \tan^2 x \end{pmatrix} $$ Substitute the trigonometric identity $1 + \tan^2 x = \sec^2 x$: $$ AA^T = \begin{pmatrix} \sec^2 x & 0 \\ 0 & \sec^2 x \end{pmatrix} $$