Question

if $p =\begin{bmatrix}1&2&1\\ 1&3&1\end{bmatrix}, Q=PP^{T} ,$ then the value of the determinant of $Q$ is equal to

• A. 2
• B. -2
• C. 1
• D. 0

Answer 1

Answer: (A)

Given, $p=\begin{bmatrix}1&2&1\\ 1&3&1\end{bmatrix}$
$\therefore Q=PP^{T}=\begin{bmatrix}1 & 2 & 1 \\ 1 & 3 & 1\end{bmatrix}\begin{bmatrix}1&1\\ 2&3\\ 1&1\end{bmatrix}$
$=\begin{bmatrix}1 \times 1+2 \times 2+1 \times 1 & 1 \times 1+2 \times 3+1 \times 1 \\ 1 \times 1+3 \times 2+1 \times 1 & 1 \times 1+3 \times 3+1 \times 1\end{bmatrix}$
$=\begin{bmatrix}1+4+1 & 1+6+1 \\ 1+6+1 & 1+9+1\end{bmatrix}=\begin{bmatrix}6 & 8 \\ 8 & 11\end{bmatrix}$
$\therefore $ The determintant of $Q=\begin{bmatrix}6&8\\ 8&11\end{bmatrix}$
$=66-64=2$