## Linear Algebra: A Modern Introduction

$\begin{bmatrix}-3 & 2\\5 &2 \\4 & 3\end{bmatrix}$
We have $B^{T}=\begin{bmatrix}4 & 0\\-2 & 2 \\1 & 3\end{bmatrix}$ Now, $C-B^{T}=\begin{bmatrix}1 & 2\\3 & 4 \\5 & 6\end{bmatrix}-\begin{bmatrix}4 & 0\\-2 & 2 \\1 & 3\end{bmatrix}$ or, $=\begin{bmatrix}1-4 & 2-0\\3+2 &4- 2 \\5-1 & 6-3\end{bmatrix}$ After simplification, we obtain: $=\begin{bmatrix}-3 & 2\\5 &2 \\4 & 3\end{bmatrix}$