C25 Doing the computations by hand, find the determinant of the matrix below.
\[\begin{bmatrix} 3 & −1 & 4 \\ 2 & 5 & 1 \\ 2 & 0 & 6 \end{bmatrix}\] \[\begin{equation} \begin{array}{|ccc|} 3 & −1 & 4 \\ 2 & 5 & 1 \\ 2 & 0 & 6 \end{array} = 3 \begin{array}{|cc|} 5 & 1 \\ 0 & 6 \end{array} - -1 \begin{array}{|cc|} 2 & 1 \\ 2 & 6 \end{array} + 4 \begin{array}{|cc|} 2 & 5 \\ 2 & 0 \end{array} \end{equation}\] \[\begin{equation}= (3)(30) + (1)(10) + (4)(-10)\end{equation}\] \[\begin{equation}= 60\end{equation}\]In R
A <- matrix(c(3 , −1 , 4, 2 , 5 , 1, 2 , 0 , 6), 3,3,byrow = T)
det(A)## [1] 60