Let A = $\begin{bmatrix}r_1 & r_2 & r_3 & r_4 & r_5\end{bmatrix}^T$ have rows $r_1$, $r_2$, $r_3$, $r_4$, $r_5$ $\in$ $\mathbb{R}^5$. Assume det(A) = -3.
Compute
det $\begin{bmatrix}2r_1 + 3r_2 + 4r_3 + 4r_4\\ r_1 + 2r_2\\ r_2+3r_3\\r_3+4r_4\\r_1\end{bmatrix}$ and justify your answer.
(the part before this was the same thing but with this matrix)
det $\begin{bmatrix}5r_1 + 5r_2 + 5r_3 + 5r_4 + 5r_5\\ 4r_1 + 4r_2 + 4r_3 + 4r_4\\ r_1\\2r_1+2r_2\\3r_1+3r_2+3r_3\end{bmatrix}$
and I think I figured this one out (you just Gauss-Jordan it and it ends up being equal to A so it's just -3)
But my main confusion is with the first matrix, because even with Gauss-Jordan it doesn't equal A so how can I even find the determinant because it's not a square matrix.
Any help is appreciated. Thanks in advance!
Best Answer
Hint 1: What happens to new matrix if you do $R_1 \to R_1 - R_2 - R_3 -R_4 -R_5$ where $R$ represents the rows of the matrix in part (i).
Hint 2: The answer for the determinant is zero.