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.


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!

  • $\begingroup$ $r_1, r_2$, $r_3$, $r_4$, and $r_5\in\mathbb R^5$ are rows of the matrix; there are $5$ rows, each with $5$ components, so it's square $\endgroup$ – J. W. Tanner May 1 at 18:38

Hint: Note that the two matrices under consideration can be written as $M_1A$ and $M_2A$ respectively, where $$ M_1 = \pmatrix{2&3&4&4&0\\ 1&2&0&0&0\\ 0&1&3&0&0\\0&0&1&4&0\\1&0&0&0&0}, \quad M_2 = \pmatrix{ 5&5&5&5&5\\ 4&4&4&4&0\\ 1&0&0&0&0\\2&2&0&0&0\\3&3&3&0&0 }. $$

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  • $\begingroup$ can they? because the matrix itself is not 5x5, it's 5x1... $\endgroup$ – user716286 May 1 at 23:26
  • $\begingroup$ @grace Yes. Recall that two matrices can be multiplied as long as the "inner dimensions" agree. In this case, we're multiplying a $5 \times \color{red}{5}$ and a $\color{red}{5} \times 5$ matrix. $\endgroup$ – Ben Grossmann May 1 at 23:30
  • $\begingroup$ @grace The fact that pretending that $A$ is a column vector still gives you the correct answer is an example of block-matrix multiplication. $\endgroup$ – Ben Grossmann May 1 at 23:32
  • $\begingroup$ Okay, but when I do that I get a completely different answer for the the other matrix (the one with 5's, 4's, 3's, 2's, and 1's). When I do it the way you suggest I get -120 but when I do it the way I initially did it I get -3, which is the same determinant as matrix A. $\endgroup$ – user716286 May 1 at 23:44
  • $\begingroup$ The correct answer is that the determinant of your second matrix is equal to $\det(M_2)\cdot \det(A) = -120 \cdot -3 = 360$. It seems that whatever you initially did is incorrect. $\endgroup$ – Ben Grossmann May 2 at 0:00

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.

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  • $\begingroup$ but there is no r5 component in the first matrix, if that's what you mean (?) $\endgroup$ – user716286 May 1 at 18:44
  • $\begingroup$ By $R_5$ I meant 5th row of the matrix in part(i). That is $R_5$ is actually $r_1$. $\endgroup$ – Shiv Tavker May 1 at 18:46
  • $\begingroup$ when you subtract all rows from R1 then it becomes 0. but we still have rows 2-5, do we do the same to them? $\endgroup$ – user716286 May 1 at 18:59
  • $\begingroup$ If one row becomes zero the determinant vanishes. The easiest way to see this is by computing the determinant via the cofactor expansion. $\endgroup$ – Shiv Tavker May 1 at 19:00
  • $\begingroup$ ah okay, thank you! $\endgroup$ – user716286 May 1 at 19:15

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