Determinant question

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Determinant question

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[A =\left| \begin{array}{ccc} v_{1} \\ v_{2} \\ v_{3}\\v_{4}\end{array} \right| \det = 8\] What is the det of \[\left| \begin{array}{ccc} 6v_{1} +2v_{4} \\ v_{2} \\ v_{3}\\v_{4} + 3v_{1}\end{array} \right|\]
Gotta love row operations ^^ Let's be general here: \[\det\left[ \begin{array}{ccc} v_{1} \\ v_{2} \\ v_{3}\\v_{4}\end{array} \right] =k\]
Three elementary row operations: switch, multiply, and "add row multiple" All these have distinct effects on the determinant of the original matrix ^^

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Yeah! So I thought the answer would be that it would just be scaled by 3, because scalar addition or subtraction doesn't affect the determinant! However, the answer says it's 0, because the first row is twice the last. That didn't make much sense to me!
When you switch one pair of rows, such as here: \[\det\left[ \begin{array}{ccc} v_{1} \\ \color{red}{v_{3}} \\ \color{blue}{v_{2}}\\v_{4}\end{array} \right] =-k\] the determinant gets negated. And will be negated for every pair of rows that you switch :D
The first row is indeed twice the last. Are you finding it hard to believe that that causes the determinant to be zero? ^^
Oops, scaled by 6! (v1 in first row is scaled by 6) Yeah, how does that change anything if the 4th row is a multiple of the first? Does that mean that the 4th row is redundant and if I rref'ed it the row would be 0, so if I tried to take the diagonals it would be 0?
Okay. Let's take a weird approach to this. First, what I said: Switching a pair of rows causes the determinant to be negated or multiplied by a factor of (-1). Next, multiplying a row by a scalar in turn scales the determinant by that scalar. ie \[\det\left[ \begin{array}{ccc} v_{1} \\ \color{green}{n}v_{2} \\ v_{3}\\v_{4}\end{array} \right] =\color{green}{n}k\]
So far, so good, yes? ^^
So suppose we have a matrix where the first row is a multiple of the last row, and suppose its determinant was d. Like this \[\Large \det\left[ \begin{array}{ccc} mu_{3} \\ u_{1} \\ u_{2}\\u_{3}\end{array} \right]=h \]
If we switch the first and fourth rows, as per the rule, the determinant gets multiplied by (-1) We get \[\Large \det\left[ \begin{array}{ccc} \color{red}{u_{3}} \\ u_{1} \\ u_{2}\\ \color{red}{mu_{3}}\end{array} \right]=\color{red}{-h}\]
If we scale the first row by m, as per the other rule, the determinant gets scaled as well by a factor of m. \[\Large \det\left[ \begin{array}{ccc} \color{red}mu_{3} \\ u_{1} \\ u_{2}\\mu_{3}\end{array} \right]=-h\color{red}{(m)}\]
Now we scale the fourth row by 1/m. Again, by the rules, this would also scale the determinant by a factor of 1/m. \[\Large \det\left[ \begin{array}{ccc} mu_{3} \\ u_{1} \\ u_{2}\\ \color{red}{\left(\frac1m\right)}mu_{3}\end{array} \right]=-h(m)\color{red}{\left(\frac1m\right)}\]
Simplifying, we get \[\Large \det\left[ \begin{array}{ccc} mu_{3} \\ u_{1} \\ u_{2}\\u_{3}\end{array} \right]=-h\] We ended up with our original matrix, but THIS time, we have shown that its determinant is also equal to -h. That can only mean h = -h Or h = 0 Done ^^
You are awesome! Thanks!
No problem ^^

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