Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

I tried gaussian elimination and ended up with:

$\begin{bmatrix} v1\\ v2\\ v3\\ \frac{5}{2}v1+v4 \end{bmatrix}$

Then I used the rule that says If B is obtained from A by adding a multiple of a row of A to another row, then det(B)=det(A) but it's saying that the answer detB=detA=8 is incorrect.

Where did I go wrong?

share|improve this question
    
You can get a subscript using an underscore, e.g. v_1 produces $v_1$. –  joriki Apr 17 '12 at 6:42

3 Answers 3

up vote 3 down vote accepted

Use the fact that the determinant is multilinear. $$\det\left(\begin{array}{c} r_1\\\vdots\\r_{i-1}\\ r_i+\alpha s_i\\r_{i+1}\\\vdots\\r_n\end{array}\right) = \det\left(\begin{array}{c}r_1\\\vdots\\r_{i-1}\\r_i\\r_{i+1}\\\vdots\\r_n\end{array}\right) + \alpha \det\left(\begin{array}{c} r_1\\\vdots\\r_{i-1}\\s_i\\r_{i+1}\\\vdots\\r_n\end{array}\right)$$ where $r_1,\ldots,r_n$ are rows, $s_i$ is a row, $\alpha$ is a scalar, and $i$ is arbitrary.

Also use the fact that a determinant of a matrix with two identical rows is equal to $0$.

(Since you don't show your Gaussian elimination, I can't tell whether you made a mistake or performed an operation that would change the value of the determinant.)

Here's my computation of this, using multilinearity; when we exchange two rows, it multiplies the determinant by $-1$: $$\begin{align*} \det\left(\begin{array}{c}4v_1+2v_4\\v_2\\v_3\\5v_1+2v_4\end{array}\right) &= \det\left(\begin{array}{c}4v_1\\v_2\\v_3\\5v_1+2v_4\end{array}\right) + \det\left(\begin{array}{c}2v_4\\v_2\\v_3\\5v_1+2v_4\end{array}\right)\\ &= \det\left(\begin{array}{c} 4v_1\\v_2\\v_3\\5v_1\end{array}\right) + \det\left(\begin{array}{c}4v_1\\v_2\\v_3\\2v_4\end{array}\right) + \det\left(\begin{array}{c}2v_4\\v_2\\v_3\\5v_1\end{array}\right) + \det\left(\begin{array}{c}2v_4\\v_2\\v_3\\2v_4\end{array}\right)\\ &= 4\cdot5\cdot\det\left(\begin{array}{c}v_1\\v_2\\v_3\\v_1\end{array}\right) + 4\cdot2\cdot\det\left(\begin{array}{c}v_1\\v_2\\v_3\\v_4\end{array}\right) \\ &\qquad\mathop{+} 2\cdot 5\cdot\det\left(\begin{array}{c}v_4\\v_2\\v_3\\v_1\end{array}\right) + 2\cdot2\cdot\det\left(\begin{array}{c}v_4\\v_2\\v_3\\v_4\end{array}\right)\\ &=20(0) + 8\det(A) +10(-1)\det\left(\begin{array}{c}v_1\\v_2\\v_3\\v_4\end{array}\right) + 4(0)\\ &= 8\det(A) - 10\det(A)\\ &= -2\det(A)\\ &= -2(8)\\ &= -16. \end{align*}$$

share|improve this answer
    
Ok I tried the gaussian elimination again like so: add -1*row1 to row4, add -4*row4 to row1, divide row1 by 2, swapped row4 and row1. So I ended up with [v1, v2, v3, v4] and I figured the operations I made would result in the determinant being -$\frac{1}{2}det(A)=-4$ but it says that answer is incorrect. –  StickFigs Apr 17 '12 at 16:35
1  
@StickFigs: You should have put that information on the main question, not buried here in a comment. I think you have the constant wrong. Note that $$\det\left(\begin{array}{c}2v_4\\v_2\\v_3\\v_1\end{array}\right) = 2\det\left(\begin{array}{c}v_4\\v_2\\v_3\\v_1\end{array}\right) = -2\det\left(\begin{array}{c}v_1\\v_2\\v_3\\v_4\end{array}\right) = -2\det(A).$$ That is, when you multiply the first row by $\frac{1}{2}$, you don't get a $\frac{1}{2}$ "outside", you get a $2$ (because it must cancel the factor of $\frac{1}{2}$ you had). –  Arturo Magidin Apr 17 '12 at 16:42
    
Ah, I see what I did wrong now. I confused B and A and thought B was obtained from A by multiplying a row by 1/2 but B is actually obtained from A by multiplying a row by 2. –  StickFigs Apr 17 '12 at 16:50
    
@StickFigs: Yes; in my experience, this is the common "pitfall" in computing determinants like this. It's best to keep track of the operations as you do them (as I did in the comment above), rather than try to figure out the appropriate constant to multiply by after the fact by going over the list of operations done. –  Arturo Magidin Apr 17 '12 at 16:58

Gaussian elimination may involve swapping rows, and swapping rows changes the sign of the determinant; thus Gaussian elimination doesn't leave the determinant unchanged. However, it seems you also performed other operations beyond the ones minimally required for Gaussian elimination.

The most direct way to arrive at your result would be to subtract the fourth row from the first, then multiply the first row by $-1$ and the last row by $1/2$. Since the determinant is multilinear, multiplying a row by a factor multiplies the determinant by that factor. Thus the determinant of your result is $-1/2$ times the determinant of the matrix you started from.

share|improve this answer

Starting from $A$, to get to your target, you might first multiply the first row by $4$, then add $2$ times the fourth row to the first row. I'll leave it to you to decide how to get from there to the target. Adding $2$ times the fourth row to the first row doesn't change the determinant, but multiplying the first row by $4$ multiplies the determinant by $4$.

Another way to do it is to note that your target matrix is $BA$ where $$B = \pmatrix{ 4 & 0 & 0 & 2\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 5 & 0 & 0 & 2\cr}$$ (do you see why?), and that $\det(BA) = \det(B) \det(A)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.