I think I'm getting the hang of these type of problems involving finding the determinant of a matrix using the properties of determinants. So i'll post the question here and i'll enter the matrix for you guys to see:

Given $\det(A)= 2$ and $\det (B) = 3$, where $$A =\begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix}$$

And where B is a 3x3 matrix Find the Determinant of the following (explain the properties used):

$$det =\begin{pmatrix} a_1+2b_1 & a_2+2b_2 & a_3+2b_3 \\-b_1 & -b_2 & -b_3 \\ c_1+2b_1 & c_2+ 2b_2 & c_3+2b_3 \end {pmatrix}$$

alright so once I was workin gon the problem I ended up first doing taking R1'=R1+2R2

which got me to the following:

$$det =\begin{pmatrix} a_1 & a_2 & a_3 \\-b_1 & -b_2 & -b_3 \\ c_1+2b_1 & c_2+ 2b_2 & c_3+2b_3 \end {pmatrix}$$

Next I saw I can factor out the 2 in row 3 so I think I might end up with something like this alongside the -1 I factor out in row 2:

$$det =\begin{pmatrix}a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \\ c_1+b_1 & c_2+ b_2 & c_3+b_3 \end {pmatrix}$$

Next I just took row R3=R3-R and go to the following

$$det =\begin{pmatrix}a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end {pmatrix}$$

I'm trying to put -2 beside but I'm not sure how to do that with the matrix syntax I'm using. But anyway I ended with this in the end:


-2*2= -4

am I in the right direction or you guys think I'm wrong?

  • $\begingroup$ What is B, and it seems like you jumped a lot of steps in your explanation $\endgroup$
    – Quality
    Mar 14, 2015 at 2:57
  • $\begingroup$ ah crap I forgot to add what B was. B is is just a 3x3 matrix according to my instructions. Let me add that in. And do you want me to finish fully? I can with no problem? $\endgroup$ Mar 14, 2015 at 2:58
  • $\begingroup$ No thats okay as long as you understand, but it might be helpful for yourself and others to see what you have $\endgroup$
    – Quality
    Mar 14, 2015 at 2:59
  • $\begingroup$ Do you know some of the basic rules, such as that adding a scalar multiple does not change the determinant, etc? $\endgroup$
    – Quality
    Mar 14, 2015 at 2:59
  • $\begingroup$ From what I was taught in class and the statement you just said, I wasn't. I just know from GJ elimination method of the steps I can take perhaps $\endgroup$ Mar 14, 2015 at 3:01

1 Answer 1


You have $$A= \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{pmatrix}$$

and Det(A)=2

You are looking for the determinant of some matrix, say B,

that is $$ B= \begin{pmatrix} a_1 + 2b_1 & a_2+2b_2 & a_3+2b_3 \\ -b_1 & -b_2 & -b_3 \\ c_1+b_1 & c_2+b_2 & c_3+b_3 \\ \end{pmatrix}$$

Well the way you do this is by taking account of the following,

If you add or subtract a scalar multiple of one row or column to another,the determinant does not change. If you swap two rows, the sign of the determinant switches, and if you multiply one row or column by a constant, you multiply the determinant by that.

So in your case, adding the rows of B to row 1 and 3 changed nothing so the deterimant would have stayed 2, but you multiplied row 2 by -1, so the det must be multiplied by minus one as well,

thus det(B)=-det(A)=-2

  • $\begingroup$ thanks I'm still a little fuzzy on it. But i'll review it more to make sure I'll understand it better. Thanks for the help i'll make sure I understand it by mid-term $\endgroup$ Mar 14, 2015 at 3:21
  • $\begingroup$ No worries, yea just practice with these and you'll get the hang of it. good luck $\endgroup$
    – Quality
    Mar 14, 2015 at 3:22

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