# Find the determinant after a certain row operation is applied to a matrix with known determinant

$A = \begin{bmatrix} --a--\\ --b--\\ --c-- \end{bmatrix}$

$A$ is a $3\times3$ matrix. The rows are all different, and the first row is called $a$. The second row is called $b$. The third row is called $c$. I'm not saying that all the values in the first row are the same, I am simply saying that the first row is called $a$. The determinant is $not$ $0$.

If the $\det(A)=3$, what is the determinant of:

$\begin{bmatrix} --a+b--\\ --b+c--\\ --c+a-- \end{bmatrix}$

I thought about this, but I recalled that whenever you have a matrix, and you add a row multiple of another, the determinant does not change. However in this case, the answer in the back of my textbook is $6$ and I don't understand how?

• And... what matrix is $\;A\;$ itself ? – DonAntonio Mar 14 '17 at 11:50
• It doesn't matter – K Split X Mar 14 '17 at 11:52
• A is a $3\times 3$ matrix – K Split X Mar 14 '17 at 11:53
• @K Either you don't understand the question, or you're kidding...or I don't understand the question: how come it isn't relevant?! – DonAntonio Mar 14 '17 at 11:53
• I think, K, that you had better write out the entire question as you have found it, not leaving out the least little bit, since what you have written so far is nonsense. – Gerry Myerson Mar 14 '17 at 11:58

$$\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{bmatrix}\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \begin{bmatrix}a+b \\ b+c \\ c+a\end{bmatrix}$$ Therefore, $$\begin{vmatrix}a+b \\ b+c \\ c+a\end{vmatrix} = \begin{vmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{vmatrix}\begin{vmatrix}a \\ b \\ c\end{vmatrix} = 2 \times 3 = 6$$

• Where did the get the $\times 2$ from? – K Split X Mar 14 '17 at 12:26
• @KSplitX $\begin{vmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{vmatrix} = 2$ – TenaliRaman Mar 14 '17 at 12:28
• Determinant is 2. Okay thanks – K Split X Mar 14 '17 at 12:30

Lets do it with rows instead of columns and remember that the determinant is multilinear:

\begin{align} \det(a+b,b+c,c+a) = \quad &\det(a,b,c)+\det(a,b,a)\\ +&\det(a,c,c)+ \det(a,c,a)\\ +&\det(b,b,c)+ \det(b,b,a)\\ +&\det(b,c,c)+ \det(b,c,a). \end{align}

All terms except the first and the last vanish because there is the same entry twice. And because rearranging the rows does not change the determinant you will find

$$\det(a+b,b+c,c+a)=\det(a,b,c)+\det(b,c,a)=2\det(a,b,c).$$

In your case $\det(A)=3$, so $2\cdot\det(A)=6$ is the result.

Well, with the explantion you added (and the one you also deleted...) it is finally clearer what you meant, but then, since the determinant of a matrix with two identical rows is zero, and if we intechange two rows we multiply the determinant by $\;-1\;$ , by multilinearity we get:$${}$$

$$\det\begin{pmatrix}-a+b-\\-b+c-\\-c+a-\end{pmatrix}=\color{red}{\det\begin{pmatrix}-a-\\-b-\\-c-\end{pmatrix}}+\underbrace{\det\begin{pmatrix}-a-\\-b-\\-a-\end{pmatrix}+\det\begin{pmatrix}-a-\\-c-\\-c-\end{pmatrix}+\ldots}_{=0}+\color{red}{\det\begin{pmatrix}-b-\\-c\\-a-\end{pmatrix}}$$$${}$$

$$=\color{red}{\det A}+\color{red}{\det A}=6$$

I will use slightly different notation, write

$$A = \begin{bmatrix} v_{1}\\ v_{2}\\ v_{3}\end{bmatrix}$$ where $v_{i}$ are the rows of $A$ and $\operatorname{det}(A) = 3.$

Then

$$\operatorname{det}\begin{bmatrix} v_{1} + v_{2}\\ v_{2} + v_{3}\\ v_{3} + v_{1} \end{bmatrix} = \operatorname{det} \left(\begin{bmatrix} v_{1}\\ v_{2} \\v_{3} \end{bmatrix} + \begin{bmatrix} 0 &1 &0\\ 0 & 0 & 1\\ 1& 0& 0\end{bmatrix}\begin{bmatrix} v_{1}\\ v_{2}\\v_{3}\end{bmatrix}\right) = \operatorname{det}\left(A + \begin{bmatrix} 0 &1 &0\\ 0 & 0 & 1\\ 1& 0& 0\end{bmatrix}A \right)$$ So, $$\operatorname{det}(A)\operatorname{det}\left(I+\begin{bmatrix} 0 &1 &0\\ 0 & 0 & 1\\ 1& 0& 0\end{bmatrix}\right) = 6$$

• Yup another answer just did this! Thanks for the indepth version tho – K Split X Mar 14 '17 at 12:32
• How did you arrive at the last step from the third step? Where did the $\det(A)$ on the outside come from? – K Split X Mar 14 '17 at 12:34
• Split the determinant over multiplication – JessicaK Mar 14 '17 at 12:36