Question

The value of the determinant $\left|\begin{array}{ccc}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|$ is

• A. $a+b+c$
• B. $1+a+b+c$
• C. $ab+bc+ca$
• D. $0$

Answer 1

Answer: (D)

$0$

We know that for a $3\times 3$ matrix $A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$, the determinant is:

$\left|A\right|=a(ei-fh)-b(di-fg)+c(dh-eg)$

Here, the given matrix is $A=\left[\begin{array}{ccc}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right]$, then the the determinant of $A$ is:

$\left|A\right|=1\left[b\right(a+b)-c(c+a\left)\right]-a\left[1\right(a+b)-1(c+a\left)\right]+(b+c)\left[1\right(c)-1(b\left)\right]=1(ab+b^2-c^2-ac)-a(a+b-c-a)+(b+c)(c-b)=ab+b^2-c^2-ac-a^2-ab+ac+a^2+bc-b^2+c^2-bc=ab-ab+b^2-b^2-c^2+c^2-ac+ac-a^2+a^2+bc-bc=0$

Hence, $\left|\begin{array}{ccc}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|=0$.