By using properties of determinants, show that
Question:

By using properties of determinants, show that

$\left|\begin{array}{lll}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

Solution:

$\Delta=\left|\begin{array}{lll}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|$

Taking out common factors ab, and c from R1, R2, and Rrespectively, we have:

$\Delta=a b c\left|\begin{array}{lll}a+\frac{1}{a} & b & c \\ a & b+\frac{1}{b} & c \\ a & b & c+\frac{1}{c}\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we have:

$\Delta=a b c\left|\begin{array}{ccc}a+\frac{1}{a} & b & c \\ -\frac{1}{a} & \frac{1}{b} & 0 \\ -\frac{1}{a} & 0 & \frac{1}{c}\end{array}\right|$

Applying $\mathrm{C}_{1} \rightarrow a \mathrm{C}_{1}, \mathrm{C}_{2} \rightarrow b \mathrm{C}_{2}$, and $\mathrm{C}_{3} \rightarrow c \mathrm{C}_{3}$, we have:

\begin{aligned} \Delta &=a b c \times \frac{1}{a b c}\left|\begin{array}{lll}a^{2}+1 & b^{2} & c^{2} \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right| \\ &=\left|\begin{array}{lll}a^{2}+1 & b^{2} & c^{2} \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right| \end{aligned}

Expanding along R3, we have:

\begin{aligned} \Delta &=-1\left|\begin{array}{ll}b^{2} & c^{2} \\ 1 & 0\end{array}\right|+1\left|\begin{array}{ll}a^{2}+1 & b^{2} \\ -1 & 1\end{array}\right| \\ &=-1\left(-c^{2}\right)+\left(a^{2}+1+b^{2}\right)=1+a^{2}+b^{2}+c^{2} \end{aligned}

Hence, the given result is proved.