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By using properties of determinants, show that
$\left|\begin{array}{ccc}a^2+1 & a b & a c \\ a b & b^2+1 & b c \\ c a & c b & c^2+1\end{array}\right|=1+a^2+b^2+c^2$
$\left|\begin{array}{ccc}a^2+1 & a b & a c \\ a b & b^2+1 & b c \\ c a & c b & c^2+1\end{array}\right|=1+a^2+b^2+c^2$
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Verified Answer
Let
$\begin{aligned} \Delta &=\left|\begin{array}{ccc}\mathrm{a}^2+1 & \mathrm{ab} & \mathrm{ac} \\ \mathrm{ab} & \mathrm{b}^2+1 & \mathrm{bc} \\ \mathrm{ca} & \mathrm{cb} & \mathrm{c}^2+1\end{array}\right| \\ &=\left|\begin{array}{ccc}\mathrm{a}^2+1 & \mathrm{ab}+0 & \mathrm{ac}+0 \\ \mathrm{ab}+0 & \mathrm{~b}^2+1 & \mathrm{bc}+0 \\ \mathrm{ca}+0 & \mathrm{cb}+0 & \mathrm{c}^2+1\end{array}\right| \end{aligned}$
This may be expressed as the sum of 8 determinants
$\therefore \Delta=\left|\begin{array}{lll}
\mathrm{a}^2 & \mathrm{ab} & \mathrm{ac} \\
\mathrm{ab} & \mathrm{b}^2 & \mathrm{bc} \\
\mathrm{ca} & \mathrm{bc} & \mathrm{c}^2
\end{array}\right|+\left|\begin{array}{lll}
1 & \mathrm{ab} & \mathrm{ac} \\
0 & \mathrm{~b}^2 & \mathrm{bc} \\
0 & \mathrm{bc} & \mathrm{c}^2
\end{array}\right|$
$+\left|\begin{array}{ccc}\mathrm{a}^2 & 0 & \mathrm{ac} \\ \mathrm{ab} & 1 & \mathrm{bc} \\ \mathrm{ca} & 0 & \mathrm{c}^2\end{array}\right|+\left|\begin{array}{ccc}\mathrm{a}^2 & \mathrm{ab} & 0 \\ \mathrm{ab} & \mathrm{b}^2 & 0 \\ \mathrm{ca} & \mathrm{bc} & 1\end{array}\right|$
$+\left|\begin{array}{lll}\mathrm{a}^2 & 0 & 0 \\ \mathrm{ab} & 1 & 0 \\ \mathrm{ca} & 0 & 1\end{array}\right|+\left|\begin{array}{lll}1 & \mathrm{ab} & 0 \\ 0 & \mathrm{~b}^2 & 0 \\ 0 & \mathrm{bc} & 1\end{array}\right|+\left|\begin{array}{ccc}1 & 0 & \mathrm{ac} \\ 0 & 1 & \mathrm{bc} \\ 0 & 0 & \mathrm{c}^2\end{array}\right|+\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$
$=\left|\begin{array}{lll}\mathrm{a} & \mathrm{a} & \mathrm{a} \\ \mathrm{b} & \mathrm{b} & \mathrm{b} \\ \mathrm{c} & \mathrm{c} & \mathrm{c}\end{array}\right| \mathrm{abc}+\left(\mathrm{b}^2 \mathrm{c}^2-\mathrm{b}^2 \mathrm{c}^2\right)+\left(\mathrm{a}^2 \mathrm{c}^2-\mathrm{a}^2 \mathrm{c}^2\right)$
$+\left(a^2 b^2-a^2 b^2\right)+a^2+b^2+c^2+1$
$=a^2+b^2+c^2+1$
$\begin{aligned} \Delta &=\left|\begin{array}{ccc}\mathrm{a}^2+1 & \mathrm{ab} & \mathrm{ac} \\ \mathrm{ab} & \mathrm{b}^2+1 & \mathrm{bc} \\ \mathrm{ca} & \mathrm{cb} & \mathrm{c}^2+1\end{array}\right| \\ &=\left|\begin{array}{ccc}\mathrm{a}^2+1 & \mathrm{ab}+0 & \mathrm{ac}+0 \\ \mathrm{ab}+0 & \mathrm{~b}^2+1 & \mathrm{bc}+0 \\ \mathrm{ca}+0 & \mathrm{cb}+0 & \mathrm{c}^2+1\end{array}\right| \end{aligned}$
This may be expressed as the sum of 8 determinants
$\therefore \Delta=\left|\begin{array}{lll}
\mathrm{a}^2 & \mathrm{ab} & \mathrm{ac} \\
\mathrm{ab} & \mathrm{b}^2 & \mathrm{bc} \\
\mathrm{ca} & \mathrm{bc} & \mathrm{c}^2
\end{array}\right|+\left|\begin{array}{lll}
1 & \mathrm{ab} & \mathrm{ac} \\
0 & \mathrm{~b}^2 & \mathrm{bc} \\
0 & \mathrm{bc} & \mathrm{c}^2
\end{array}\right|$
$+\left|\begin{array}{ccc}\mathrm{a}^2 & 0 & \mathrm{ac} \\ \mathrm{ab} & 1 & \mathrm{bc} \\ \mathrm{ca} & 0 & \mathrm{c}^2\end{array}\right|+\left|\begin{array}{ccc}\mathrm{a}^2 & \mathrm{ab} & 0 \\ \mathrm{ab} & \mathrm{b}^2 & 0 \\ \mathrm{ca} & \mathrm{bc} & 1\end{array}\right|$
$+\left|\begin{array}{lll}\mathrm{a}^2 & 0 & 0 \\ \mathrm{ab} & 1 & 0 \\ \mathrm{ca} & 0 & 1\end{array}\right|+\left|\begin{array}{lll}1 & \mathrm{ab} & 0 \\ 0 & \mathrm{~b}^2 & 0 \\ 0 & \mathrm{bc} & 1\end{array}\right|+\left|\begin{array}{ccc}1 & 0 & \mathrm{ac} \\ 0 & 1 & \mathrm{bc} \\ 0 & 0 & \mathrm{c}^2\end{array}\right|+\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$
$=\left|\begin{array}{lll}\mathrm{a} & \mathrm{a} & \mathrm{a} \\ \mathrm{b} & \mathrm{b} & \mathrm{b} \\ \mathrm{c} & \mathrm{c} & \mathrm{c}\end{array}\right| \mathrm{abc}+\left(\mathrm{b}^2 \mathrm{c}^2-\mathrm{b}^2 \mathrm{c}^2\right)+\left(\mathrm{a}^2 \mathrm{c}^2-\mathrm{a}^2 \mathrm{c}^2\right)$
$+\left(a^2 b^2-a^2 b^2\right)+a^2+b^2+c^2+1$
$=a^2+b^2+c^2+1$
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