How do I prove this?

[Guest]

Hello

The question reads: Prove that
\(\displaystyle \L
\left| {\begin{array}{c}
b & { - c} & 1 \\
1 & a & { - b} \\
{ - a} & 1 & c \\
\end{array}} \right| = (a + b + c)^2 - 2(ab + bc + ca) + 1\)

Does anyone have any suggestions about how you would prove this?

Thank you.

 

[pka]

What do you mean: “How do I prove this?”
Do you know how to expand a 3x3 determinant?
To do the problem you must know how to do that
Look it up in your text, or do an internet search.

 

[Guest]

Hi

I've worked out the determinant but I'm not getting the required answer. Can you check and tell me where I am going wrong?

Prove that
\(\displaystyle \L
\left| {\begin{array}{c}
b & { - c} & 1 \\
1 & a & { - b} \\
{ - a} & 1 & c \\
\end{array}} \right| = (a + b + c)^2 - 2(ab + bc + ca) + 1\)

I worked out

\(\displaystyle \L

\begin{array}{l}
{\rm = b[(a}{\rm .c) - (1}{\rm . - b)]- ( - c)[(1}{\rm .c) - ( - a}{\rm . - b)] +[(1x1) - ( - a}{\rm .a)]} \\
\\
{\rm = b[ac + b] + c[c - ab] + [1 + a}^{\rm 2} {\rm ]} \\
\\
{\rm = abc + b}^{\rm 2} {\rm + c}^{\rm 2} {\rm - abc + 1 + a}^{\rm 2} \\
\\
{\rm = a}^{\rm 2} {\rm + b}^{\rm 2} {\rm + c}^{\rm 2} {\rm + 1} \\
\\
{\rm = [(a + b + c) (a + b + c)] + 1 = a}^{\rm 2} {\rm + ab + ac + ab + b}^{\rm 2} {\rm + bc + ac + bc + c}^{\rm 2} {\rm + 1} \\
\\
{\rm = (a + b + c) }^{\rm 2} {\rm + 2(ab + bc + ca) + 1} \\
\end{array}\)


I'm trying to get \(\displaystyle \L = (a + b + c)^2 - 2(ab + bc + ca) + 1\)

Please can you help?

 

[tkhunny]

You're there. You just messed up the last step. Think about it a little harder.

Note: (a+b+c)² ≠ (a²+b²+c²)

 

[pka]

\(\displaystyle \L
\begin{array}{l}
(a + b + c)^2 & = & a^2 + b^2 + c^2 + 2ab + 2ac + 2bc \\
a^2 + b^2 + c^2 & = & (a + b + c)^2 - 2(ab + ac + bc) \\
\end{array}\) .

 

[Guest]

Note: (a+b+c)² ≠ (a²+b²+c²)

Try again:

\(\displaystyle \L

\begin{array}{l}
= a^2 + b^2 + c^2 + 1
= [(a + b + c) + (a + b + c)] + 1 \\ \end{array}\) ... Is this right?

\(\displaystyle \L \begin{array}{l}
= a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2 + 1 \\
= (a^2 + b^2 + c^2 ) + 2(ab + bc + ca) + 1 \\
\end{array}\)


But I'm still getting "+2(ab+bc+ca) + 1" instead of "-2(ab+bc+ca) + 1".

I think my algebra is letting me down. Can anyone tell me where I'm going wrong now?

Thanks

 

[pka]

\(\displaystyle \L
a^2 + b^2 + c^2 + 1
= [(a + b + c) + (a + b + c)] + 1 \\) ... This is incorrect!

Look at my posting above.
You need to square a trinominal.