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Let $a,b,c \in\mathbb{R}$. Find the determinant of the matrix $$A=\begin{bmatrix} b+c&a&a\\ b&c+a&b\\ c&c&a+b\end{bmatrix}$$

I got the answer that the determinant is $A = 9abc.$ Is this correct?

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    $\begingroup$ I think you should find $4abc$. $\endgroup$
    – StackTD
    Jan 11, 2018 at 12:44

4 Answers 4

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It is a polynomial of $3$rd degree in $a,b,c$. Put $a=0$:

$$\left|\begin{matrix}b+c & 0 & 0\\b & c & b\\c & c & b\end{matrix}\right|=0$$

This means that the polynomial is divisible by $a$. Similarly, it is divisible by $b$ and $c$, so it must be of the form $Cabc$, where $C$ is a constant. Replacing $a=b=c=1$ gives:

$$C=\left|\begin{matrix}2 & 1 & 1\\1 & 2 & 1\\1 & 1 & 2\end{matrix}\right|=\left|\begin{matrix}4 & 4 & 4\\1 & 2 & 1\\1 & 1 & 2\end{matrix}\right|=4\left|\begin{matrix}1 & 1 & 1\\1 & 2 & 1\\1 & 1 & 2\end{matrix}\right|=4\left|\begin{matrix}1 & 1 & 1\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right|=4$$

(elementary operations on rows that preserve determinants have been used throughout).

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  • $\begingroup$ thanks a lots @user8734617 $\endgroup$
    – user476275
    Jan 11, 2018 at 15:01
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You can always use for $3\times 3$ matrices the following trick, add the first and the second column in this order after the 3rd column: \begin{array}[ccccc] &c_1 & c_2 & c_3 & c_1& c_2\\ d_1 & d_2 & d_3 & d_1 & d_2\\ e_1 & e_2 & e_3 & e_1 & e_2 \end{array} Now, take $c_3$ and go diagonally to the right and multiply these elements: $$ c_3\cdot d_1\cdot e_2 $$ continue the same manner with $c_2$: $$ c_2\cdot d_3\cdot e_1 $$ and finally for $c_1$. These results you add together, let us call it $C$. After this you start with $c_3$ again but this time you go to the left diagonally and multiply by the same manner, then you take $c_2$ and $c_1$. The results from these multiplications you subtract from $C$ and you have the determinant.

This only wotks with $3\times 3$ matrices.

Have fun!

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  • $\begingroup$ thanks a lots vinyl-coat-jawa $\endgroup$
    – user476275
    Jan 11, 2018 at 15:02
  • $\begingroup$ You are welcome $\endgroup$ Jan 11, 2018 at 15:29
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You can use Laplace formula on the first row to solve it like:
$$ (b+c)\begin{vmatrix} c+a&b\\ c&a+b\end{vmatrix}- a\begin{vmatrix} b&b\\ c&a+b\end{vmatrix}+ a\begin{vmatrix} b&c+a\\ c&c\end{vmatrix}=$$ $$=(b+c)((c+a)(a+b)-cb)-a(b(a+b)-cb)+a(bc-c(c+a))=$$ $$=(b+c)(ca+cb+a^2+ab-cb)-a(ba+b^2-cb)+a(bc-c^2-ca)=$$ $$=(abc+a^2b+ab^2+ac^2+a^2c+abc)+(-a^2b-ab^2+abc)+(abc-ac^2-a^2c)=$$ $$=2abc+abc+abc=4abc$$

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  • $\begingroup$ The first line should be $$(b+c)\begin{vmatrix} c+a&b\\ c&a+b\end{vmatrix} -a\begin{vmatrix} b&b\\ c&a+b\end{vmatrix}+ a\begin{vmatrix} b&c+a\\ c&c\end{vmatrix}$$ $\endgroup$
    – Robert Z
    Jan 11, 2018 at 13:50
  • $\begingroup$ I was about to post the same comment. And changing the sign of that term will lead to the correct $4abc$ instead of $2ab(a+b+c)$. $\endgroup$
    – StackTD
    Jan 11, 2018 at 13:51
  • $\begingroup$ Thank you for the remark, it slipped... $\endgroup$
    – Plexus
    Jan 11, 2018 at 14:13
  • $\begingroup$ thanks a lots Plexus $\endgroup$
    – user476275
    Jan 11, 2018 at 15:02
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No, you are wrong. By using the rule of Sarrus (a scheme which works for $3\times 3$ matrices), the determinant should be $$(a+b)(b+c)(c+a)+abc+abc-ab(a+b)-bc(b+c)-ac(c+a).$$ Can you take it form here and revise your evaluation?

P.S. As remarked above by StackTD, the final result should be $4abc$.

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  • $\begingroup$ thanks alots Robert Z $\endgroup$
    – user476275
    Jan 11, 2018 at 15:02

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