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I'm asked to find the determinant of a matrix $B$ if:

$$A=\left |\begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{bmatrix}\right|=-k$$ $$B=\begin{bmatrix} a_3 & a_2 & a_1 \\ b_3-2a_3 & b_2-2a_2 & b_1-2a_1 \\ c_3 & c_2 & c_1 \\ \end{bmatrix}$$

I used rule of Sarrus in order to find the determinant and finally I got:

For $A$: $$-k=a_1b_2c_3+a_2b_3c_1+a_3b_1c_2-a_3b_2c_1-a_1b_3c_2-a_2b_1c_3$$ $$k=-a_1b_2c_3-a_2b_3c_1-a_3b_1c_2+a_3b_2c_1+a_1b_3c_2+a_2b_1c_3$$

I also factorised to get this: $$k=b_2(a_3c_1-a_1c_3)+b_1(a_2c_3-a_3c_2)+b_3(a_1c_2-a_2c_1) \tag{1}$$

For $B$:

$$(b_2-2a_2)(a_3c_1-a_1c_3)+(b_1-2a_1)(a_2c_3-a_3c_2)+(b_3-2a_3)(a_1c_2-a_2c_1) \tag{2}$$

I don't know what else to do here, I've thought maybe about stating $1=2$ and that would mean $$(b_2-2a_2)=b_2\\ (b_1-2a_1)=b_1\\ (b_3-2a_3)=b_3$$

Therefore:

$$B= \left|\begin{bmatrix} a_3 & a_2 & a_1 \\ b_3 & b_2 & b_1 \\ c_3 & c_2 & c_1 \\ \end{bmatrix}\right|=k $$

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\begin{bmatrix} instead of \begin{matrix}, do the same to \end. –  Git Gud Mar 25 '13 at 20:13
    
On the last equality do you mean determinants or what? –  Git Gud Mar 25 '13 at 20:14
    
Yes, (1)=(2). Thank you, I fixed it. –  ChairOTP Mar 25 '13 at 20:15

1 Answer 1

Note that $$B=\begin{pmatrix} 1 & 0 & 0\\ -2 & 1 & 0\\ 0 & 0 & 1\\\end{pmatrix}A\begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\\\end{pmatrix}$$ and so we have $$\det(B)=\det\begin{pmatrix} 1 & 0 & 0\\ -2 & 1 & 0\\ 0 & 0 & 1\\\end{pmatrix}\det(A)\det\begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\\\end{pmatrix}.$$ Can you find the determinants of these two matrices?

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Yes, I got 1 and -1 respectively. I don't see how's that different from what I found? –  ChairOTP Mar 25 '13 at 20:20
    
It's the same answer, but your reasoning is not correct (as (1) and (2) are not equal in general). –  Alex Becker Mar 25 '13 at 20:42
    
Oh! So how could I explain it without an example? –  ChairOTP Mar 25 '13 at 20:46

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