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$$\begin{vmatrix} 1 & 1 & 1 \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c} \\ bc & ac & ab \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ b+c & a+c & a+b \end{vmatrix}$$ (a, b, c are non-zero constants)

Someone who can help me with the following proof, the truth is that I don't understand what I should look for if I should check that the determinant is the same in both matrices or I don't know if I'm honest, I'm sorry for my question and I hope I don't bother anyone.

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  • $\begingroup$ You might explore this by finding the determinants for some example distinct values of $a,b,c$ to see the issue. $\endgroup$
    – Henry
    Commented Feb 10 at 2:00

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I assume that "Gauss's method" is referring to the fact that you can row-reduce a matrix, then evaluate the determinant by taking the product along the main diagonal.

It looks like the question is asking you to prove that the two determinants are equal. I also assume that they don't want you to brute-force it by expanding the $3 \times 3$ determinant.

You can use the fact that adding a multiple of one row to another doesn't change the determinant of a matrix.

So in this case, let's start with the LHS and subtract $(abc)$ times the second row from the third row (i.e., $R3 - (abc) R2$): $$\begin{align} \begin{bmatrix} 1 & 1 & 1\\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\\ bc & ac & ab \end{bmatrix} &\xrightarrow{R_3-(abc)R_2} \begin{bmatrix} 1 & 1 & 1\\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\\ 0 & 0 & 0 \end{bmatrix} \end{align}$$ which implies that the determinant is zero.

Similarly, you can row-reduce the second matrix to show that the determinant of the second matrix is also zero. Observe that if you add the second row to the third row, then $$\begin{align} \begin{bmatrix} 1 & 1 & 1\\ a & b & c\\ b + c & a + c & a + b \end{bmatrix} &\xrightarrow{R_3 + R_2} \begin{bmatrix} 1 & 1 & 1\\ a & b & c\\ a + b + c & a + b + c & a + b + c \end{bmatrix} \end{align}$$ since the third row is just $(a + b + c)$ times the first row, the determinant of this matrix is also zero.

Altogether, we've shown that $$\begin{align} \det \begin{bmatrix} 1 & 1 & 1\\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\\ bc & ac & ab \end{bmatrix} &= 0 = \det\begin{bmatrix} 1 & 1 & 1\\ a & b & c\\ b + c & a + c & a + b \end{bmatrix} \end{align}$$

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