Compute the determinant of the following matrix

Question

$$A = \left[\begin{matrix} 1 & b & c \\ b & c & 1 \\ c & 1 & b \end{matrix}\right]$$ $R_2-bR_1$ & $R_3-cR_1$

$$ = \left[\begin{matrix} 1 & b & c \\ 0 & c-b^2 & 1-bc \\ 0 & 1-cb & b-c^2 \end{matrix}\right]$$ $R_3(c-b^2)$

$$ = \left[\begin{matrix} 1 & b & c \\ 0 & c-b^2 & 1-bc \\ 0 & (1-cb)(c-b^2) & (b-c^2)(c-b^2) \end{matrix}\right]$$ $R_3-(1-cb)R_2$

$$ = \left[\begin{matrix} 1 & b & c \\ 0 & c-b^2 & 1-bc \\ 0 & 0 & D \end{matrix}\right]$$

$D=(b-c^2)(c-b^2)-(1-bc)(1-bc)(1-cb)$
$=bc-b^3-c^3+b^2c^2-(1-cb-bc+b^2c^2)$
$=bc-b^3-c^3+b^2c^2-1+cb+bc-b^2c^2$
$=-b^3-c^3+3cb-1$

So then I'd have to multiply the main diagonal which gives me the equation: $Det(A)=1(c-b^2)(-b^3-c^3+3bc-1)$

In class my professor did an example similar to this where he used the lower triangular method.

I know with a different method that the answer should be $-b^3-c^3+3bc-1$, so I'm really confused. If someone could tell me what I'm doing wrong or needs further explanation please let me know. Thanks in advance!

Answer 1

After this step $$\left[\begin{matrix} 1 & b & c \\ 0 & c-b^2 & 1-bc \\ 0 & 1-cb & b-c^2 \end{matrix}\right]$$ you should multiply row 2 by $-\frac{1-bc}{c-b^2}$ and add it to row 3 to get $$\left[\begin{matrix} 1 & b & c \\ 0 & c-b^2 & 1-bc \\ 0 & 0 & -\frac{(1-bc)^2}{c-b^2}+b-c^2 \end{matrix}\right].$$

Multiplying the diagonal entries you get $-b^3-c^3+3bc-1$. Alternatively, when you multiplied a row by $(c-b^2)$ you could note that this multiplies the determinant by $c-b^2$. Thus the determinant of the original matrix is what you found divided by $c-b^2$.

Answer 2

The determinant of a $3\times 3$ matrix $M$ :

$$M= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$

is given by :

$$\det(M) = a_{11} \Bigg|\begin{matrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix}\Bigg| - a_{12}\Bigg| \begin{matrix} a_{21} & a_{23} \\ a_{31} & a_{33}\end{matrix} \Bigg| + a_{13} \Bigg| \begin{matrix} a_{21} & a_{22} \\ a_{31} & a_{32}\end{matrix} \Bigg|$$

Applying this to your given matrix $A$, should yield you the result :

$$\det(A) = -b^3 + 3bc - c^3 - 1$$

If you need help to find where you're wrong with your solution, please make sure you edit your question and contain it in it.

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