Use row reduction to show that the determinant is equal to this variable.

Question

Show determinant of: \begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}

is equal to $(b - a)(c - a)(c - b)$

I'm not sure if you can use squares or square roots hmmm.. please help me. I'm sure it's a simple question. Much appreciated.

Answer 1

We see immmediatley that the determinant is a polynomial in $a,b,c$, homgenous of degree $3$. And if $a=b$ or $b=c$ or $c=a$, two columns are identical and make the determiniant zero. On the other hand, if $a,b,c$ are pairwise distinct, the determinant must be nonzero as any vanishing linear combination of rows would result in a quadratic polynomial having three roots $a,b,c$. If these arguments are not convincing enough, just employ Gauss's algorithm: $$\begin{align}\det\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix} &=\det\begin{pmatrix}1&1&1\\0&b-a&c-a\\0&b^2-a^2&c^2-a^2\end{pmatrix}\\ &=\det\begin{pmatrix}1&1&1\\0&b-a&c-a\\0&0&c^2-a^2-(b+a)(c-a)\end{pmatrix}\\ &=(b-a)\cdot(c^2-a^2-(b+a)(c-a))\\ &=(b-a)(c-a)(c+a-(b+a))\\&=(b-a)(c-a)(c-b) \end{align} $$

Answer 2

The determinant of such a matrix is a third-degree homogeneous polynomial in $a,b,c$ that vanishes when $a=b,a=c$ or $b=c$, hence: $$ \det M = K (b-a)(c-a)(c-b) $$ for some constant $K$. To find the value of such a constant, it is sufficient to compute the determinant of $V$ for $(a,b,c)=(1,2,3)$, for instance.

Answer 3

Subtracting $\DeclareMathOperator{Col}{Col}\Col_1$ from $\Col_2$ and $\Col_3$ gives $$ \begin{pmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^2 & b^2-a^2 & c^2-a^2 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^2 & (b+a)(b-a) & (c+a)(c-a) \end{pmatrix} $$ The determinant is then \begin{align*} (b-a)(c+a)(c-a)-(c-a)(b+a)(b-a) &=(b-a)(c-a)(c+a-b-a) \\ &=(b-a)(c-a)(c-b) \end{align*} Of course, if you'd rather use row operations then you could consider the transpose and use the fact that $\det(A^\top)=\det(A)$.

Answer 4

Just expand over the first row. Call your matrix A.

|A| = $1(bc^2-cb^2) - 1(ac^2-ca^2) + 1(ab^2-ba^2)$

Show that it is equivalent to (b-a) etc.