Factorising determinantsAQA A-Level Further Maths: Revision notes

For a $3\times3$ matrix the determinant can be expanded along any row or column, using the alternating sign pattern $+\ -\ +$ / $-\ +\ -$ / $+\ -\ +$: $\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31}).$ Expanding along a row or column that contains zeros saves work, which is why row and column operations are used to create zeros before expanding.

These rules apply equally to rows and columns:

• Adding a multiple of one row (column) to another leaves the determinant unchanged, e.g. $R_2\to R_2-3R_1$ or $C_1\to C_1+C_2+C_3$.
• Swapping two rows (columns) changes the sign of the determinant.
• Multiplying one row (column) by $k$ multiplies the determinant by $k$. Read in reverse, a common factor of a row or column can be taken outside the determinant.
• A determinant with two identical rows (columns), or one row a multiple of another, is zero.
• A row or column of zeros gives determinant zero. The first type is the one you use to simplify, because it changes the entries without changing the value.

To factorise a determinant containing a variable $x$, use operations that make a whole row or column share a factor.

1. Look for rows or columns that add up to the same expression, such as $x+2$ or $x+6$. Replace one column by the sum of all of them, then take the factor out.
2. Or subtract one row (column) from another to create zeros and common factors, as in $b^2-a^2=(b-a)(b+a)$.
3. After taking out the factors, create zeros in a row or column of what is left and expand. Write every operation next to the line it produces, so each step can be followed.

If substituting $x=k$ makes two rows or two columns identical (or proportional), the determinant is zero at $x=k$, so by the factor theorem $(x-k)$ is a factor. Example: in $\begin{vmatrix} 1 & 1 & 1 \\ x & 2 & 3 \\ x^2 & 4 & 9 \end{vmatrix}$, columns 1 and 2 are identical when $x=2$, and columns 1 and 3 are identical when $x=3$. So $(x-2)$ and $(x-3)$ are both factors. The determinant is quadratic in $x$, so it equals $k(x-2)(x-3)$, and the coefficient of $x^2$, found by expanding along the first column, is $1\times(3-2)=1$, so $k=1$.

Factorise $\Delta=\begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$. $C_1\to C_1+C_2+C_3$ gives a first column of $x+2$ in every row, so $\Delta=(x+2)\begin{vmatrix} 1 & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}.$ Now $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$: $\Delta=(x+2)\begin{vmatrix} 1 & 1 & 1 \\ 0 & x-1 & 0 \\ 0 & 0 & x-1 \end{vmatrix}=(x+2)(x-1)^2.$ Check: at $x=1$ all entries are 1, so $\Delta=0$, matching the repeated factor $(x-1)$.

Show that $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix}=(a-b)(b-c)(c-a)$. $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$ put zeros in the first row. Taking out $(b-a)$ and $(c-a)$ leaves $\begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 1 \\ a^2 & b+a & c+a \end{vmatrix}=(c+a)-(b+a)=c-b$. So the determinant is $(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)$. The result is zero whenever $a=b$, $b=c$ or $c=a$, because two columns would then be identical. That is a quick check of the factors.