C9 - Factorising determinants

C9 - Factorising determinants

Expanding a determinant immediately can hide its structure. You will use row and column operations to create zeros, keep track of changes to the determinant and expose polynomial factors before expanding. The aim is to make the calculation shorter while preserving exactly what the determinant equals.

Row and column replacement

Can you factorise a determinant without first expanding a cubic? A good first move is often to subtract one row from another. Similar-looking rows can then reveal a common factor. We start with numerical matrices so that you can see which operations preserve the determinant.

Write RiR_i for row ii and CjC_j for column jj. The instruction R2←R2−2R1R_2\leftarrow R_2-2R_1 means replace row 2 by “old row 2 minus twice row 1”. The other rows stay as they were.

Replacement leaves the determinant unchanged

Adding a multiple of one row to a different row leaves the determinant unchanged:

Ri←Ri+kRj(i≠j)R_i\leftarrow R_i+kR_j\qquad(i\ne j)

The corresponding column operation Ci←Ci+kCjC_i\leftarrow C_i+kC_j also leaves it unchanged. The multiplier kk may be a number or an algebraic expression. These rules are not in the formula booklet.

Why does this work? A determinant is linear in each individual row when the others are fixed. After the replacement, its expansion is the old determinant plus kk times a determinant with two equal rows. The latter is zero: swapping those equal rows both leaves it unchanged and reverses its sign. The same argument works for columns.

Worked example 1: creating a simple first column

Question. Find D=∣123257113∣D=\begin{vmatrix}1&2&3\\2&5&7\\1&1&3\end{vmatrix} using row operations.

Plan. Subtract multiples of row 1 to make the other entries in column 1 zero, then expand down that column.

D=∣1230110−10∣R2←R2−2R1, R3←R3−R1=1∣11−10∣expand down column 1=0−(−1)=1\begin{aligned} D&=\begin{vmatrix}1&2&3\\0&1&1\\0&-1&0\end{vmatrix} && R_2\leftarrow R_2-2R_1,\ R_3\leftarrow R_3-R_1\\ &=1\begin{vmatrix}1&1\\-1&0\end{vmatrix} && \text{expand down column 1}\\ &=0-(-1)=1 \end{aligned}

Answer. D=1D=1.

Check. Direct top-row expansion of the original gives 1(15−7)−2(6−7)+3(2−5)=8+2−9=11(15-7)-2(6-7)+3(2-5)=8+2-9=1.

Keep the original matrix and change only the route: use columns instead of rows. Row and column operations offer the same kind of simplification.

Worked example 2: creating a simple first row

Question. Find the same determinant using column operations.

Plan. Subtract multiples of column 1 to make the other entries in row 1 zero.

D=∣1002111−10∣C2←C2−2C1, C3←C3−3C1=1∣11−10∣expand across row 1=1\begin{aligned} D&=\begin{vmatrix}1&0&0\\2&1&1\\1&-1&0\end{vmatrix} && C_2\leftarrow C_2-2C_1,\ C_3\leftarrow C_3-3C_1\\ &=1\begin{vmatrix}1&1\\-1&0\end{vmatrix} && \text{expand across row 1}\\ &=1 \end{aligned}

Answer. D=1D=1.

Check. Expanding the original down column 3 gives 3(2−5)−7(1−2)+3(5−4)=−9+7+3=13(2-5)-7(1-2)+3(5-4)=-9+7+3=1.

The equality signs refer to determinants, not to equal matrices. Replacing a row changes the matrix even when it preserves the determinant. If several operations are written together, make clear which rows or columns are unchanged; above, every operation uses the unchanged first row or column.

Scaling and sign changes

Not every convenient operation preserves the determinant. A row replacement, a row scaling and a row swap have different effects. Keeping an equality true means recording each effect.

Determinant changes

For a square matrix of order nn:

  • swapping two rows reverses the determinant's sign;
  • multiplying one row by kk multiplies the determinant by kk;
  • the same rules hold for columns;
  • multiplying the whole matrix by kk gives det⁡(kA)=kndet⁡A\det(k\mathbf{A})=k^n\det\mathbf{A}, since all nn rows are scaled.

These rules are not in the formula booklet.

Linearity in one row explains the single-row factor. Swapping rows reverses the alternating signs in the determinant expansion. In particular, equal rows force a determinant to be zero; proportional rows do too. A zero row makes every term of an expansion along that row zero. These statements also hold for columns.

Worked example 3: scaling one row

Question. Given A=(1235)\mathbf{A}=\begin{pmatrix}1&2\\3&5\end{pmatrix}, find the determinant of B=(2435)\mathbf{B}=\begin{pmatrix}2&4\\3&5\end{pmatrix}.

Plan. Find det⁡A\det\mathbf{A}, then account for doubling only its first row.

det⁡A=1(5)−2(3)=−1det⁡B=2det⁡A=−2one row was doubled\begin{aligned} \det\mathbf{A}&=1(5)-2(3)=-1\\ \det\mathbf{B}&=2\det\mathbf{A}=-2 && \text{one row was doubled} \end{aligned}

Answer. −2-2.

Check. Direct evaluation gives 2(5)−4(3)=10−12=−22(5)-4(3)=10-12=-2.

Now change only the row order in B\mathbf{B}. Its entries and scale stay fixed, but the determinant changes sign.

Worked example 4: swapping two rows

Question. Find the determinant of C=(3524)\mathbf{C}=\begin{pmatrix}3&5\\2&4\end{pmatrix}, using B\mathbf{B} above.

Plan. Identify one row swap and negate the previous determinant.

det⁡C=−det⁡B=−(−2)=2\det\mathbf{C}=-\det\mathbf{B}=-(-2)=2

Answer. 22.

Check. Direct evaluation gives 3(4)−5(2)=12−10=23(4)-5(2)=12-10=2.

When you extract a common factor from a row, write it outside the determinant. For example,

∣2a2bcd∣=2∣abcd∣\begin{vmatrix}2a&2b\\c&d\end{vmatrix} =2\begin{vmatrix}a&b\\c&d\end{vmatrix}

Extracting a 2 from each of three rows gives 23=82^3=8, not 2. Also, R2←2R2+R1R_2\leftarrow2R_2+R_1 doubles the determinant: first double row 2, then add row 1. It is not a determinant-preserving replacement of the form R2+kR1R_2+kR_1.

Watch out

“Divide a row by x−1x-1” is unsafe at x=1x=1. Instead, factor each entry algebraically and extract the common factor. The resulting polynomial identity then remains valid even where that factor is zero.

Track a scaling followed by a swap.

Exposing polynomial factors

Polynomial entries make row and column operations especially useful. Rather than expand everything, create a row or column whose entries share a factor. Extract it, simplify again and leave the result factorised.

Worked example 5: extracting a row factor

Question. Factorise fully

D(x)=∣1111xx2124∣D(x)=\begin{vmatrix}1&1&1\\1&x&x^2\\1&2&4\end{vmatrix}

Plan. Subtract row 1 from rows 2 and 3, extract x−1x-1 from row 2, then expand.

D(x)=∣1110x−1x2−1013∣R2←R2−R1, R3←R3−R1=(x−1)∣11101x+1013∣x2−1=(x−1)(x+1)=(x−1){3−(x+1)}expand down column 1=(x−1)(2−x)\begin{aligned} D(x)&=\begin{vmatrix}1&1&1\\0&x-1&x^2-1\\0&1&3\end{vmatrix} && R_2\leftarrow R_2-R_1,\ R_3\leftarrow R_3-R_1\\ &=(x-1)\begin{vmatrix}1&1&1\\0&1&x+1\\0&1&3\end{vmatrix} && x^2-1=(x-1)(x+1)\\ &=(x-1)\{3-(x+1)\} && \text{expand down column 1}\\ &=(x-1)(2-x) \end{aligned}

Answer. D(x)=(x−1)(2−x)=−(x−1)(x−2)D(x)=(x-1)(2-x)=-(x-1)(x-2).

Check. At x=1x=1 or x=2x=2, two original rows coincide, giving zero as the factors predict. Direct expansion gives (4x−2x2)−(4−x2)+(2−x)=−x2+3x−2(4x-2x^2)-(4-x^2)+(2-x)=-x^2+3x-2, confirming both factors and the leading sign.

Transpose the arrangement: the former rows now appear as columns. This changes only the direction of the operations, and shows how column factorisation works.

Worked example 6: extracting a column factor

Question. Factorise fully

E(x)=∣1111x21x24∣E(x)=\begin{vmatrix}1&1&1\\1&x&2\\1&x^2&4\end{vmatrix}

Plan. Subtract column 1 from columns 2 and 3, then extract x−1x-1 from column 2.

E(x)=∣1001x−111x2−13∣C2←C2−C1, C3←C3−C1=(x−1)∣1001111x+13∣extract the common column factor=(x−1){3−(x+1)}expand across row 1=(x−1)(2−x)\begin{aligned} E(x)&=\begin{vmatrix}1&0&0\\1&x-1&1\\1&x^2-1&3\end{vmatrix} && C_2\leftarrow C_2-C_1,\ C_3\leftarrow C_3-C_1\\ &=(x-1)\begin{vmatrix}1&0&0\\1&1&1\\1&x+1&3\end{vmatrix} && \text{extract the common column factor}\\ &=(x-1)\{3-(x+1)\} && \text{expand across row 1}\\ &=(x-1)(2-x) \end{aligned}

Answer. E(x)=−(x−1)(x−2)E(x)=-(x-1)(x-2).

Check. Direct expansion of the original gives (4x−2x2)−(4−2)+(x2−x)=−x2+3x−2(4x-2x^2)-(4-2)+(x^2-x)=-x^2+3x-2. It agrees with the factorisation, including at x=1,2x=1,2.

No division by x−1x-1 was used: the equalities are polynomial identities for every real xx. The opening question is answered: row and column operations can reveal factors before a full expansion. They also explain why particular values make the determinant zero: two rows or columns become dependent.

Quick Recap

  • Adding a multiple of one row or column to another preserves a determinant.
  • A swap reverses its sign; scaling one row or column supplies one factor.
  • Scaling an entire order-nn matrix by kk supplies knk^n.
  • Create zeros and extract common factors before expanding polynomial determinants.

A determinant's factors can be visible in its rows or columns long before they appear in an expanded polynomial.