How to Evaluate a Determinant Without Expanding It

A determinant turns a square matrix into a single number, and that number reveals a lot: whether the matrix has an inverse, whether a system of equations has a unique solution, and how areas change under a transformation.

This part covers evaluating determinants up to order 3, minors and cofactors, the properties of determinants, and problems solved with those properties.

**A $2\times 2$ determinant is $ad-bc$ for rows $(a,b)$ and $(c,d)$, and a $3\times 3$ determinant is expanded along any row or column as the sum of each entry times its cofactor, giving the same value whichever row or column is chosen.

Expansion along the first row:**

$$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=a_1(b_2{c}_3-b_3{c}_2)-b_1(a_2{c}_3-a_3{c}_2)+c_1(a_2{b}_3-a_3{b}_2)$$

Worked example (order 2). $\left|\begin{array}{cc}4& -3\\ 2& 5\end{array}\right|=20-(-6)=26$.

Worked example (order 3).

$$\left|\begin{array}{ccc}2& 1& 3\\ 0& 4& -1\\ 5& 2& 1\end{array}\right|=2(4+2)-1(0+5)+3(0-20)=12-5-60=-53$$

Expanding along the first column instead gives $2(4+2)-0+5(-1-12)=12-65=-53$, the same value.

An everyday example. Checking whether three price equations for three vegetables can be solved uniquely begins by evaluating the determinant of their coefficients.

The substance. Expand along the row or column with the most zeros — each zero removes a whole term from the work.

**The minor $M_{ij}$ of an element is the determinant left after deleting its row and column, and its cofactor is $C_{ij}=(-1{)}^{i+j}{M}_{ij}$, so cofactors follow a checkerboard pattern of signs starting with plus in the top-left corner.

Worked example.** For $\left|\begin{array}{ccc}2& 1& 3\\ 0& 4& -1\\ 5& 2& 1\end{array}\right|$:

- $M_{11}=\left|\begin{array}{cc}4& -1\\ 2& 1\end{array}\right|=6$, so $C_{11}=6$
- $M_{12}=\left|\begin{array}{cc}0& -1\\ 5& 1\end{array}\right|=5$, so $C_{12}=-5$
- $M_{13}=\left|\begin{array}{cc}0& 4\\ 5& 2\end{array}\right|=-20$, so $C_{13}=-20$
- $M_{23}=\left|\begin{array}{cc}2& 1\\ 5& 2\end{array}\right|=-1$, so $C_{23}=(-1{)}^5(-1)=1$

Two key facts:

- A row times its own cofactors gives the determinant: $2(6)+1(-5)+3(-20)=-53$
- A row times the cofactors of a different row gives 0: $0(6)+4(-5)+(-1)(-20)=0$

An everyday example. Covering one row and one column of a seating chart with your hands leaves a smaller block — exactly how a minor is formed.

The substance. A minor and its cofactor differ at most in sign — they are equal wherever $i+j$ is even.

A determinant is unchanged by transposing or by adding a multiple of one row to another, changes sign when two rows are swapped, is zero when two rows are identical or proportional, and is multiplied by k when one row is multiplied by k.

The properties, true for columns as well:

- $|A^T|=|A|$
- Swapping two rows changes the sign
- Two identical or proportional rows give 0
- Multiplying one row by k multiplies the determinant by k, so $|kA|=k^n|A|$ for order n
- $R_i\to R_i+k{R}_j$ leaves the value unchanged

Worked example. Evaluate $\left|\begin{array}{ccc}102& 18& 36\\ 1& 3& 4\\ 17& 3& 6\end{array}\right|$ without expanding. Row 1 is 6 times row 3, since $6\times 17=102$, $6\times 3=18$ and $6\times 6=36$. Proportional rows give a determinant of 0.

An everyday example. Swapping two rows of a data table of coefficients changes the sign of its determinant but never its size.

The substance. **$|kA|=k|A|$ is wrong for any matrix larger than $1\times 1$** — every row picks up the factor k.

Problems are solved by using row and column operations to create zeros or common factors, taking those factors outside, and reducing the determinant to a simple product before expanding.

Worked example. Prove that

$$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)(c-a)$$

- Apply $R_1\to R_1-R_2$ and then $R_2\to R_2-R_3$: the first two rows become $(0,a-b,a^2-b^2)$ and $(0,b-c,b^2-c^2)$
- Take out $(a-b)$ and $(b-c)$: those rows become $(0,1,a+b)$ and $(0,1,b+c)$
- Expand along the first column: $(a-b)(b-c)[(b+c)-(a+b)]=(a-b)(b-c)(c-a)$

Numerical check. With $a=1$, $b=2$ and $c=4$, the determinant is 6, and $(1-2)(2-4)(4-1)=6$.

An everyday example. Checking whether three boundary stones on a farm lie on one straight line reduces to showing that a determinant is zero, which these properties make quick.

The substance. Apply operations one after another, not simultaneously — replacing $R_1$ by $R_1-R_2$ and $R_2$ by $R_2-R_1$ at the same moment makes the new rows negatives of each other and wrongly gives zero.