Determinant of order 1 and 2

Order 1: the trivial base case. A $1\times 1$ matrix has exactly one entry, and its determinant is simply that entry, sign included.

where: $a$ is the single entry of the matrix. For example, $A=[-7]$ gives $|A|=-7$. This looks like nothing to compute, but it is the base case every recursive determinant expansion eventually bottoms out at.

Order 2: cross-multiply and subtract. For a $2\times2$ array of numbers, the determinant is defined directly, not by any deeper procedure — it is simply given a name.

where: $a,b$ are the entries of the first row and $c,d$ the entries of the second row, read left to right, top to bottom. In words: multiply along the diagonal you keep (top-left times bottom-right), multiply along the diagonal you cross out (top-right times bottom-left), and subtract the second product from the first.

Why the rows matter, not just the numbers. The quantity $ad-bc$ is comparing the two diagonal products of the array, and this has a geometric consequence worth noticing early: if the second row is a scalar multiple of the first, $(c,d) = k(a,b)$, then $ad-bc = a(kb) - b(ka) = 0$ always. Proportional rows force a zero determinant — the first hint that a determinant of zero signals a matrix that is singular (non-invertible), a theme that runs through the rest of the chapter.

Numbers versus algebraic entries. Nothing about the formula $ad-bc$ requires $a,b,c,d$ to be plain numbers — they can just as well be algebraic expressions in a variable, as long as you can still multiply and subtract them. This is exactly what makes determinants useful for solving equations: set up the $2\times2$ array with entries involving $x$, expand using $ad-bc$, and the result is an ordinary algebraic expression in $x$ that you can then set equal to a target value and solve. The worked examples below show both flavours — a purely numeric determinant and an algebraic one used to find an unknown.

Why speed matters here specifically. Every $3\times3$ determinant that appears later in this chapter is expanded, by minors and cofactors along a chosen row or column, into exactly three $2\times2$ determinants of this type, each multiplied by a coefficient and a sign. A single arithmetic slip inside one of those $2\times2$ pieces — a dropped minus sign, a swapped row — silently corrupts the entire $3\times3$ answer even if every other step of the expansion is done correctly. So the payoff of mastering $ad-bc$ now is not really about order-2 problems themselves; it is about removing the single most common source of error from every larger determinant calculation still to come.

Confirming that "diagonal product minus cross product" really is the right rule, and not just a name attached to two random operations, is worth doing once because the whole chapter leans on this single computation being both fast and correct — every later $3\times3$ expansion writes a $2\times2$ minor exactly in this form and applies exactly this formula, so any hesitation or sign slip here propagates straight through the harder problems later.

Evaluate $\begin{vmatrix} 5 & -2 \\ 3 & 4 \end{vmatrix}$, and then find $x$ if $\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix} = 3$.

Step 1. For the first determinant, read off $a=5,\ b=-2,\ c=3,\ d=4$.

Step 2. Apply $ad-bc$: $ad=(5)(4)=20$ and $bc=(-2)(3)=-6$, so $|A| = 20-(-6) = 20+6 = 26$.

Step 3. For the second determinant, $a=x,\ b=x+1,\ c=x-1,\ d=x$, so $|A| = x\cdot x - (x+1)(x-1) = x^2 - (x^2-1) = 1$.

Step 4. The determinant simplifies to the constant $1$ for every value of $x$ — the unknowns cancel out entirely. So the equation $|A|=3$ becomes $1=3$, which is never true: there is no value of $x$ that makes this determinant equal $3$; it is always $1$.

Competitive stretch

Change the array slightly and the outcome changes completely: solve $\begin{vmatrix} x & 2 \\ 3 & x \end{vmatrix} = 4$. Here $|A| = x^2 - 6$, so $x^2-6=4$ gives $x^2=10$, hence $\boxed{x=\pm\sqrt{10}}$. Comparing this with the example above is the real lesson: you cannot tell in advance whether the unknown will cancel out or survive — always expand and simplify fully before deciding what the equation is actually asking.

• Writing $ad+bc$ instead of $ad-bc$ — dropping the minus sign is the single most common slip in this topic.
• Misreading the array by columns instead of rows, so $a,b,c,d$ get assigned to the wrong positions.
• Treating $|[a]|$ as "nothing to compute" and forgetting to keep the sign of $a$.
• Sign errors when entries are negative, such as computing $-(-2)$ as $-2$ instead of $+2$.
• Trying to "solve for $x$" before fully simplifying the algebraic expression, and missing that the unknown has already cancelled.

Ask the tutor: Want a few more mixed numeric-and-algebraic $2\times2$ determinant problems to drill the sign rule before moving on to $3\times3$ expansions?