Determinants and Cramer's Rule

Learning goals

Find $ad - bc$ appearing as the denominator in elimination
Compute a $2 \times 2$ determinant as main minus anti-diagonal
Apply Cramer's rule, replacing a column with the constants
Expand a $3 \times 3$ determinant by cofactors along a row
Tie a unique solution to $D \ne 0$
Choose row reduction for large systems instead

Solving the general two-variable system

Take the most general system of two linear equations in two unknowns. Write the coefficients as letters so the result works for every system at once:

\begin{aligned} ax + by &= e, \\ cx + dy &= f. \end{aligned}

Solve it by elimination, exactly as you always have. To clear $y$ , multiply the first equation by $d$ and the second by $b$ , so the $y$ terms match:

\begin{aligned} adx + bdy &= ed, \\ bcx + bdy &= bf. \end{aligned}

Subtract the second line from the first. The $bdy$ terms cancel, and $x$ is left with the coefficient $ad - bc$ :

(ad - bc)\,x = ed - bf.

Now do the mirror image to clear $x$ . Multiply the first equation by $c$ and the second by $a$ , then subtract the first from the second, and the same coefficient $ad - bc$ reappears in front of $y$ :

(ad - bc)\,y = af - ce.

Look at what these two results share. The quantity $ad - bc$ sits in front of both $x$ and $y$ , which means it becomes the denominator the moment you divide to solve:

x = \frac{ed - bf}{ad - bc}, \qquad y = \frac{af - ce}{ad - bc}.

One combination of the four coefficients controls both answers. If it is not zero, you can divide and the system has a clean solution. If it is zero, the division is illegal and something has gone wrong with the system itself. A number that important should not stay nameless.

The determinant and its bar notation

The shared quantity $ad - bc$ is the determinant of the four coefficients, arranged in the square grid in which they appear:

\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.

The straight bars are the determinant symbol. The rule they stand for is short: multiply down the main diagonal ( $a$ times $d$ ), then subtract the product up the other diagonal ( $b$ times $c$ ). We write the coefficient determinant of our system as $D$ , so $D = ad - bc$ .

The 2 by 2 determinant is the product down the main diagonal minus the product up the anti-diagonal.

Check your understanding

Evaluate the determinant $\begin{vmatrix} 2 & 5 \\ 3 & 4 \end{vmatrix}$ .

-7

7

23

-23

Cramer’s rule for a two-variable system

Look again at the answers we derived,

x = \frac{ed - bf}{ad - bc}, \qquad y = \frac{af - ce}{ad - bc},

and notice that every numerator is itself a determinant. The denominator is $D$ in both. The numerator for $x$ is $ed - bf$ , which is the determinant you get from $D$ by a single column swap. You take $D$ and replace its first column (the $x$ -column, $a$ and $c$ ) with the constants $e$ and $f$ . The numerator for $y$ is $af - ce$ , the determinant from replacing the second column (the $y$ -column, $b$ and $d$ ) with the same constants. Name these column-swapped determinants $D_x$ and $D_y$ :

D_x = \begin{vmatrix} e & b \\ f & d \end{vmatrix} = ed - bf, \qquad D_y = \begin{vmatrix} a & e \\ c & f \end{vmatrix} = af - ce.

With that naming, the whole solution collapses into a pattern worth memorizing. This is Cramer’s rule:

x = \frac{D_x}{D}, \qquad y = \frac{D_y}{D} \qquad (D \ne 0).

Cramer's rule builds each numerator by replacing one column of D with the constant column: the x-numerator swaps the x-column, the y-numerator swaps the y-column.

Why Cramer's rule gives the exact solution when $D \ne 0$ #

Look back at the elimination. Starting from the two original equations, honest algebra produced $D\,x = D_x$ and $D\,y = D_y$ , so every solution of the system must satisfy those two relations. When $D \ne 0$ we may divide, and the relations force

x = \frac{D_x}{D}, \qquad y = \frac{D_y}{D}.

No other values are possible, so the system has at most this one solution. It remains to check that this pair actually solves the system, since forcing the values does not by itself guarantee they work. Substitute them into the first equation and clear the denominator:

a\cdot\frac{D_x}{D} + b\cdot\frac{D_y}{D} = \frac{a(ed - bf) + b(af - ce)}{ad - bc} = \frac{aed - bce}{ad - bc} = \frac{e(ad - bc)}{ad - bc} = e.

The middle step used $-abf + baf = 0$ , so those terms drop and $e$ factors out of what remains. The same computation on $cx + dy$ returns $f$ . The pair therefore satisfies both equations, and it is the only pair that can, so it is the unique solution.

Worked example 1 Solve $3x + 2y = 16$ and $5x - 3y = 14$ by Cramer's rule

Read the coefficients straight off the equations. The coefficient determinant is

D = \begin{vmatrix} 3 & 2 \\ 5 & -3 \end{vmatrix} = (3)(-3) - (2)(5) = -9 - 10 = -19.

Since $D \ne 0$ , the system has a unique solution and Cramer’s rule applies. For $D_x$ , replace the $x$ -column with the constants $16$ and $14$ :

D_x = \begin{vmatrix} 16 & 2 \\ 14 & -3 \end{vmatrix} = (16)(-3) - (2)(14) = -48 - 28 = -76.

For $D_y$ , replace the $y$ -column with the constants instead:

D_y = \begin{vmatrix} 3 & 16 \\ 5 & 14 \end{vmatrix} = (3)(14) - (16)(5) = 42 - 80 = -38.

Now divide:

x = \frac{D_x}{D} = \frac{-76}{-19} = 4, \qquad y = \frac{D_y}{D} = \frac{-38}{-19} = 2.

The solution is $(x, y) = (4, 2)$ . A quick check confirms it: $3(4) + 2(2) = 16$ and $5(4) - 3(2) = 14$ . Notice $D$ was negative the whole way through, which never caused trouble, because a determinant is allowed to be negative.

The determinant test for a unique solution

The derivation quietly answered a deeper question than “what is the solution.” It told us exactly when a solution exists and is unique. Everything hinged on whether we were allowed to divide by $D$ , so $D$ is the switch that decides the whole character of the system.

A two-variable system has a unique solution exactly when $D \ne 0$ #

We prove both directions, since “exactly when” is a claim in both. Throughout, recall the two relations that hold for any solution: $D\,x = D_x$ and $D\,y = D_y$ .

First, if $D \ne 0$ , the previous proof already produced one and only one solution, $x = D_x/D$ and $y = D_y/D$ . So $D \ne 0$ gives a unique solution.

Now the other direction, in its contrapositive form: suppose $D = 0$ , and show the solution cannot be unique. With $D = 0$ the two relations read $0 = D_x$ and $0 = D_y$ , and both must hold for any solution to exist at all. Two cases cover it.

If either $D_x \ne 0$ or $D_y \ne 0$ , then one of the relations says $0 = (\text{a nonzero number})$ , which nothing can make true. The system then has no solution.

If instead $D_x = 0$ and $D_y = 0$ as well, look at the two equations directly, assuming each has a nonzero coefficient on at least one variable. Excluding that case costs nothing, since a degenerate equation like $0x + 0y = 1$ has no solution by itself, which is still not a unique solution. Since $D = 0$ , the coefficient rows $(a, b)$ and $(c, d)$ are proportional. Because $D_x = D_y = 0$ forces the constants to scale by that same factor as well, one equation is a constant multiple of the other. The two therefore impose at most one line’s worth of constraint, so the solution set is never a single point. In the consistent case that set is a whole line, giving infinitely many solutions.

Either way, a system with $D = 0$ has no solution or infinitely many, never exactly one. Combined with the first direction, a system has a unique solution precisely when $D \ne 0$ .

This is the tidy end of the one-none-infinitely-many trichotomy you met at the start of the chapter. The determinant is a single number that detects which case you are in:

$D \ne 0$ : the two lines cross at one point, so there is exactly one solution.
$D = 0$ : the lines are parallel (no solution) or the same line (infinitely many). The determinant cannot tell those two apart by itself, only that the system is degenerate.

Be careful with that last point, because it is the easiest thing to overstate. A zero determinant does not mean “no solution.” It means “not a unique solution,” which is either no solution or infinitely many. You need one more look at the constants to decide which.

Check your understanding

For which values of $k$ does the system $\begin{cases} kx + y = 1 \\ x + ky = 3 \end{cases}$ have exactly one solution?

k = 1

k = -1

k \ne 1

and

k \ne -1

C only

k = 0

D every real number

k

Determinants of three-variable systems

For three unknowns the coefficients form a $3 \times 3$ array, and its determinant is built out of $2 \times 2$ determinants. Expand along the first row: multiply each top entry by the $2 \times 2$ determinant left after crossing out that entry’s own row and column. Then attach the alternating signs to those products in order: plus, minus, plus.

\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a\begin{vmatrix} e & f \\ h & i \end{vmatrix} - b\begin{vmatrix} d & f \\ g & i \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix}.

Each small determinant is the minor of the entry in front of it. The middle term carries a minus sign, which is the single most common slip, so watch for it. This method is called cofactor expansion, and it reduces any $3 \times 3$ determinant to three easy $2 \times 2$ ones.

Worked example 2 Evaluate a 3 by 3 determinant by cofactor expansion

Expand along the first row of

\begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{vmatrix}.

Take the three top entries $1$ , $2$ , $3$ in turn, each with its minor and its sign:

1\begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3\begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}.

Evaluate each $2 \times 2$ determinant:

\begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} = -24, \qquad \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} = -20, \qquad \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} = -5.

Now combine with the signs, and mind the minus on the middle term:

1(-24) - 2(-20) + 3(-5) = -24 + 40 - 15 = 1.

The determinant equals $1$ .

Check your understanding

Evaluate $\begin{vmatrix} 2 & 1 & 0 \\ 3 & 2 & 1 \\ 1 & 4 & 2 \end{vmatrix}$ by expanding along the first row.

5

0

-5

-9

Cramer’s rule for a three-variable system

Cramer’s rule extends to three unknowns with no new ideas, only more columns. Let $D$ be the coefficient determinant of

\begin{aligned} a_1 x + b_1 y + c_1 z &= d_1, \\ a_2 x + b_2 y + c_2 z &= d_2, \\ a_3 x + b_3 y + c_3 z &= d_3. \end{aligned}

Form $D_x$ , $D_y$ , and $D_z$ by replacing the $x$ -, $y$ -, or $z$ -column with the constant column $d_1, d_2, d_3$ . Then, as long as $D \ne 0$ ,

x = \frac{D_x}{D}, \qquad y = \frac{D_y}{D}, \qquad z = \frac{D_z}{D}.

Worked example 3 Solve a 3 by 3 system with Cramer's rule

Solve

\begin{aligned} x + y + z &= 6, \\ 2x - y + z &= 6, \\ x + 2y - z &= 1. \end{aligned}

Start with the coefficient determinant, expanding along the first row:

D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = 1(1 - 2) - 1(-2 - 1) + 1(4 + 1) = -1 + 3 + 5 = 7.

Because $D = 7 \ne 0$ , the system has a unique solution. Replace the $x$ -column with the constants $6, 6, 1$ :

D_x = \begin{vmatrix} 6 & 1 & 1 \\ 6 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = 6(-1) - 1(-7) + 1(13) = -6 + 7 + 13 = 14.

Replace the $y$ -column instead:

D_y = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 6 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 1(-7) - 6(-3) + 1(-4) = -7 + 18 - 4 = 7.

And the $z$ -column:

D_z = \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 6 \\ 1 & 2 & 1 \end{vmatrix} = 1(-13) - 1(-4) + 6(5) = -13 + 4 + 30 = 21.

Divide each by $D$ :

x = \frac{14}{7} = 2, \qquad y = \frac{7}{7} = 1, \qquad z = \frac{21}{7} = 3.

The solution is $(x, y, z) = (2, 1, 3)$ , which checks in all three equations. Had you wanted only $z$ , you could have computed just $D$ and $D_z$ and skipped the other two determinants entirely.

Cramer’s rule or row reduction?

Cramer’s rule and the row reduction from the previous lesson solve the same systems, so which should you reach for? They have opposite strengths, and the honest answer is that it depends on what you want.

Cramer’s rule is a closed formula: it writes each unknown as one determinant over another, with no back-substitution and no bookkeeping of row swaps. That makes it excellent in two situations. First, when you need only one unknown, you compute just $D$ and that variable’s numerator and stop, while row reduction would make you solve for everything. Second, for theory, the formula shows plainly how the answer depends on the numbers in the system, which is why $D \ne 0$ reads off the unique-solution condition at a glance.

Row reduction wins on scale. A cofactor expansion of an $n \times n$ determinant fans out into smaller determinants and the work grows explosively with $n$ , and Cramer needs $n + 1$ of these determinants. Row reduction grows far more gently, which is why it, not Cramer, is what people and computers use to solve large systems.

Situation	Better tool	Why
You need only one of the unknowns	Cramer’s rule	Compute just $D$ and that one numerator
You want a formula in the coefficients	Cramer’s rule	Each unknown is a ratio of determinants
A full solve of a $2 \times 2$ or $3 \times 3$	Either one	The effort is about the same
A large system ( $4 \times 4$ and up)	Row reduction	Determinant cost grows explosively

Common mistakes

Reading the bars as absolute value. A determinant is a signed number and is frequently negative; $\begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = -5$ . The bars mean absolute value only around a single number, never around an array.
Reversing the diagonal subtraction. The $2 \times 2$ rule is $ad - bc$ , main diagonal minus anti-diagonal, not $bc - ad$ . Swapping the order flips the sign of every answer.
Forgetting the minus in the middle of a cofactor expansion. The first-row signs are plus, minus, plus. The middle term is subtracted; treating all three as additions is the classic $3 \times 3$ error.
Concluding $D = 0$ means no solution. A zero determinant means the solution is not unique, which is either no solution or infinitely many. Deciding which requires checking the constants.
Dividing the wrong way. Cramer’s rule is $x = D_x / D$ , the swapped-column determinant over the coefficient determinant, not $D / D_x$ .
Replacing the wrong column. For $D_x$ the constants go into the $x$ -column; for $D_y$ they go into the $y$ -column. Mixing them up silently swaps your answers.

Key takeaways

Solving the general system $ax + by = e$ , $cx + dy = f$ by elimination makes the quantity $ad - bc$ appear as the denominator of both unknowns. That quantity is the determinant $D$ .
The $2 \times 2$ determinant is $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$ , main diagonal minus anti-diagonal. The bars are not absolute value; a determinant can be negative.
Cramer’s rule: $x = D_x / D$ and $y = D_y / D$ , where $D_x$ and $D_y$ replace the $x$ - or $y$ -column of $D$ with the constants. The same pattern gives $x, y, z$ in the $3 \times 3$ case.
A $3 \times 3$ determinant is found by cofactor expansion along the first row, with signs plus, minus, plus.
A square system has a unique solution exactly when $D \ne 0$ (proved above for two variables; the same test holds for three). When $D = 0$ the system is degenerate, with either no solution or infinitely many, which ties the determinant to the chapter’s trichotomy.
Prefer Cramer’s rule for a single unknown or for theory; prefer row reduction for large systems.

Determinants and Cramer's Rule

Learning goals

Solving the general two-variable system

The determinant and its bar notation

Cramer’s rule for a two-variable system

The determinant test for a unique solution

Determinants of three-variable systems

Cramer’s rule for a three-variable system

Cramer’s rule or row reduction?

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)