Determinants and Cramer's Rule: Free Response

The coefficient determinant takes the four coefficients in the positions they already occupy: the $x$-coefficients down the first column, the $y$-coefficients down the second.

$D = \begin{vmatrix} 4 & -1 \\ 3 & 2 \end{vmatrix} = (4)(2) - (-1)(3) = 8 + 3 = 11$

Two minus signs meet in that second product. The entry itself is $-1$, and the rule subtracts the anti-diagonal product, so together they contribute $+3$ rather than $-3$.

Since $D = 11$ is not zero, the determinant test settles the count on its own: the system has exactly one solution, and Cramer's rule is allowed to run. Nothing about the constants $-8$ and $5$ was needed to reach that.

$D_x$ replaces the $x$-column of $D$, the entries $4$ and $3$, with the constants $-8$ and $5$. The $y$-column is left exactly as it was.

$D_x = \begin{vmatrix} -8 & -1 \\ 5 & 2 \end{vmatrix} = (-8)(2) - (-1)(5) = -16 + 5 = -11$

$D_y$ replaces the $y$-column instead, and now it is the $x$-column that is left alone.

$D_y = \begin{vmatrix} 4 & -8 \\ 3 & 5 \end{vmatrix} = (4)(5) - (-8)(3) = 20 + 24 = 44$

Each swapped determinant goes on top, and $D$ underneath.

$x = \frac{-11}{11} = -1, \qquad y = \frac{44}{11} = 4$

A pair is worth reporting only once it has been checked in every equation it came from, not just the first one.

$4(-1) - (4) = -8, \qquad 3(-1) + 2(4) = -3 + 8 = 5$

Both original equations hold, so the solution is $(x, y) = (-1, 4)$.

Look at what each determinant is made of. $D$ is assembled from the four coefficients alone, one column of $x$-coefficients beside one column of $y$-coefficients, and no constant ever enters it.

$D = \begin{vmatrix} 4 & -1 \\ 3 & 2 \end{vmatrix} = 11 \quad \text{for every choice of the two constants}$

So $D$ cannot change. Each numerator is a different object: one of its columns has been replaced by the constants, so $D_x$ and $D_y$ generally move when the constants move. Generally, not always. A particular new pair could leave a numerator at its old value by coincidence, which is why the honest statement is that they CAN change while $D$ cannot.

The consequence is the useful part. The determinant test reads $D$ and nothing else, and $D = 11$ is nonzero for the whole family, so every system with these four coefficients has exactly one solution no matter what the two constants are. Geometrically, changing the constants slides the two lines without turning either of them, so the crossing point moves while the fact that they cross does not.

The coefficient determinant collects the three columns of coefficients and leaves the constants out.

$D = \begin{vmatrix} 1 & 2 & -1 \\ 3 & 1 & 2 \\ 2 & -1 & 1 \end{vmatrix}$

Expand along the first row. Each top entry is multiplied by the $2 \times 2$ determinant left after deleting that entry's own row and column, and the three signs run plus, minus, plus.

$1\begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} - 2\begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} + (-1)\begin{vmatrix} 3 & 1 \\ 2 & -1 \end{vmatrix}$

Evaluate the three minors on their own first.

$\begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} = 3, \qquad \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = -1, \qquad \begin{vmatrix} 3 & 1 \\ 2 & -1 \end{vmatrix} = -5$

Now combine them with their signs, minding the minus on the middle term.

$1(3) - 2(-1) + (-1)(-5) = 3 + 2 + 5 = 10$

Since $D = 10$ is not zero, the system has exactly one solution, so the rule is allowed to run.

The numerator for $z$ takes the coefficient determinant and replaces the $z$-column, the third one, with the constant column $-9$, $13$, $10$. The first two columns are untouched.

$D_z = \begin{vmatrix} 1 & 2 & -9 \\ 3 & 1 & 13 \\ 2 & -1 & 10 \end{vmatrix}$

Expand along the first row again, signs plus, minus, plus.

$1\begin{vmatrix} 1 & 13 \\ -1 & 10 \end{vmatrix} - 2\begin{vmatrix} 3 & 13 \\ 2 & 10 \end{vmatrix} + (-9)\begin{vmatrix} 3 & 1 \\ 2 & -1 \end{vmatrix}$

The three minors are $10 + 13 = 23$, then $30 - 26 = 4$, then $-3 - 2 = -5$.

$1(23) - 2(4) + (-9)(-5) = 23 - 8 + 45 = 60$

Divide the swapped determinant by the coefficient determinant.

$z = \frac{60}{10} = 6$

The numerators for $x$ and $y$ were never needed. Cramer's rule builds each unknown from its own numerator over the same denominator, so a request for one unknown costs two determinants rather than four.

The three first-row products are each top entry times its own minor, before any sign is attached.

$1(3) = 3, \qquad 2(-1) = -2, \qquad (-1)(-5) = 5$

Those three numbers are all that either version needs; the two totals disagree about nothing except how they are combined. The correct expansion alternates the signs, while the all-additions version adds every product as it stands.

$3 - (-2) + 5 = 10, \qquad 3 + (-2) + 5 = 6$

The gap is exactly twice the middle product, since that one term changed sides, and the wrong total is small and unremarkable: nothing about $6$ looks wrong on its own.

That is what makes this slip dangerous. The determinant test only asks whether the value is zero, and $6$ is not zero, so the wrong total still reports exactly one solution. The verdict survives, and it survives for the wrong reason. Every VALUE, though, is now wrong, because this number is the denominator under all three unknowns.

$z = \frac{60}{6} = 10$

A check against the original equations is what exposes it. With the wrong denominator the whole triple comes out as $x = \frac{5}{3}$, $y = -\frac{10}{3}$, $z = 10$, and the first equation returns

$\frac{5}{3} + 2\left(-\frac{10}{3}\right) - 10 = -5 - 10 = -15,$

not the required $-9$. Notice what that check costs. Part B skipped the numerators for $x$ and $y$ on purpose, and substituting into an original equation needs all three values, so speed and verifiability pull against each other here: buying the check back means computing the two numerators the fast route left out.

Substitution is also not the only thing that could have caught the slip. Expanding the same determinant along a different row or column, or solving the system by elimination, would each disagree with $6$. What substitution has going for it is that it is a reliable final check on the values actually reported, tested against the equations they came from, so it cannot repeat the sign convention that produced the error.

Name a variable for each count, since the counts are exactly what the club does not know. Let $x$ be the number of starter kits, $y$ the number of standard kits, and $z$ the number of deluxe kits.

Each total stated in the situation becomes one equation. The kits themselves come first: the three versions together make up all $12$ of them.

$x + y + z = 12$

Brackets next. A starter takes $1$, a standard $2$, a deluxe $2$, and the total came to $20$, so each per-kit rate multiplies its own count.

$x + 2y + 2z = 20$

Screw packs last, at $2$, $1$ and $3$ per kit respectively, totalling $22$.

$2x + y + 3z = 22$

The variables appear in the same order in all three equations and the constants sit alone on the right. That is what lets the coefficients be read straight down into the columns of a determinant without any rearranging.

Read the coefficients out of the three equations in the column order $x$, $y$, $z$.

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

The first-row minors used there are $2(3) - 2(1) = 4$, then $1(3) - 2(2) = -1$, then $1(1) - 2(2) = -3$. Since the determinant is not zero, the three counts are pinned down uniquely, which a physical build had better be.

Only $z$ is wanted, so replace the third column with the constants $12$, $20$, $22$ and leave the rest alone.

$D_z = \begin{vmatrix} 1 & 1 & 12 \\ 1 & 2 & 20 \\ 2 & 1 & 22 \end{vmatrix} = 1(24) - 1(-18) + 12(-3) = 24 + 18 - 36 = 6$

Here the minors are $2(22) - 20(1) = 24$, then $1(22) - 20(2) = -18$, then $1(1) - 2(2) = -3$.

$z = \frac{6}{2} = 3$

The answer is a count, so it should be reported as one: the club built $3$ deluxe kits, and $3$ lithium batteries are what the separate order needs.

Count what each method actually spends on the club's request.

Cramer's rule spent two determinants: the coefficient determinant, which had to be computed anyway to know the counts were pinned down, and the one swapped-column numerator. Each is a first-row expansion into three $2 \times 2$ minors, the original array is never altered, and the wanted count then falls out of one quotient.

$z = \frac{D_z}{D} \quad \text{needs} \quad D \ \text{and} \ D_z, \ \text{and nothing else}$

Row reduction is not forced to produce the other two counts here, and it is worth being exact about that rather than assuming the worst of it. The deluxe count is the LAST variable, so it sits at the bottom of the triangle and the forward pass alone delivers it. Subtract the first row of the augmented array from the second, and twice the first row from the third.

$R_2 - R_1 = [0,\ 1,\ 1 \mid 8], \qquad R_3 - 2R_1 = [0,\ -1,\ 1 \mid -2]$

Adding those two new rows clears the second column as well.

$[0,\ 1,\ 1 \mid 8] + [0,\ -1,\ 1 \mid -2] = [0,\ 0,\ 2 \mid 6]$

That last row reads $2z = 6$, so $z = 3$, with no starter count and no standard count computed at any point. Had the club needed the starter count instead, the climb back up through the other two would have been unavoidable, so which unknown is wanted decides how much of the reduction you can stop short of.

The honest contrast is therefore not that one method wastes answers. It is the shape of the work. Cramer's rule is a formula: two determinants, each evaluated on its own, with the uniqueness test included in the price. Row reduction is a procedure: every entry of the augmented array is rewritten step by step, each new row depending on the one before it, and in a system less friendly than this one those steps bring in fractions the determinants never see. For a $3 \times 3$ the two costs are close enough that this is a genuine judgement call.

The verdict does turn decisively in two ways. If the club wanted all three counts, Cramer's rule would need four $3 \times 3$ determinants, one denominator and three numerators, while a single row reduction delivers all three from one pass and a short climb back. And if the system grew, say to six equations in six unknowns, the cost of even one determinant grows explosively while a row reduction grows far more gently, so row reduction would be the sane choice even for a single unknown.

The honest summary is that Cramer's rule is at its cheapest in exactly the situation it was used in here: a small system, one unknown wanted, and a formula rather than a procedure to reach it.

The coefficients are $k$ and $3$ in the first equation, $3$ and $k$ in the second, so they go into the array in that order.

$D = \begin{vmatrix} k & 3 \\ 3 & k \end{vmatrix} = (k)(k) - (3)(3) = k^2 - 9$

The determinant test is a statement in both directions: exactly one solution precisely when the determinant is nonzero. So the values worth finding are the ones that make it zero, and everything else is settled by default.

$k^2 - 9 = 0 \;\Longrightarrow\; (k - 3)(k + 3) = 0 \;\Longrightarrow\; k = 3 \ \text{or} \ k = -3$

Those two values, and no others, are where a unique solution fails. For every other real $k$ the determinant is nonzero and the system has exactly one solution. What that solution IS still varies with $k$; only the count has been settled.

Replace one column at a time with the constant column, which is $6$ above $6$.

$D_x = \begin{vmatrix} 6 & 3 \\ 6 & k \end{vmatrix} = 6k - 18, \qquad D_y = \begin{vmatrix} k & 6 \\ 3 & 6 \end{vmatrix} = 6k - 18$

The two numerators come out equal, which is not a coincidence. Swapping $x$ with $y$ in this system turns the first equation into the second and the second into the first, so the system as a whole is unchanged, and the two unknowns must therefore behave identically.

Divide each numerator by the coefficient determinant and factor both parts before cancelling.

$x = y = \frac{6k - 18}{k^2 - 9} = \frac{6(k - 3)}{(k - 3)(k + 3)} = \frac{6}{k + 3}$

The cancellation of $k - 3$ is legal only because this part is restricted to values where the rule applies, and $k = 3$ is not among them: dividing by $k - 3$ requires that factor to be nonzero.

As a check, $k = 0$ turns the system into $3y = 6$ and $3x = 6$, giving $x = y = 2$, and the formula returns $\frac{6}{0 + 3} = 2$.

The larger of the two values is $k = 3$, and the simplified formula gives $\frac{6}{3 + 3} = 1$, so it offers the pair $(1, 1)$. Now put $k = 3$ into the original system instead.

$3x + 3y = 6, \qquad 3x + 3y = 6$

The two equations are identical, and each reduces to $x + y = 2$. The offered pair does satisfy it, since $1 + 1 = 2$, so the formula did not return nonsense. What it returned is not THE solution, though. It is merely one of them.

One equation in two unknowns leaves a whole line of pairs. Set $x = t$ for an arbitrary real $t$ and read $y$ off the equation.

$(x, y) = (t,\ 2 - t) \quad \text{for every real } t$

The family checks in both equations at once: $3t + 3(2 - t) = 6$ whatever $t$ is. The offered pair is the case $t = 1$, and $(0, 2)$ and $(5, -3)$ are just as valid.

The reason the formula still produced a number is the cancellation in part B. Dividing out $k - 3$ discarded the very information that numerator and denominator vanish together at $k = 3$: the true quotient there is $\frac{0}{0}$, which is not a value at all, and the simplified expression only looks as though it has an opinion.

Both values make the determinant zero, so the rule claims no solution at each of them. Test them one at a time, because the whole point is that they need not agree.

At $k = 3$, part C already did the work: both equations become $x + y = 2$ and the system has infinitely many solutions. A system with infinitely many solutions is not a system with no solution, so this one case refutes the rule as stated. A universal claim falls to a single counterexample.

At $k = -3$ the determinant is zero again and the outcome is completely different.

$-3x + 3y = 6, \qquad 3x - 3y = 6$

Add the two equations. Every variable term cancels, because the second equation's coefficients are exactly the negatives of the first's.

$0 = 12$

That is false whatever $x$ and $y$ are, so no pair satisfies both equations: at $k = -3$ the system genuinely has no solution. The rule is right here, which is precisely why a rule like it survives so long. It is right for the wrong reason, though. Nothing about the determinant produced the emptiness; the constants did, and a different pair of constants at the same $k$ would have produced a line of solutions instead.

So a zero determinant is compatible with none and with infinitely many, and it never chooses between them. What is true is a statement about the count being exactly one, and it holds in both directions.

$D \ne 0 \;\Longrightarrow\; \text{exactly one solution}, \qquad \text{exactly one solution} \;\Longrightarrow\; D \ne 0$

Equivalently, $D = 0$ says exactly that the number of solutions is not one, which leaves both of the remaining cases open until the constants are examined.

Substitute $x = 0$ and $y = 0$ into each equation and see what is left.

$a(0) + b(0) = 0, \qquad c(0) + d(0) = 0$

Both statements read $0 = 0$, which is true, and nothing in the check used any property of $a$, $b$, $c$ or $d$: any real coefficients whatsoever produce the same two lines. So this pair solves every homogeneous two-variable system, not merely some of them.

A linear system has exactly one of three outcomes: no solution, exactly one solution, or infinitely many. A solution has just been produced, so the first of the three is gone. A homogeneous system is always consistent, and only exactly one and infinitely many remain available to it.

That is the whole of what this part needed, and it is the fact the rest of the question turns on. The pair is often called the trivial solution, precisely because it costs nothing to find.

$D = \begin{vmatrix} 7 & 3 \\ 2 & 5 \end{vmatrix} = (7)(5) - (3)(2) = 35 - 6 = 29$

Since $29$ is not zero, the determinant test says the system has exactly one solution. That alone does not say WHICH one, and Cramer's rule would ordinarily be the next step. Here it barely needs to run, because every numerator inherits a column of zeros.

$D_x = \begin{vmatrix} 0 & 3 \\ 0 & 5 \end{vmatrix} = 0, \qquad D_y = \begin{vmatrix} 7 & 0 \\ 2 & 0 \end{vmatrix} = 0$

So $x = \frac{0}{29} = 0$ and $y = \frac{0}{29} = 0$. The shorter route reaches the same place: a solution is already in hand, and the determinant test guarantees there is only one, so the one in hand is it.

The complete solution set is the single pair $(0, 0)$, and nothing else. Note how the two facts had to be combined. The determinant supplied the count and the visible solution supplied the location; neither of them alone would have named the set.

$D = \begin{vmatrix} 6 & -9 \\ -4 & 6 \end{vmatrix} = (6)(6) - (-9)(-4) = 36 - 36 = 0$

So the solution is not unique. Part A has already ruled out no solution for a system of this kind, which leaves infinitely many as the only outcome available, and the job is now to describe them rather than to count them.

Divide the first equation by $3$ and the second by $-2$. Both collapse to the same relation.

$2x - 3y = 0 \quad \text{in both cases, that is} \quad 2x = 3y$

The two equations were multiples of one another all along, which is what the zero determinant was reporting. To name every pair satisfying $2x = 3y$, set $x = 3t$; then $2(3t) = 3y$ gives $y = 2t$.

$(x, y) = (3t,\ 2t) \quad \text{for every real } t$

Check the family in the original equations: $6(3t) - 9(2t) = 18t - 18t = 0$ and $-4(3t) + 6(2t) = -12t + 12t = 0$, for every $t$ at once. Taking $t = 0$ returns the trivial solution, and every other $t$ gives a different pair, so there really are infinitely many.

The family is not everything, though, which is worth checking: $(1, 1)$ gives $6 - 9 = -3$, not $0$, so it is not a solution.

The claim joins two implications, and an 'exactly when' is a promise about both, so each gets its own argument.

Direction 1: a zero determinant produces a solution other than the trivial one. The determinant test, read in the direction 'exactly one solution implies a nonzero determinant', says a system whose determinant is zero does not have exactly one solution. Its count is therefore none or infinitely many. Part A ruled out none for every homogeneous system.

$D = 0 \;\Longrightarrow\; \text{not exactly one} \;\Longrightarrow\; \text{infinitely many}$

Infinitely many solutions cannot all be the single pair $(0, 0)$, so at least one of them is a different pair. That is the first direction.

Direction 2: a solution other than the trivial one forces a zero determinant. Suppose some pair $(p, q)$ solves the system and is not $(0, 0)$. Part A says $(0, 0)$ solves it as well, so the system has at least two different solutions and therefore does not have exactly one.

$\text{at least two solutions} \;\Longrightarrow\; \text{not exactly one} \;\Longrightarrow\; D = 0$

The last step is the determinant test read the other way: a nonzero determinant would force exactly one solution, which has just been contradicted, so the determinant must be zero.

Both directions hold, so the two conditions are equivalent and the sharpened statement stands. Parts B and C are its two sides in the concrete: a nonzero determinant there left only the trivial pair, while a zero determinant produced a whole line of pairs beyond it.

One caution about how far this reaches. It works because a homogeneous system arrives with a solution already in hand, and that is what removes the no-solution branch. Put a nonzero constant anywhere in the system and the branch is back, so a zero determinant decides nothing on its own again.