Matrices and Systems of Equations: Free Response

Fix the column order before writing anything, because every row will follow it: column 1 for $x$, column 2 for $y$, column 3 for $z$, and the constants alone behind the bar.

The first equation has no $y$ term, so its $y$ column holds a $0$, and the third equation has no $x$ term, so its $x$ column holds a $0$. Those zeros are not filler. They keep every other entry in the column of the variable it multiplies, which is the only thing telling a later reader what the number $5$ in the last row is a coefficient of.

$\left[\begin{array}{ccc|c}3 & 0 & -1 & 7 \\ 1 & 4 & 2 & 1 \\ 0 & 5 & -1 & 6\end{array}\right]$

A variable written with no number in front of it has coefficient $1$, and a subtracted variable has coefficient $-1$, which is where the $1$ in row 2 and the two entries of $-1$ come from. Reading the first row back gives $3x + 0y - z = 7$, the equation it was built from.

Each row is one equation, each of the first three columns belongs to one variable, and the entry behind the bar is the right-hand side.

Row 1 has entries $2$, $0$ and $5$, so it says $2x + 0y + 5z = -4$, and the term carrying the zero coefficient can be dropped.

$2x + 5z = -4$

Row 2 has a $1$ in the $y$ column and a $-3$ in the $z$ column.

$y - 3z = 8$

Row 3 carries two zeros doing quite different jobs. The $0$ in the $z$ column says $z$ is absent from this equation; the $0$ behind the bar says the equation's right-hand side is the number zero. Confusing the two is exactly what the bar is drawn to prevent.

$x + y = 0$

A row of an augmented matrix records one equation by putting each coefficient in the column reserved for its variable. Nothing inside the row says which variable that is. The assignment lives in an agreement made before a single number is written down.

If the second student reserves column 1 for $z$ and column 3 for $x$, then each of their rows is the first student's row with those two entries exchanged, and the exchange happens in every row alike.

$[\,a \ \ b \ \ c \mid d\,] \quad \text{becomes} \quad [\,c \ \ b \ \ a \mid d\,]$

Decoding the second matrix under its own agreement puts $c$ back on $z$, $b$ back on $y$ and $a$ back on $x$, which is the same equation the first student encoded. So the two grids differ while the systems they stand for do not, and both encodings are faithful.

What neither grid carries is that agreement itself: which variable each column stands for, and that the last column holds constants rather than a fourth variable. That has to be stated alongside the numbers. It is why the bar is drawn at all, and why a matrix handed to someone else comes with its column order attached.

Hold the left-hand sides fixed and move only the numbers behind the bar. Both systems below have coefficient matrix $\left[\begin{array}{cc}1 & 3 \\ 2 & 7\end{array}\right]$.

Take the constants $5$ and $11$ first, and reduce with $R_2 \to R_2 - 2R_1$, applied to the constant as well: $11 - 2(5) = 1$.

$\left[\begin{array}{cc|c}1 & 3 & 5 \\ 0 & 1 & 1\end{array}\right]$

The bottom row says $y = 1$, so the top row gives $x = 5 - 3(1) = 2$. Checking in the second original equation, $2(2) + 7(1) = 11$.

Now keep every entry to the left of the bar and take the constants $5$ and $13$. The same operation sends the constant to $13 - 2(5) = 3$.

$\left[\begin{array}{cc|c}1 & 3 & 5 \\ 0 & 1 & 3\end{array}\right]$

Now $y = 3$ and $x = 5 - 3(3) = -4$. Checking, $2(-4) + 7(3) = -8 + 21 = 13$.

Two systems, every coefficient shared, and the solutions $(2, 1)$ and $(-4, 3)$ are different points. So the coefficient matrix cannot by itself determine a solution set, and the constant column is not decoration: it is the only place the difference between these two systems is recorded. That is why every row operation has to reach it.

Write the grid first, one row per equation, constants behind the bar.

$\left[\begin{array}{ccc|c}1 & 2 & 1 & 3 \\ 2 & 5 & 3 & 8 \\ 3 & 4 & 4 & 14\end{array}\right]$

The top left entry is already a $1$, a convenient pivot, so use it to clear the rest of column 1. Each operation runs across all four entries of its row, the constant included: $8 - 2(3) = 2$ in row 2, and $14 - 3(3) = 5$ in row 3.

$R_2 \to R_2 - 2R_1, \qquad R_3 \to R_3 - 3R_1$

$\left[\begin{array}{ccc|c}1 & 2 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & -2 & 1 & 5\end{array}\right]$

Row 2's leading $1$ is now the pivot for column 2, and the $-2$ below it clears by adding twice row 2. The constant follows the same multiplier: $5 + 2(2) = 9$.

$R_3 \to R_3 + 2R_2$

$\left[\begin{array}{ccc|c}1 & 2 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 3 & 9\end{array}\right]$

One scaling turns the last pivot into a $1$, and it is legal because $\tfrac{1}{3}$ is not zero.

$R_3 \to \tfrac{1}{3}R_3$

$\left[\begin{array}{ccc|c}1 & 2 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 3\end{array}\right]$

Each pivot sits strictly to the right of the one above it and every entry below a pivot is zero, so the matrix is in row-echelon form.

The bottom row of the echelon matrix has a single nonzero coefficient, so it is already an equation in one unknown.

$z = 3$

Climb one row. Row 2 reads $y + z = 2$, and $z$ is now known.

$y + 3 = 2 \;\Longrightarrow\; y = -1$

Climb again. Row 1 reads $x + 2y + z = 3$, and both other values are in hand.

$x + 2(-1) + 3 = 3 \;\Longrightarrow\; x = 2$

Now check the triple $(2, -1, 3)$ in the equations as they were given, not in the reduced ones. The reduced matrix is where the triple came from, so checking against it cannot catch a slip made during the reduction: a wrong reduced matrix and the triple climbed out of it agree with each other by construction. The original equations were never part of that chain, so they can catch such a slip.

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

$2(2) + 5(-1) + 3(3) = 4 - 5 + 9 = 8$

$3(2) + 4(-1) + 4(3) = 6 - 4 + 12 = 14$

All three balance, so the triple solves the system that was actually asked about.

Start with the climb. Row-echelon form puts each row's pivot strictly to the right of the pivot above it, so every entry below a pivot is zero. Reading a row back as an equation, its variables are the one in its own pivot column together with any in columns further right. When every variable column carries a pivot, as it does here, those further-right columns are exactly the pivot columns of the rows below, so their variables are already in hand by the time the climb reaches this row.

$[\,0 \ \ 0 \ \ 1 \mid 3\,] \;\longrightarrow\; z = 3$

So the bottom nonzero row mentions one unknown and settles it outright. Moving up one row brings in exactly one column that was not available before, so that row is again an equation in one unknown once the known values are substituted. That is what makes the climb never stall, and it is also why it cannot start at the top: the first row mentions all three unknowns at once and settles none of them.

The pivot in every variable column is doing real work in that argument. Take it away and a bottom row reading $y - z = 4$ still holds two unknowns, so zeros below the pivots are not on their own enough. The general rule keeps the shape but needs one preliminary step: give a parameter to each variable whose column carries no pivot, and then every pivot row determines exactly one unknown on the way up, namely its own pivot variable.

Now the downward pass. A stage picks a pivot column, secures a nonzero entry in the pivot position, swapping a lower row up if the position holds a zero, or moving on to the next column when no lower row has a nonzero entry there either, and then clears every entry below whatever pivot it secured. What has been settled is never disturbed afterwards, because every later operation adds a multiple of a row that already carries zeros in the cleared columns, and adding zeros puts nothing back.

$\text{stage } 1 \to \text{column } 1, \qquad \text{stage } 2 \to \text{column } 2, \qquad \text{stage } 3 \to \text{column } 3$

The quantity that advances is the pivot column the pass has reached. Each stage moves it at least one column to the right, and it can never move past the last variable column, so a system in $n$ variables is finished in at most $n$ stages, each of them a fixed, finite amount of arithmetic. That bound, rather than experience with small examples, is what makes row reduction something you could hand to a machine and know it would stop.

A row operation is one instruction carried out on a whole row. The row is $R_2 = [\,2 \ \ 5 \mid 13\,]$, the instruction is to subtract twice $R_1 = [\,1 \ \ 3 \mid 8\,]$ from it, and there are three entries to subtract from, not two.

$2 - 2(1) = 0, \qquad 5 - 2(3) = -1, \qquad 13 - 2(8) = -3$

Devon's first two entries, $0$ and $-1$, match. His third is the $13$ he started with, unchanged, so the constant is the entry the operation never reached.

$[\,0 \ \ -1 \mid -3\,]$

Applied properly, the operation is reversible, and its inverse is the same move with the opposite sign: $R_2 \to R_2 + 2R_1$ adds back exactly what was taken away and returns the matrix to the one he started with. That reversibility is what guarantees a correctly applied step neither gains nor loses solutions.

Take the reported matrix at face value first.

$\left[\begin{array}{cc|c}1 & 3 & 8 \\ 0 & -1 & 13\end{array}\right]$

Its bottom row says $-y = 13$, so $y = -13$, and the top row then gives

$x + 3(-13) = 8 \;\Longrightarrow\; x = 47.$

Test the pair $(47, -13)$ in the two equations as they were originally written.

$47 + 3(-13) = 47 - 39 = 8$

$2(47) + 5(-13) = 94 - 65 = 29$

The first equation is satisfied exactly. The second demands $13$ and gets $29$, so the pair is not a solution of the system, and one test out of two was enough to say so.

Now reduce correctly, carrying the constant along, so the bottom right entry is $13 - 2(8) = -3$.

$\left[\begin{array}{cc|c}1 & 3 & 8 \\ 0 & -1 & -3\end{array}\right]$

The bottom row gives $-y = -3$, so $y = 3$, and the top row gives $x + 3(3) = 8$, so $x = -1$. The pair $(-1, 3)$ satisfies both originals: $-1 + 9 = 8$ and $-2 + 15 = 13$.

Devon's operation rewrote row 2 and left row 1 exactly as it was. Whatever the new second row says, the top row is still the equation $x + 3y = 8$, and the climb reads $x$ off that very row, so any pair produced this way satisfies the first equation automatically. The first equation was never at risk and cannot detect the slip. Only the row that was rewritten can disagree with the system it came from, which is why the failure surfaces at the second equation and nowhere else.

Now ask what the reported row actually asserts.

$-y = 13$

A legal step replaces $E_2$ by a combination $aE_1 + bE_2$, so ask which combination could put $-y$ alone on the left. Matching the $x$ terms needs $a + 2b = 0$ and matching the $y$ terms needs $3a + 5b = -1$, and those two conditions force $a = -2$ and $b = 1$. There is exactly one candidate, namely $E_2 - 2E_1$, and its right-hand side is

$13 - 2(8) = -3.$

So no combination of the two original equations yields $-y = 13$. Devon's row is not a consequence of his system at all, which is precisely why a pair satisfying it need not satisfy the system, and why the check in part B failed. The moral is one line long: the multiplier that scales a row's coefficients scales its constant by the same amount, because the two are halves of a single equation.

Name the unknowns as the stem does: $a$, $b$ and $c$ kilograms of the three beans. Each condition becomes one equation.

The batch weight adds the three amounts.

$a + b + c = 30$

The cost multiplies each amount by its own price per kilogram and adds the results.

$9a + 12b + 15c = 348$

The last condition compares two of the amounts and never mentions bean A.

$b - c = 2$

Encoding with the columns in the order $a$, $b$, $c$, the third row needs a $0$ in the $a$ column. Bean A is absent from that condition, and the column has to stay open so that the $1$ and the $-1$ land under $b$ and $c$ rather than sliding left.

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 30 \\ 9 & 12 & 15 & 348 \\ 0 & 1 & -1 & 2\end{array}\right]$

Scaling the cost row first keeps the arithmetic small, and the move is legal because $\tfrac{1}{3}$ is not zero.

$R_2 \to \tfrac{1}{3}R_2 \;\Longrightarrow\; [\,3 \ \ 4 \ \ 5 \mid 116\,]$

Now clear column 1 with the top pivot, constant included: $116 - 3(30) = 26$.

$R_2 \to R_2 - 3R_1 \;\Longrightarrow\; [\,0 \ \ 1 \ \ 2 \mid 26\,]$

Column 2's pivot is that new $1$, and the $1$ below it in row 3 clears by subtraction, with $2 - 26 = -24$ behind the bar.

$R_3 \to R_3 - R_2 \;\Longrightarrow\; [\,0 \ \ 0 \ \ -3 \mid -24\,]$

One more scaling makes the last pivot a $1$.

$R_3 \to -\tfrac{1}{3}R_3 \;\Longrightarrow\; [\,0 \ \ 0 \ \ 1 \mid 8\,]$

The matrix is triangular.

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 30 \\ 0 & 1 & 2 & 26 \\ 0 & 0 & 1 & 8\end{array}\right]$

Back-substitute. The bottom row gives $c = 8$. Row 2 reads $b + 2c = 26$, so $b = 26 - 16 = 10$. Row 1 reads $a + b + c = 30$, so $a = 30 - 10 - 8 = 12$.

Check against the conditions as stated, not against the reduced matrix. The weights give $12 + 10 + 8 = 30$ kilograms. The cost gives $9(12) + 12(10) + 15(8) = 108 + 120 + 120 = 348$ dollars. And bean B beats bean C by $10 - 8 = 2$ kilograms.

Write the new matrix beside the old one. The three prices are unchanged, the weight condition still adds the three amounts, and the rule relating beans B and C is untouched, so every entry to the left of the bar is identical. Only the numbers behind it move, and the third one does not even do that.

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 45 \\ 9 & 12 & 15 & 510 \\ 0 & 1 & -1 & 2\end{array}\right]$

Every multiplier used in part B was chosen by looking at coefficient entries: $\tfrac{1}{3}$ because the cost row divides by $3$, then $-3$ to knock out the $3$ sitting under the leading $1$, then $-1$ to knock out the $1$ under the second pivot, and finally $-\tfrac{1}{3}$ because the last pivot had come out as $-3$. Not one of those four choices consulted a constant, so the same four operations apply unchanged, and only the constant column has any new arithmetic in it.

$R_2 \to \tfrac{1}{3}R_2 \;\Longrightarrow\; [\,3 \ \ 4 \ \ 5 \mid 170\,]$

$R_2 \to R_2 - 3R_1 \;\Longrightarrow\; [\,0 \ \ 1 \ \ 2 \mid 35\,]$

$R_3 \to R_3 - R_2 \;\Longrightarrow\; [\,0 \ \ 0 \ \ -3 \mid -33\,]$

$R_3 \to -\tfrac{1}{3}R_3 \;\Longrightarrow\; [\,0 \ \ 0 \ \ 1 \mid 11\,]$

Back-substituting gives $c = 11$, then $b = 35 - 2(11) = 13$, then $a = 45 - 13 - 11 = 21$.

Checking against next month's order: $21 + 13 + 11 = 45$ kilograms, $9(21) + 12(13) + 15(11) = 189 + 156 + 165 = 510$ dollars, and $13 - 11 = 2$ kilograms.

There is no guarantee. Run the recorded sequence on the coefficients with bean B at $9$ dollars, so the cost row starts as $[\,9 \ \ 9 \ \ 15\,]$.

$R_2 \to \tfrac{1}{3}R_2 \;\Longrightarrow\; [\,3 \ \ 3 \ \ 5\,]$

$R_2 \to R_2 - 3R_1 \;\Longrightarrow\; [\,0 \ \ 0 \ \ 2\,]$

The multiplier $-3$ was chosen to clear the entry under the leading $1$, and it still clears that entry, but this time it clears the second column too, so the pivot the third operation was counting on is not there. Carrying on regardless, $R_3 \to R_3 - R_2$ gives $[\,0 \ \ 1 \ \ -3\,]$ and $R_3 \to -\tfrac{1}{3}R_3$ gives $[\,0 \ \ -\tfrac{1}{3} \ \ 1\,]$, leaving the coefficient part as

$\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & -\tfrac{1}{3} & 1\end{array}\right]$

Row 3's leading entry sits in column 2, to the left of row 2's leading entry in column 3, so this is not row-echelon form. Swapping the last two rows repairs it, but a swap is a fifth operation, not one of the four that were recorded.

Be careful which way this cuts. Leaving every coefficient alone is enough to guarantee the recorded sequence still reduces the matrix, which is what part C relied on. The converse is false: a coefficient can change and leave the sequence perfectly good. Price bean C at $18$ dollars instead, and the same four operations give the rows $[\,1 \ \ 1 \ \ 1\,]$, $[\,0 \ \ 1 \ \ 3\,]$ and $[\,0 \ \ 0 \ \ \tfrac{4}{3}\,]$, which is still triangular. So the safe statement runs one way only: leave the coefficients alone and the sequence is guaranteed, change one and it has to be checked.

The third equation has no $z$ term, so its $z$ column holds a $0$.

$\left[\begin{array}{ccc|c}1 & 2 & -1 & 3 \\ 2 & 5 & 1 & 7 \\ 3 & 7 & 0 & 10\end{array}\right]$

Clear column 1 with the top pivot, carrying the constants: $7 - 2(3) = 1$ and $10 - 3(3) = 1$.

$R_2 \to R_2 - 2R_1, \qquad R_3 \to R_3 - 3R_1$

$\left[\begin{array}{ccc|c}1 & 2 & -1 & 3 \\ 0 & 1 & 3 & 1 \\ 0 & 1 & 3 & 1\end{array}\right]$

Rows 2 and 3 have come out identical, constants included, so clearing column 2 empties the last row completely.

$R_3 \to R_3 - R_2$

$\left[\begin{array}{ccc|c}1 & 2 & -1 & 3 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 0 & 0\end{array}\right]$

The $x$ column carries a pivot in row 1 and the $y$ column carries one in row 2. The $z$ column carries none, so two of the three variable columns have pivots. Counting by column rather than by row is what matters here, since the count of pivots is what will decide the shape of the answer.