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Matrices and Systems of Equations: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra II. You can skip it.

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Problem 1 of 10

Problem 1 Reordered variables

Write the augmented matrix of $2z-x=7$ , $y=3x+1$ , and $z-y=4$ , using columns in the order $x,y,z$ .
1. Hint 1
  Put all variable terms on the left in the named order.
2. Hint 2
  A missing variable still needs a zero in its column.
Answer

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

Full solution

Rewrite the equations as $-x+2z=7$ , $-3x+y=1$ , and $-y+z=4$ .
The constant column stays on the right.
$\left[\begin{array}{ccc|c} -1 & 0 & 2 & 7 \\ -3 & 1 & 0 & 1 \\ 0 & -1 & 1 & 4 \end{array}\right]$
Each zero keeps the absent variable in its correct place.

Answer

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

Key idea

Keep a fixed variable order when translating equations into matrix rows.
Problem 1: Mark complete
Problem 2 An encoded equation

The row $\left[\begin{array}{ccc|c}2&-1&0&6\end{array}\right]$ uses variable order $p,q,r$ . Write the equation it records.
1. Hint 1
  Each coefficient multiplies the variable in its named column.
2. Hint 2
  The entry after the bar is the right side.
Answer

$2p-q=6$ .

Full solution

The row records $2p-q+0r=6$ .
Removing the zero term gives
$2p-q=6$
The absent $r$ term does not shift the $q$ coefficient into another column.

Answer

$2p-q=6$ .

Key idea

A zero entry records an absent variable without changing column order.
Problem 2: Mark complete
Problem 3 Scaling a record

Replace $R_1$ by $-\frac12R_1$ in the augmented matrix shown. Give the resulting matrix.
$\left[\begin{array}{cc|c} 2 & -4 & 6 \\ -1 & 3 & 2 \end{array}\right]$
1. Hint 1
  Apply the multiplier to every entry in the named row.
2. Hint 2
  Leave the other row unchanged.
Answer

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

Full solution

Multiplying the top row by $-1/2$ gives $(-1,2\mid-3)$ .
Hence the new matrix is
$\left[\begin{array}{cc|c} -1 & 2 & -3 \\ -1 & 3 & 2 \end{array}\right]$
Multiplying its first row by $-2$ recovers the original record.

Answer

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

Key idea

Scaling a row multiplies its constant along with its coefficients.
Problem 3: Mark complete
Problem 4 Three linked records

Write an augmented matrix, using columns in the order $x,y,z$ , and solve $y+2z=5$ , $x-y=1$ , and $2x+z=8$ by row reduction.
1. Hint 1
  A row beginning with $x$ makes a useful top row.
2. Hint 2
  After clearing the $x$ column, clear the $y$ column below its pivot.
Answer

Augmented matrix $\left[\begin{array}{ccc|c} 0 & 1 & 2 & 5 \\ 1 & -1 & 0 & 1 \\ 2 & 0 & 1 & 8 \end{array}\right]$ ; $(x,y,z)=(\tfrac{10}{3},\tfrac73,\tfrac43)$ .

Full solution

In variable order $x,y,z$ , the matrix is
$\left[\begin{array}{ccc|c} 0 & 1 & 2 & 5 \\ 1 & -1 & 0 & 1 \\ 2 & 0 & 1 & 8 \end{array}\right]$

Swap the first two rows.
Replace $R_3$ by $R_3-2R_1$ , then by $R_3-2R_2$ .
$\left[\begin{array}{ccc|c} 1 & -1 & 0 & 1 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & -3 & -4 \end{array}\right]$

The bottom row gives $z=4/3$ .
Then
$\begin{aligned} y&=5-2z \\ &=7/3 \\ x&=y+1 \\ &=10/3 \end{aligned}$
Check: $7/3+8/3=5$ , $10/3-7/3=1$ , and
$20/3+4/3=8.$

Answer

Augmented matrix $\left[\begin{array}{ccc|c} 0 & 1 & 2 & 5 \\ 1 & -1 & 0 & 1 \\ 2 & 0 & 1 & 8 \end{array}\right]$ ; $(x,y,z)=(\tfrac{10}{3},\tfrac73,\tfrac43)$ .

Key idea

A row swap can place a useful pivot before the forward elimination sweep.
Problem 4: Mark complete
Problem 5 A third measurement

The augmented matrix below uses columns $x,y$ . Reduce it and describe its solution set.
$\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 1 & 5 \end{array}\right]$
1. Hint 1
  There are more rows than variables, so one equation may conflict rather than repeat.
2. Hint 2
  Combine the first two rows and compare with the third.
Answer

No solution; the system is inconsistent.

Full solution

Replace $R_3$ by $R_3-R_1-R_2$ .
$\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{array}\right]$

The bottom row reads $0=2$ , which is false for every $x,y$ .
The first two equations alone give $x=1$ and $y=2$ , but the third equation $x+y=5$ requires $1+2=5$ , which fails, so the system has no solution.

Answer

No solution; the system is inconsistent.

Key idea

A zero row on the left with a nonzero constant is a contradiction: the system has no solution regardless of the other rows.
Problem 5: Mark complete
Problem 6 A short record

A system in $x,y,z$ has the augmented matrix below. Find all its solutions and name the free variable.
$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 0 & 0 & 2 & 6 \end{array}\right]$
1. Hint 1
  The second row fixes a variable even though its pivot skips a column.
2. Hint 2
  Let the variable without a pivot be a real parameter.
Answer

$y$ is free; $(x,y,z)=(1-2t,t,3)$ for real $t$ .

Full solution

The bottom row gives $2z=6$ , hence $z=3$ .
There is no pivot in the $y$ column, so let $y=t$ .
The first row then gives
$x=1-2t$
Substitution gives $(1-2t)+2t+3=4$ and $2(3)=6$ for all $t$ .

Answer

$y$ is free; $(x,y,z)=(1-2t,t,3)$ for real $t$ .

Key idea

In a consistent system's row-echelon form, a variable column without a pivot identifies a free variable.
Problem 6: Mark complete
Problem 7 Completing the constants

For real $c$ , the augmented matrix below uses columns $x,y,z$ in that order. Determine all $c$ giving a consistent system and describe the solution set for those $c$ .
$\left[\begin{array}{ccc|c} 1 & -1 & 0 & 2 \\ 0 & 0 & 1 & 3 \\ 2 & -2 & 1 & c \end{array}\right]$
1. Hint 1
  Compare the third row with a combination of the first two.
2. Hint 2
  Keep the constant column in the same subtraction as the coefficients.
Answer

$c=7$ ; $(x,y,z)=(2+t,t,3)$ for real $t$ .

Full solution

Replace $R_3$ by $R_3-2R_1-R_2$ , understood as two successive legal operations.
$\left[\begin{array}{ccc|c} 1 & -1 & 0 & 2 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & c-7 \end{array}\right]$

The last row is false unless $c=7$ .
At $c=7$ , it vanishes; $z=3$ and $x-y=2$ remain.
Letting $y=t$ gives the stated family, whose third left side is
$2(2+t)-2t+3=7.$

Answer

$c=7$ ; $(x,y,z)=(2+t,t,3)$ for real $t$ .

Key idea

A parameter in the constant column can decide whether the pivot-free variable is free or the system is inconsistent.
Problem 7: Mark complete
Problem 8 A proposed update

For the equations $x=2$ and $y=5$ , a student simultaneously replaces both rows by their sum and keeps two copies of $x+y=7$ . They say the solution set is unchanged. Is that correct? Justify with a pair of values.
1. Hint 1
  A permitted addition keeps the row being added available elsewhere.
2. Hint 2
  Test a point on the new line that fails one original equation.
Answer

No; for example, $(0,7)$ solves the new system but not the original.

Full solution

The originals force the unique pair $(2,5)$ .
The simultaneous replacements both say
$x+y=7$
The pair $(0,7)$ satisfies this twice but fails $x=2$ .
Replacing both rows in this way loses the information needed to recover the original rows.

Answer

No; for example, $(0,7)$ solves the new system but not the original.

Key idea

Adding one row to another preserves information when the source row remains available.
Problem 8: Mark complete
Problem 9 Checking an update

A student replaces $R_3$ by $R_3+4R_1$ while leaving $R_1$ and $R_2$ fixed. They claim the original $R_3$ can be recovered by subtracting $4R_1$ from the new $R_3$ . Is this correct? Explain.
1. Hint 1
  Check whether the row needed to undo the change is still present.
2. Hint 2
  Write the change followed by the proposed reverse operation.
Answer

Yes.

Full solution

The first row is unchanged and still available.
The proposed reverse gives
$(R_3+4R_1)-4R_1=R_3$
Every entry, including the constant, returns to its original value.
Thus the original and changed systems have exactly the same solutions.

Answer

Yes.

Key idea

A reversible row operation preserves the entire system’s solution set.
Problem 9: Mark complete
Problem 10 Counting available information

A consistent system has five equations in three variables. Its row-echelon form has two zero rows and a pivot in each variable column. Another consistent system has two equations in three variables and no zero rows. Explain which system has a unique solution.
1. Hint 1
  The decisive count concerns variable columns, not the number of zero rows.
2. Hint 2
  Two nonzero rows can supply at most two pivots.
Answer

The five-equation system has a unique solution; the two-equation system has infinitely many solutions.

Full solution

The first system has three variable pivots, so back-substitution fixes all three variables despite the redundant rows.
The second has at most two pivots for three variables, so at least one variable is free.
It is consistent, hence that free variable supplies infinitely many solutions.

Answer

The five-equation system has a unique solution; the two-equation system has infinitely many solutions.

Key idea

For a consistent system, every variable needs a pivot for the solution to be unique.
Problem 10: Mark complete

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