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Determinants and Cramer's Rule: 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 A missing corner entry

Find the real number $k$ such that $\begin{vmatrix} 2&-3\\4&k\end{vmatrix}=10$ .
1. Hint 1
  The determinant is a signed combination of the two diagonal products.
2. Hint 2
  Write a linear equation involving the unknown entry.
Answer

$k=-1$ .

Full solution

Main diagonal minus the other diagonal gives
$\begin{aligned} 2k-(-3)(4)&=10 \\ 2k+12&=10 \\ k&=-1 \end{aligned}$

Check:
$2(-1)-(-3)(4)=10.$

Answer

$k=-1$ .

Key idea

A determinant value can determine an unknown matrix entry.
Problem 1: Mark complete
Problem 2 Elimination on a specific system

For $3x-2y=8$ and $5x+y=9$ , eliminate $y$ by adding the first equation to twice the second. Identify the coefficient of $x$ that remains.
1. Hint 1
  Track the coefficient arithmetic separately from the constants.
2. Hint 2
  Use main diagonal minus the other diagonal on the original coefficient array.
Answer

$13$ .

Full solution

The stated combination gives
$13x=26$
The coefficient determinant in the original equation order is
$D=3(1)-(-2)(5)$
Thus $D=13$ , matching the surviving coefficient, because scaling by $d=1$ and $b=-2$ is exactly the elimination that produces $D$ .
Since it is nonzero, it permits division and forces one $x$ value.

Answer

$13$ .

Key idea

The determinant appears as the surviving variable coefficient in elimination.
Problem 2: Mark complete
Problem 3 Three short rows

Evaluate $\begin{vmatrix}0&-2&1\\3&1&0\\1&4&2\end{vmatrix}$ by expanding along its first row.
1. Hint 1
  Each first-row entry multiplies the determinant formed by removing its row and column.
2. Hint 2
  The signs across that row alternate plus, minus, plus.
Answer

$23$ .

Full solution

The first entry contributes zero.
The minor under the middle entry is $3(2)-0(1)=6$ , and the last minor is $3(4)-1(1)=11$ .
$\begin{aligned} D&=0-(-2)(6)+1(11) \\ D&=23 \end{aligned}$

Answer

$23$ .

Key idea

A negative middle entry still receives the subtraction in cofactor expansion.
Problem 3: Mark complete
Problem 4 A straight-line model

A model $f(t)=at+b$ satisfies $f(-1)=6$ and $f(2)=-3$ . Use Cramer’s rule to find $a$ and $b$ .
1. Hint 1
  The two function values form a system for the model’s coefficients.
2. Hint 2
  Replace the coefficient column for the unknown you are finding with the two outputs.
Answer

$a=-3$ , $b=3$ .

Full solution

The equations are $-a+b=6$ and $2a+b=-3$ .
In variable order $a,b$ ,
$\begin{aligned} D&=(-1)(1)-(1)(2) \\ D&=-3 \\ D_a&=(6)(1)-(1)(-3) \\ D_a&=9 \end{aligned}$

The other replacement gives
$\begin{aligned} D_b&=(-1)(-3)-(6)(2) \\ D_b&=-9 \\ a&=\frac9{-3} \\ b&=\frac{-9}{-3} \end{aligned}$

So $a=-3$ and $b=3$ .
Check $f(-1)$ :
$-3(-1)+3=6.$
Check $f(2)$ :
$-3(2)+3=-3.$

Answer

$a=-3$ , $b=3$ .

Key idea

Cramer’s rule applies when the unknowns are model coefficients as well as coordinates.
Problem 4: Mark complete
Problem 5 An isolated column

Use Cramer’s rule to find $x$ in $x+z=4$ , $2y-z=3$ , and $x+y+z=7$ .
1. Hint 1
  Keep the zero coefficient in each missing-variable position.
2. Hint 2
  Compute the original determinant and the one with the first column replaced.
Answer

$x=1$ .

Full solution

In order $x,y,z$ ,
$D=\begin{vmatrix}1&0&1\\0&2&-1\\1&1&1\end{vmatrix}$
Cofactor expansion gives
$\begin{aligned} D&=1(3)+1(-2) \\ &=1 \end{aligned}$
Replacing the first column gives
$D_x=\begin{vmatrix}4&0&1\\3&2&-1\\7&1&1\end{vmatrix}$
Then
$\begin{aligned} D_x&=4(3)+1(3-14) \\ D_x&=1 \\ x&=\frac11 \end{aligned}$

The check can be made with $y=3$ and $z=3$ : $1+3=4$ , $6-3=3$ , and $1+3+3=7$ .

Answer

$x=1$ .

Key idea

One desired unknown requires its replacement determinant and the shared coefficient determinant.
Problem 5: Mark complete
Problem 6 All three readings

Use Cramer’s rule to solve $x-y=1$ , $y+z=3$ , and $2x+z=7$ .
1. Hint 1
  Write zero coefficients where an equation omits a variable.
2. Hint 2
  Replace one column at a time, keeping the other two columns fixed.
Answer

$(x,y,z)=(3,2,1)$ .

Full solution

The coefficient determinant is
$D=\begin{vmatrix}1&-1&0\\0&1&1\\2&0&1\end{vmatrix}$
Its first-row expansion gives
$\begin{aligned} D&=1(1)-(-1)(-2) \\ &=-1 \end{aligned}$
The replacements are
$D_x=\begin{vmatrix}1&-1&0\\3&1&1\\7&0&1\end{vmatrix}$ $D_y=\begin{vmatrix}1&1&0\\0&3&1\\2&7&1\end{vmatrix}$ $D_z=\begin{vmatrix}1&-1&1\\0&1&3\\2&0&7\end{vmatrix}$

Expanding along the top row gives
$\begin{aligned} D_x&=1+(3-7) \\ D_x&=-3 \\ D_y&=(3-7)-(-2) \\ D_y&=-2 \\ D_z&=7-6-2 \\ D_z&=-1 \end{aligned}$

Dividing each by the nonzero $D=-1$ gives $(3,2,1)$ .
Check: $3-2=1$ , $2+1=3$ , and $6+1=7$ .

Answer

$(x,y,z)=(3,2,1)$ .

Key idea

Each coordinate uses the determinant with its own coefficient column replaced.
Problem 6: Mark complete
Problem 7 Which inputs distinguish settings

The equations $au+b=5$ and $av+b=8$ have unknowns $a,b$ and real parameters $u,v$ . Determine when they have a unique solution, and describe what happens otherwise.
1. Hint 1
  Arrange the coefficients of $a,b$ in a two-row array.
2. Hint 2
  Consider what the two left sides become when its determinant is zero.
Answer

Unique when $u\ne v$ ; no solution when $u=v$ .

Full solution

The coefficient determinant is
$D=u-v$
It is nonzero exactly when $u\ne v$ .
If $u=v$ , the left sides are identical, yet the right sides are $5$ and $8$ ; subtraction gives $0=3$ .
Thus this particular zero-determinant case is inconsistent.

Answer

Unique when $u\ne v$ ; no solution when $u=v$ .

Key idea

The coefficient determinant tests uniqueness, while the constants decide a degenerate case.
Problem 7: Mark complete
Problem 8 One vanished numerator

A two-variable system has $D=-6$ and $D_x=0$ . A student says the system has no solution since one determinant is zero. Is that correct? State what is known about $x$ and uniqueness.
1. Hint 1
  Distinguish the coefficient determinant from a replacement determinant.
2. Hint 2
  Identify which determinant is the divisor in Cramer’s rule.
Answer

No; the system has a unique solution and $x=0$ .

Full solution

The coefficient determinant is nonzero, so the system is uniquely solvable.
The first coordinate is
$x=\frac0{-6}$
Thus $x=0$ .
A zero numerator permits a zero coordinate; it does not block division.

Answer

No; the system has a unique solution and $x=0$ .

Key idea

A zero replacement determinant can give a zero coordinate in a unique solution.
Problem 8: Mark complete
Problem 9 Unchanged coefficients

A square system has coefficient determinant zero. A student changes just one right side and claims this can make the solution unique. Is this possible? Explain.
1. Hint 1
  Changing a right side does not alter the coefficient matrix.
2. Hint 2
  Use the uniqueness condition for a square system.
Answer

No; the system remains nonunique.

Full solution

The coefficient matrix is unchanged, so
$D_{\mathrm{new}}=0$
A square system has a unique solution exactly when its coefficient determinant is nonzero.
Changing a right side may switch between no solutions and infinitely many, but cannot produce a unique solution here.

Answer

No; the system remains nonunique.

Key idea

A change to constants cannot repair a zero coefficient determinant.
Problem 9: Mark complete
Problem 10 One unknown, not the whole solve

A solver has a dense system of six equations in six unknowns, known to have a unique solution, but needs only the value of one particular unknown, not a full solution. Recommend whether to use Cramer's rule or row reduction, and explain.
1. Hint 1
  Row reduction's cost for one unknown depends on where it falls in the elimination order; Cramer's rule's cost does not.
2. Hint 2
  Cramer's rule needs only the coefficient determinant and one replacement determinant for a single unknown, whichever one it is.
Answer

Use Cramer's rule.

Full solution

Row reduction's cost for one unknown depends on where it falls in the elimination order, and can require back-substituting through most of the other five, even though only one value is wanted.

Cramer's rule pays a fixed cost no matter which unknown is wanted: the coefficient determinant $D$ and the one replacement determinant for that variable, with no back-substitution through the others.
That fixed cost does not grow with how far the wanted unknown sits from the end of the elimination order, unlike row reduction's.

Answer

Use Cramer's rule.

Key idea

Needing only one unknown, not a full solve, is when Cramer's rule's fixed cost per unknown matters most against row reduction's cost, which depends on elimination order.
Problem 10: Mark complete

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