12 multiple-choice questions, progressively harder.
Use Cramer's rule to find xxx for {x+y=5x−y=1\begin{cases} x + y = 5 \\ x - y = 1 \end{cases}{x+y=5x−y=1.
Solution
Correct answer: B
The coefficient determinant is D=∣111−1∣=−2D = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -2D=111−1=−2. Replacing the xxx-column with the constants 555 and 111 gives Dx=∣511−1∣=−6D_x = \begin{vmatrix} 5 & 1 \\ 1 & -1 \end{vmatrix} = -6Dx=511−1=−6.
x=DxD=−6−2=3x = \frac{D_x}{D} = \frac{-6}{-2} = 3x=DDx=−2−6=3
Use Cramer's rule to find yyy for {2x+y=7x+3y=11\begin{cases} 2x + y = 7 \\ x + 3y = 11 \end{cases}{2x+y=7x+3y=11.
Correct answer: C
Here D=∣2113∣=5D = \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = 5D=2113=5. Replacing the yyy-column with the constants 777 and 111111 gives Dy=∣27111∣=22−7=15D_y = \begin{vmatrix} 2 & 7 \\ 1 & 11 \end{vmatrix} = 22 - 7 = 15Dy=21711=22−7=15.
y=DyD=155=3y = \frac{D_y}{D} = \frac{15}{5} = 3y=DDy=515=3
Use Cramer's rule to find yyy for {x+2y=113x−y=5\begin{cases} x + 2y = 11 \\ 3x - y = 5 \end{cases}{x+2y=113x−y=5.
Correct answer: A
Compute D=∣123−1∣=−1−6=−7D = \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = -1 - 6 = -7D=132−1=−1−6=−7, and, replacing the yyy-column, Dy=∣11135∣=5−33=−28D_y = \begin{vmatrix} 1 & 11 \\ 3 & 5 \end{vmatrix} = 5 - 33 = -28Dy=13115=5−33=−28.
y=DyD=−28−7=4y = \frac{D_y}{D} = \frac{-28}{-7} = 4y=DDy=−7−28=4
For what value(s) of kkk is ∣k41k∣=0\begin{vmatrix} k & 4 \\ 1 & k \end{vmatrix} = 0k14k=0?
Evaluate the determinant and set it equal to zero.
∣k41k∣=k2−4=0\begin{vmatrix} k & 4 \\ 1 & k \end{vmatrix} = k^2 - 4 = 0k14k=k2−4=0
So k2=4k^2 = 4k2=4, giving k=2k = 2k=2 or k=−2k = -2k=−2.
A two-variable system has coefficient determinant D=0D = 0D=0. How many solutions is it possible for the system to have?
Correct answer: D
A zero determinant rules out a unique solution but does not by itself decide which degenerate case holds.
D=0 ⇒ no solution or infinitely manyD = 0 \;\Rightarrow\; \text{no solution or infinitely many}D=0⇒no solution or infinitely many
Checking the constants is what settles which of the two it is.
Evaluate ∣120312041∣\begin{vmatrix} 1 & 2 & 0 \\ 3 & 1 & 2 \\ 0 & 4 & 1 \end{vmatrix}130214021.
Expand along the first row with signs plus, minus, plus.
1(1⋅1−2⋅4)−2(3⋅1−2⋅0)+0=−7−6+0=−131(1 \cdot 1 - 2 \cdot 4) - 2(3 \cdot 1 - 2 \cdot 0) + 0 = -7 - 6 + 0 = -131(1⋅1−2⋅4)−2(3⋅1−2⋅0)+0=−7−6+0=−13
Evaluate ∣1−12031211∣\begin{vmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 2 & 1 & 1 \end{vmatrix}102−131211.
Expand along the first row, carrying the sign of the entry −1-1−1 through the middle term.
1(3⋅1−1⋅1)−(−1)(0⋅1−1⋅2)+2(0⋅1−3⋅2)=2−2−12=−121(3 \cdot 1 - 1 \cdot 1) - (-1)(0 \cdot 1 - 1 \cdot 2) + 2(0 \cdot 1 - 3 \cdot 2) = 2 - 2 - 12 = -121(3⋅1−1⋅1)−(−1)(0⋅1−1⋅2)+2(0⋅1−3⋅2)=2−2−12=−12
Use Cramer's rule to find xxx for {2x−3y=−4x+y=3\begin{cases} 2x - 3y = -4 \\ x + y = 3 \end{cases}{2x−3y=−4x+y=3.
Compute D=∣2−311∣=2+3=5D = \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = 2 + 3 = 5D=21−31=2+3=5 and Dx=∣−4−331∣=−4+9=5D_x = \begin{vmatrix} -4 & -3 \\ 3 & 1 \end{vmatrix} = -4 + 9 = 5Dx=−43−31=−4+9=5.
x=DxD=55=1x = \frac{D_x}{D} = \frac{5}{5} = 1x=DDx=55=1
Use Cramer's rule to find xxx for {x+y+z=6x−y+z=22x+y−z=1\begin{cases} x + y + z = 6 \\ x - y + z = 2 \\ 2x + y - z = 1 \end{cases}⎩⎨⎧x+y+z=6x−y+z=22x+y−z=1.
Expand DDD along the first row.
D=1(1−1)−1(−1−2)+1(1+2)=0+3+3=6D = 1(1 - 1) - 1(-1 - 2) + 1(1 + 2) = 0 + 3 + 3 = 6D=1(1−1)−1(−1−2)+1(1+2)=0+3+3=6
Replace the xxx-column with the constants 6,2,16, 2, 16,2,1 and expand again.
Dx=6(1−1)−1(−2−1)+1(2+1)=0+3+3=6D_x = 6(1 - 1) - 1(-2 - 1) + 1(2 + 1) = 0 + 3 + 3 = 6Dx=6(1−1)−1(−2−1)+1(2+1)=0+3+3=6
x=DxD=66=1x = \frac{D_x}{D} = \frac{6}{6} = 1x=DDx=66=1
Evaluate ∣023104210∣\begin{vmatrix} 0 & 2 & 3 \\ 1 & 0 & 4 \\ 2 & 1 & 0 \end{vmatrix}012201340.
Expand along the first row; the first term vanishes since its entry is 000.
0−2(1⋅0−4⋅2)+3(1⋅1−0⋅2)=−2(−8)+3(1)=190 - 2(1 \cdot 0 - 4 \cdot 2) + 3(1 \cdot 1 - 0 \cdot 2) = -2(-8) + 3(1) = 190−2(1⋅0−4⋅2)+3(1⋅1−0⋅2)=−2(−8)+3(1)=19
For what value of kkk does {2x+ky=3x+2y=1\begin{cases} 2x + ky = 3 \\ x + 2y = 1 \end{cases}{2x+ky=3x+2y=1 fail to have a unique solution?
The unique solution is lost exactly when the coefficient determinant is zero.
D=∣2k12∣=4−k=0D = \begin{vmatrix} 2 & k \\ 1 & 2 \end{vmatrix} = 4 - k = 0D=21k2=4−k=0
So k=4k = 4k=4 is the value that makes the system degenerate.
A two-variable linear system has coefficient determinant D=0D = 0D=0. Which outcome is impossible?
A zero determinant forbids a unique solution, while still allowing no solution or infinitely many.
D=0 ⇒ not exactly one solutionD = 0 \;\Rightarrow\; \text{not exactly one solution}D=0⇒not exactly one solution
So 'exactly one solution' is the outcome that cannot occur.
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