12 multiple-choice questions, progressively harder.
Use Cramer's rule to find zzz for {x+2y+z=72x+y+3z=13x−y+z=4\begin{cases} x + 2y + z = 7 \\ 2x + y + 3z = 13 \\ x - y + z = 4 \end{cases}⎩⎨⎧x+2y+z=72x+y+3z=13x−y+z=4.
Solution
Correct answer: B
Expand DDD along the first row.
D=1(1+3)−2(2−3)+1(−2−1)=4+2−3=3D = 1(1 + 3) - 2(2 - 3) + 1(-2 - 1) = 4 + 2 - 3 = 3D=1(1+3)−2(2−3)+1(−2−1)=4+2−3=3
Replace the zzz-column with the constants 7,13,47, 13, 47,13,4 and expand again.
Dz=1(4+13)−2(8−13)+7(−2−1)=17+10−21=6D_z = 1(4 + 13) - 2(8 - 13) + 7(-2 - 1) = 17 + 10 - 21 = 6Dz=1(4+13)−2(8−13)+7(−2−1)=17+10−21=6
z=DzD=63=2z = \frac{D_z}{D} = \frac{6}{3} = 2z=DDz=36=2
Evaluate ∣312241123∣\begin{vmatrix} 3 & 1 & 2 \\ 2 & 4 & 1 \\ 1 & 2 & 3 \end{vmatrix}321142213.
Correct answer: A
Expand along the first row with signs plus, minus, plus.
3(4⋅3−1⋅2)−1(2⋅3−1⋅1)+2(2⋅2−4⋅1)=3(10)−1(5)+2(0)=253(4 \cdot 3 - 1 \cdot 2) - 1(2 \cdot 3 - 1 \cdot 1) + 2(2 \cdot 2 - 4 \cdot 1) = 3(10) - 1(5) + 2(0) = 253(4⋅3−1⋅2)−1(2⋅3−1⋅1)+2(2⋅2−4⋅1)=3(10)−1(5)+2(0)=25
For what value(s) of kkk is ∣k28k∣=0\begin{vmatrix} k & 2 \\ 8 & k \end{vmatrix} = 0k82k=0?
Correct answer: C
Evaluate the determinant and set it equal to zero.
∣k28k∣=k2−16=0\begin{vmatrix} k & 2 \\ 8 & k \end{vmatrix} = k^2 - 16 = 0k82k=k2−16=0
So k2=16k^2 = 16k2=16, giving k=4k = 4k=4 or k=−4k = -4k=−4.
How many solutions does {2x+3y=44x+6y=8\begin{cases} 2x + 3y = 4 \\ 4x + 6y = 8 \end{cases}{2x+3y=44x+6y=8 have?
The determinant ∣2346∣=0\begin{vmatrix} 2 & 3 \\ 4 & 6 \end{vmatrix} = 02436=0, so check the equations themselves. The second is exactly twice the first.
4x+6y=2(2x+3y)=2(4)=84x + 6y = 2(2x + 3y) = 2(4) = 84x+6y=2(2x+3y)=2(4)=8
The two equations are the same line, so there are infinitely many solutions.
Use Cramer's rule to find yyy for {x+y+z=32x−y+z=−2x+2y+3z=4\begin{cases} x + y + z = 3 \\ 2x - y + z = -2 \\ x + 2y + 3z = 4 \end{cases}⎩⎨⎧x+y+z=32x−y+z=−2x+2y+3z=4.
D=1(−3−2)−1(6−1)+1(4+1)=−5−5+5=−5D = 1(-3 - 2) - 1(6 - 1) + 1(4 + 1) = -5 - 5 + 5 = -5D=1(−3−2)−1(6−1)+1(4+1)=−5−5+5=−5
Replace the yyy-column with the constants 3,−2,43, -2, 43,−2,4 and expand again.
Dy=1(−6−4)−3(6−1)+1(8+2)=−10−15+10=−15D_y = 1(-6 - 4) - 3(6 - 1) + 1(8 + 2) = -10 - 15 + 10 = -15Dy=1(−6−4)−3(6−1)+1(8+2)=−10−15+10=−15
y=DyD=−15−5=3y = \frac{D_y}{D} = \frac{-15}{-5} = 3y=DDy=−5−15=3
Use Cramer's rule to find yyy for {3x+2y=45x+3y=7\begin{cases} 3x + 2y = 4 \\ 5x + 3y = 7 \end{cases}{3x+2y=45x+3y=7.
Compute D=∣3253∣=9−10=−1D = \begin{vmatrix} 3 & 2 \\ 5 & 3 \end{vmatrix} = 9 - 10 = -1D=3523=9−10=−1 and Dy=∣3457∣=21−20=1D_y = \begin{vmatrix} 3 & 4 \\ 5 & 7 \end{vmatrix} = 21 - 20 = 1Dy=3547=21−20=1.
y=DyD=1−1=−1y = \frac{D_y}{D} = \frac{1}{-1} = -1y=DDy=−11=−1
If ∣abcd∣=5\begin{vmatrix} a & b \\ c & d \end{vmatrix} = 5acbd=5, what is ∣2a2bcd∣\begin{vmatrix} 2a & 2b \\ c & d \end{vmatrix}2ac2bd?
Multiplying a single row by 222 multiplies the whole determinant by 222.
∣2a2bcd∣=2ad−2bc=2(ad−bc)=2(5)=10\begin{vmatrix} 2a & 2b \\ c & d \end{vmatrix} = 2ad - 2bc = 2(ad - bc) = 2(5) = 102ac2bd=2ad−2bc=2(ad−bc)=2(5)=10
To find only zzz in a three-variable system using Cramer's rule, which determinants must you compute?
Correct answer: D
Cramer's rule writes zzz as a ratio of just two determinants.
z=DzDz = \frac{D_z}{D}z=DDz
So you compute DDD and DzD_zDz only, skipping DxD_xDx and DyD_yDy.
Use Cramer's rule to find yyy for {7x+2y=13x+y=0\begin{cases} 7x + 2y = 1 \\ 3x + y = 0 \end{cases}{7x+2y=13x+y=0.
Compute D=∣7231∣=7−6=1D = \begin{vmatrix} 7 & 2 \\ 3 & 1 \end{vmatrix} = 7 - 6 = 1D=7321=7−6=1 and Dy=∣7130∣=0−3=−3D_y = \begin{vmatrix} 7 & 1 \\ 3 & 0 \end{vmatrix} = 0 - 3 = -3Dy=7310=0−3=−3.
y=DyD=−31=−3y = \frac{D_y}{D} = \frac{-3}{1} = -3y=DDy=1−3=−3
For what value(s) of kkk is ∣kk3k∣=0\begin{vmatrix} k & k \\ 3 & k \end{vmatrix} = 0k3kk=0?
Expand the determinant and factor.
∣kk3k∣=k2−3k=k(k−3)=0\begin{vmatrix} k & k \\ 3 & k \end{vmatrix} = k^2 - 3k = k(k - 3) = 0k3kk=k2−3k=k(k−3)=0
The product is zero when k=0k = 0k=0 or k=3k = 3k=3.
Use Cramer's rule to find xxx for {x+y+z=22x+y−z=−1x−y+2z=−1\begin{cases} x + y + z = 2 \\ 2x + y - z = -1 \\ x - y + 2z = -1 \end{cases}⎩⎨⎧x+y+z=22x+y−z=−1x−y+2z=−1.
D=1(2−1)−1(4+1)+1(−2−1)=1−5−3=−7D = 1(2 - 1) - 1(4 + 1) + 1(-2 - 1) = 1 - 5 - 3 = -7D=1(2−1)−1(4+1)+1(−2−1)=1−5−3=−7
Replace the xxx-column with the constants 2,−1,−12, -1, -12,−1,−1 and expand again.
Dx=2(2−1)−1(−2−1)+1(1+1)=2+3+2=7D_x = 2(2 - 1) - 1(-2 - 1) + 1(1 + 1) = 2 + 3 + 2 = 7Dx=2(2−1)−1(−2−1)+1(1+1)=2+3+2=7
x=DxD=7−7=−1x = \frac{D_x}{D} = \frac{7}{-7} = -1x=DDx=−77=−1
Use Cramer's rule to find xxx for {5x−2y=13x+y=5\begin{cases} 5x - 2y = 1 \\ 3x + y = 5 \end{cases}{5x−2y=13x+y=5.
Compute D=∣5−231∣=5+6=11D = \begin{vmatrix} 5 & -2 \\ 3 & 1 \end{vmatrix} = 5 + 6 = 11D=53−21=5+6=11 and Dx=∣1−251∣=1+10=11D_x = \begin{vmatrix} 1 & -2 \\ 5 & 1 \end{vmatrix} = 1 + 10 = 11Dx=15−21=1+10=11.
x=DxD=1111=1x = \frac{D_x}{D} = \frac{11}{11} = 1x=DDx=1111=1
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