12 multiple-choice questions, progressively harder.
Use Cramer's rule to find xxx for {4x−3y=52x+y=5\begin{cases} 4x - 3y = 5 \\ 2x + y = 5 \end{cases}{4x−3y=52x+y=5.
Solution
Correct answer: D
Compute D=∣4−321∣=4+6=10D = \begin{vmatrix} 4 & -3 \\ 2 & 1 \end{vmatrix} = 4 + 6 = 10D=42−31=4+6=10 and Dx=∣5−351∣=5+15=20D_x = \begin{vmatrix} 5 & -3 \\ 5 & 1 \end{vmatrix} = 5 + 15 = 20Dx=55−31=5+15=20.
x=DxD=2010=2x = \frac{D_x}{D} = \frac{20}{10} = 2x=DDx=1020=2
Evaluate ∣213041521∣\begin{vmatrix} 2 & 1 & 3 \\ 0 & 4 & 1 \\ 5 & 2 & 1 \end{vmatrix}205142311.
Correct answer: C
Expand along the first row with signs plus, minus, plus.
2(4⋅1−1⋅2)−1(0⋅1−1⋅5)+3(0⋅2−4⋅5)=2(2)+5−60=−512(4 \cdot 1 - 1 \cdot 2) - 1(0 \cdot 1 - 1 \cdot 5) + 3(0 \cdot 2 - 4 \cdot 5) = 2(2) + 5 - 60 = -512(4⋅1−1⋅2)−1(0⋅1−1⋅5)+3(0⋅2−4⋅5)=2(2)+5−60=−51
Use Cramer's rule to find zzz for {x+y+z=2x−y+2z=92x+y+z=3\begin{cases} x + y + z = 2 \\ x - y + 2z = 9 \\ 2x + y + z = 3 \end{cases}⎩⎨⎧x+y+z=2x−y+2z=92x+y+z=3.
Correct answer: A
Expand DDD along the first row.
D=1(−1−2)−1(1−4)+1(1+2)=−3+3+3=3D = 1(-1 - 2) - 1(1 - 4) + 1(1 + 2) = -3 + 3 + 3 = 3D=1(−1−2)−1(1−4)+1(1+2)=−3+3+3=3
Replace the zzz-column with the constants 2,9,32, 9, 32,9,3 and expand again.
Dz=1(−3−9)−1(3−18)+2(1+2)=−12+15+6=9D_z = 1(-3 - 9) - 1(3 - 18) + 2(1 + 2) = -12 + 15 + 6 = 9Dz=1(−3−9)−1(3−18)+2(1+2)=−12+15+6=9
z=DzD=93=3z = \frac{D_z}{D} = \frac{9}{3} = 3z=DDz=39=3
For what value(s) of kkk is ∣3kk12∣=0\begin{vmatrix} 3 & k \\ k & 12 \end{vmatrix} = 03kk12=0?
Correct answer: B
Evaluate the determinant and set it equal to zero.
∣3kk12∣=36−k2=0\begin{vmatrix} 3 & k \\ k & 12 \end{vmatrix} = 36 - k^2 = 03kk12=36−k2=0
So k2=36k^2 = 36k2=36, giving k=6k = 6k=6 or k=−6k = -6k=−6.
How many solutions does {x−2y=3−2x+4y=−6\begin{cases} x - 2y = 3 \\ -2x + 4y = -6 \end{cases}{x−2y=3−2x+4y=−6 have?
The determinant ∣1−2−24∣=4−4=0\begin{vmatrix} 1 & -2 \\ -2 & 4 \end{vmatrix} = 4 - 4 = 01−2−24=4−4=0, so inspect the equations. The second is −2-2−2 times the first.
−2(x−2y)=−2x+4y=−2(3)=−6-2(x - 2y) = -2x + 4y = -2(3) = -6−2(x−2y)=−2x+4y=−2(3)=−6
The two equations are the same line, so there are infinitely many solutions.
Evaluate ∣102310241∣\begin{vmatrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 2 & 4 & 1 \end{vmatrix}132014201.
Expand along the first row; the middle entry is 000, so its term drops.
1(1⋅1−0⋅4)−0+2(3⋅4−1⋅2)=1+2(10)=211(1 \cdot 1 - 0 \cdot 4) - 0 + 2(3 \cdot 4 - 1 \cdot 2) = 1 + 2(10) = 211(1⋅1−0⋅4)−0+2(3⋅4−1⋅2)=1+2(10)=21
If ∣abcd∣=9\begin{vmatrix} a & b \\ c & d \end{vmatrix} = 9acbd=9, what is ∣acbd∣\begin{vmatrix} a & c \\ b & d \end{vmatrix}abcd?
This second determinant is the transpose, with rows and columns swapped, and the transpose leaves the value unchanged.
∣acbd∣=ad−cb=ad−bc=9\begin{vmatrix} a & c \\ b & d \end{vmatrix} = ad - cb = ad - bc = 9abcd=ad−cb=ad−bc=9
Use Cramer's rule to find yyy for {x+y+z=4x+2y+z=72x+y+3z=4\begin{cases} x + y + z = 4 \\ x + 2y + z = 7 \\ 2x + y + 3z = 4 \end{cases}⎩⎨⎧x+y+z=4x+2y+z=72x+y+3z=4.
D=1(6−1)−1(3−2)+1(1−4)=5−1−3=1D = 1(6 - 1) - 1(3 - 2) + 1(1 - 4) = 5 - 1 - 3 = 1D=1(6−1)−1(3−2)+1(1−4)=5−1−3=1
Replace the yyy-column with the constants 4,7,44, 7, 44,7,4 and expand again.
Dy=1(21−4)−4(3−2)+1(4−14)=17−4−10=3D_y = 1(21 - 4) - 4(3 - 2) + 1(4 - 14) = 17 - 4 - 10 = 3Dy=1(21−4)−4(3−2)+1(4−14)=17−4−10=3
y=DyD=31=3y = \frac{D_y}{D} = \frac{3}{1} = 3y=DDy=13=3
A student needs only yyy from a three-variable system and chooses Cramer's rule. The fastest correct approach is to compute which determinants?
Cramer's rule writes yyy as a ratio of just two determinants.
y=DyDy = \frac{D_y}{D}y=DDy
Computing DDD and DyD_yDy and dividing is the shortcut; the other two determinants are unnecessary.
For what value(s) of kkk is ∣k+123k∣=0\begin{vmatrix} k+1 & 2 \\ 3 & k \end{vmatrix} = 0k+132k=0?
Expand the determinant, then solve the quadratic.
∣k+123k∣=(k+1)k−6=k2+k−6=(k+3)(k−2)=0\begin{vmatrix} k+1 & 2 \\ 3 & k \end{vmatrix} = (k+1)k - 6 = k^2 + k - 6 = (k+3)(k-2) = 0k+132k=(k+1)k−6=k2+k−6=(k+3)(k−2)=0
So k=−3k = -3k=−3 or k=2k = 2k=2.
Evaluate ∣3000−20005∣\begin{vmatrix} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 5 \end{vmatrix}3000−20005.
Expand along the first row; the two zeros in the top row kill the second and third terms.
3(−2⋅5−0⋅0)−0+0=3(−10)=−303(-2 \cdot 5 - 0 \cdot 0) - 0 + 0 = 3(-10) = -303(−2⋅5−0⋅0)−0+0=3(−10)=−30
The answer is the product of the diagonal entries 3×(−2)×53 \times (-2) \times 53×(−2)×5, which always happens for a diagonal array.
Use Cramer's rule to find xxx for {8x+3y=135x+2y=8\begin{cases} 8x + 3y = 13 \\ 5x + 2y = 8 \end{cases}{8x+3y=135x+2y=8.
Compute D=∣8352∣=16−15=1D = \begin{vmatrix} 8 & 3 \\ 5 & 2 \end{vmatrix} = 16 - 15 = 1D=8532=16−15=1 and Dx=∣13382∣=26−24=2D_x = \begin{vmatrix} 13 & 3 \\ 8 & 2 \end{vmatrix} = 26 - 24 = 2Dx=13832=26−24=2.
x=DxD=21=2x = \frac{D_x}{D} = \frac{2}{1} = 2x=DDx=12=2
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