12 multiple-choice questions, progressively harder.
For which value of kkk does the system with augmented matrix [26813k]\left[\begin{array}{cc|c}2 & 6 & 8 \\ 1 & 3 & k\end{array}\right][21638k] have infinitely many solutions?
Solution
Correct answer: B
The left sides are proportional (2:6=1:32 : 6 = 1 : 32:6=1:3). Using R1→R1−2R2R_1 \to R_1 - 2R_2R1→R1−2R2 gives (0, 0∣8−2k)(0,\ 0 \mid 8 - 2k)(0, 0∣8−2k), which is 0=00 = 00=0 only when
8−2k=0 ⇒ k=48 - 2k = 0 \;\Rightarrow\; k = 48−2k=0⇒k=4
Then the two equations coincide and there are infinitely many solutions.
The reduced matrix [10−1201350000]\left[\begin{array}{ccc|c}1 & 0 & -1 & 2 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & 0 & 0\end{array}\right]100010−130250 (columns xxx, yyy, zzz) has which solution set?
Correct answer: C
The zero row leaves zzz free; set z=tz = tz=t. Row 222 gives y+3z=5y + 3z = 5y+3z=5, so y=5−3ty = 5 - 3ty=5−3t. Row 111 gives x−z=2x - z = 2x−z=2, so x=2+tx = 2 + tx=2+t.
x=2+t,y=5−3t,z=tx = 2 + t,\quad y = 5 - 3t,\quad z = tx=2+t,y=5−3t,z=t
The system with augmented matrix [1423k1]\left[\begin{array}{cc|c}1 & 4 & 2 \\ 3 & k & 1\end{array}\right][134k21] has a unique solution unless kkk equals:
Apply R2→R2−3R1R_2 \to R_2 - 3R_1R2→R2−3R1: the row becomes (0, k−12∣−5)(0,\ k - 12 \mid -5)(0, k−12∣−5). A unique solution needs k−12≠0k - 12 \ne 0k−12=0, so it fails only when
k−12=0 ⇒ k=12k - 12 = 0 \;\Rightarrow\; k = 12k−12=0⇒k=12
Then the row is 0=−50 = -50=−5 and there is no solution.
A consistent system in 444 variables reduces to 222 pivots. How many free variables are there?
Correct answer: D
Free variables are those without a pivot.
4−2=24 - 2 = 24−2=2
So there are 222 free variables and infinitely many solutions.
Clearing column 111 of [13254139212528]\left[\begin{array}{ccc|c}1 & 3 & 2 & 5 \\ 4 & 13 & 9 & 21 \\ 2 & 5 & 2 & 8\end{array}\right]14231352925218 with row 111, what does row 222 become?
Correct answer: A
Use R2→R2−4R1R_2 \to R_2 - 4R_1R2→R2−4R1 to clear the leading 444.
(4−4, 13−12, 9−8, 21−20)=(0, 1, 1, 1)(4 - 4,\ 13 - 12,\ 9 - 8,\ 21 - 20) = (0,\ 1,\ 1,\ 1)(4−4, 13−12, 9−8, 21−20)=(0, 1, 1, 1)
So row 222 becomes [0111]\left[\begin{array}{ccc|c}0 & 1 & 1 & 1\end{array}\right][0111].
Which reduced bottom row signals a dependent system (infinitely many solutions), assuming the rows above it are consistent?
A fully zero row stands for 0=00 = 00=0, which removes one equation's worth of information and leaves a free variable.
[0000]⇒0=0\left[\begin{array}{ccc|c}0 & 0 & 0 & 0\end{array}\right] \Rightarrow 0 = 0[0000]⇒0=0
With the rows above consistent, that means infinitely many solutions. The 0=50 = 50=5 row would instead give no solution.
A triangular system gives the rows −z=2-z = 2−z=2, y+2z=1y + 2z = 1y+2z=1, and x−y+z=0x - y + z = 0x−y+z=0 (bottom to top). What is xxx?
The bottom row gives −z=2-z = 2−z=2, so z=−2z = -2z=−2. Then y+2z=1y + 2z = 1y+2z=1 gives y=1−2(−2)=5y = 1 - 2(-2) = 5y=1−2(−2)=5. Finally x−y+z=0x - y + z = 0x−y+z=0.
x=y−z=5−(−2)=7x = y - z = 5 - (-2) = 7x=y−z=5−(−2)=7
In row-echelon form, the pivots of successive rows must:
Row-echelon form requires the leading entry of each row to sit farther right than the leading entry of the row above, which creates the staircase.
pivot in row i+1 lies right of the pivot in row i\text{pivot in row } i + 1 \text{ lies right of the pivot in row } ipivot in row i+1 lies right of the pivot in row i
The pivots need not equal 111; that is the stricter reduced form.
The augmented matrix [1523156]\left[\begin{array}{cc|c}1 & 5 & 2 \\ 3 & 15 & 6\end{array}\right][1351526] describes a system that is:
Apply R2→R2−3R1R_2 \to R_2 - 3R_1R2→R2−3R1 to get (0, 0∣0)(0,\ 0 \mid 0)(0, 0∣0), a row of zeros.
[000]⇒0=0\left[\begin{array}{cc|c}0 & 0 & 0\end{array}\right] \Rightarrow 0 = 0[000]⇒0=0
The second equation was just the first tripled, so only x+5y=2x + 5y = 2x+5y=2 remains: infinitely many solutions.
Which condition would stop a matrix from being in row-echelon form?
Row-echelon form requires any all-zero rows to sit at the bottom. A zero row above a nonzero row breaks the staircase.
all-zero rows must come last\text{all-zero rows must come last}all-zero rows must come last
The other three conditions are exactly what echelon form demands.
For [125238]\left[\begin{array}{cc|c}1 & 2 & 5 \\ 2 & 3 & 8\end{array}\right][122358], what is the bottom row after R2→R2−2R1R_2 \to R_2 - 2R_1R2→R2−2R1?
Subtract 222 times row 111 from row 222.
(2−2, 3−4, 8−10)=(0, −1, −2)(2 - 2,\ 3 - 4,\ 8 - 10) = (0,\ -1,\ -2)(2−2, 3−4, 8−10)=(0, −1, −2)
So the bottom row is [0−1−2]\left[\begin{array}{cc|c}0 & -1 & -2\end{array}\right][0−1−2], meaning −y=−2-y = -2−y=−2 and y=2y = 2y=2.
Solve the system with augmented matrix [012511162117]\left[\begin{array}{ccc|c}0 & 1 & 2 & 5 \\ 1 & 1 & 1 & 6 \\ 2 & 1 & 1 & 7\end{array}\right]012111211567.
Swap R1R_1R1 and R2R_2R2 for a leading 111, giving rows (1,1,1∣6)(1, 1, 1 \mid 6)(1,1,1∣6), (0,1,2∣5)(0, 1, 2 \mid 5)(0,1,2∣5), (2,1,1∣7)(2, 1, 1 \mid 7)(2,1,1∣7). Then R3→R3−2R1R_3 \to R_3 - 2R_1R3→R3−2R1 gives (0,−1,−1∣−5)(0, -1, -1 \mid -5)(0,−1,−1∣−5), and R3→R3+R2R_3 \to R_3 + R_2R3→R3+R2 gives (0,0,1∣0)(0, 0, 1 \mid 0)(0,0,1∣0), so z=0z = 0z=0. Back-substitute: y+2z=5y + 2z = 5y+2z=5 gives y=5y = 5y=5.
x+y+z=6 ⇒ x+5+0=6 ⇒ x=1x + y + z = 6 \;\Rightarrow\; x + 5 + 0 = 6 \;\Rightarrow\; x = 1x+y+z=6⇒x+5+0=6⇒x=1
So (x,y,z)=(1,5,0)(x, y, z) = (1, 5, 0)(x,y,z)=(1,5,0).
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