12 multiple-choice questions, progressively harder.
Reducing [1152313]\left[\begin{array}{cc|c}1 & 1 & 5 \\ 2 & 3 & 13\end{array}\right][1213513] with R2→R2−2R1R_2 \to R_2 - 2R_1R2→R2−2R1 gives a bottom row that says:
Solution
Correct answer: A
Apply the operation to row 222: (2−2, 3−2, 13−10)=(0, 1, 3)(2 - 2,\ 3 - 2,\ 13 - 10) = (0,\ 1,\ 3)(2−2, 3−2, 13−10)=(0, 1, 3).
[013]⇒y=3\left[\begin{array}{cc|c}0 & 1 & 3\end{array}\right] \Rightarrow y = 3[013]⇒y=3
The matrix [123495]\left[\begin{array}{cc|c}1 & 2 & 3 \\ 4 & 9 & 5\end{array}\right][142935] became [12301−7]\left[\begin{array}{cc|c}1 & 2 & 3 \\ 0 & 1 & -7\end{array}\right][10213−7]. Which operation was used?
The first entry of row 222 went from 444 to 000, so 444 copies of row 111 (whose pivot is 111) were subtracted.
(4−4, 9−8, 5−12)=(0, 1, −7)(4 - 4,\ 9 - 8,\ 5 - 12) = (0,\ 1,\ -7)(4−4, 9−8, 5−12)=(0, 1, −7)
That matches R2→R2−4R1R_2 \to R_2 - 4R_1R2→R2−4R1.
A system in xxx, yyy, zzz reduces to the rows z=4z = 4z=4, y−z=1y - z = 1y−z=1, and x+y+z=10x + y + z = 10x+y+z=10 (bottom to top). What is yyy?
Correct answer: C
Start from the known value z=4z = 4z=4 and substitute into y−z=1y - z = 1y−z=1.
y−4=1 ⇒ y=5y - 4 = 1 \;\Rightarrow\; y = 5y−4=1⇒y=5
To turn the pivot of R2=[0−312]R_2 = \left[\begin{array}{cc|c}0 & -3 & 12\end{array}\right]R2=[0−312] into 111, multiply the row by:
Correct answer: D
The pivot is −3-3−3. Multiplying the whole row by −13-\tfrac{1}{3}−31 turns it into 111.
−13×(−3)=1-\tfrac{1}{3} \times (-3) = 1−31×(−3)=1
The row scales to [01−4]\left[\begin{array}{cc|c}0 & 1 & -4\end{array}\right][01−4].
While reducing a system you reach [132007]\left[\begin{array}{cc|c}1 & 3 & 2 \\ 0 & 0 & 7\end{array}\right][103027]. The system has:
Correct answer: B
The bottom row reads 0x+0y=70x + 0y = 70x+0y=7, that is 0=70 = 70=7, which is impossible.
[007]⇒0=7\left[\begin{array}{cc|c}0 & 0 & 7\end{array}\right] \Rightarrow 0 = 7[007]⇒0=7
So the system is inconsistent and has no solution.
With R1=[1213]R_1 = \left[\begin{array}{ccc|c}1 & 2 & 1 & 3\end{array}\right]R1=[1213] and R2=[3857]R_2 = \left[\begin{array}{ccc|c}3 & 8 & 5 & 7\end{array}\right]R2=[3857], after R2→R2−3R1R_2 \to R_2 - 3R_1R2→R2−3R1 what is the new entry in row 222, column 222?
Only row 222 changes. Its column-222 entry becomes 8−3(2)8 - 3(2)8−3(2).
8−6=28 - 6 = 28−6=2
Solve the system with augmented matrix [2151310]\left[\begin{array}{cc|c}2 & 1 & 5 \\ 1 & 3 & 10\end{array}\right][2113510].
The system is 2x+y=52x + y = 52x+y=5, x+3y=10x + 3y = 10x+3y=10. Swap to put the pivot 111 on top, then R2→R2−2R1R_2 \to R_2 - 2R_1R2→R2−2R1 gives (0, −5∣−15)(0,\ -5 \mid -15)(0, −5∣−15), so y=3y = 3y=3. Back-substitute into x+3y=10x + 3y = 10x+3y=10.
x+9=10 ⇒ x=1x + 9 = 10 \;\Rightarrow\; x = 1x+9=10⇒x=1
So (x,y)=(1,3)(x, y) = (1, 3)(x,y)=(1,3).
Multiplying an entire row of an augmented matrix by 12\tfrac{1}{2}21 is which kind of row operation?
Multiplying every entry of one row by the same nonzero number is the second row operation, scaling.
Ri→12RiR_i \to \tfrac{1}{2} R_iRi→21Ri
Since 12≠0\tfrac{1}{2} \ne 021=0, it is legal and reversible (multiply back by 222).
How many equations and unknowns does the augmented matrix [10230114]\left[\begin{array}{ccc|c}1 & 0 & 2 & 3 \\ 0 & 1 & 1 & 4\end{array}\right][10012134] represent?
Count rows for equations and coefficient columns (left of the bar) for unknowns.
2 rows, 3 coefficient columns⇒2 equations, 3 unknowns2 \text{ rows},\ 3 \text{ coefficient columns} \Rightarrow 2 \text{ equations},\ 3 \text{ unknowns}2 rows, 3 coefficient columns⇒2 equations, 3 unknowns
With more unknowns than equations, expect a free variable.
A triangular system gives the rows z=2z = 2z=2, y=−1y = -1y=−1, and x+y+z=6x + y + z = 6x+y+z=6. What is xxx?
Substitute y=−1y = -1y=−1 and z=2z = 2z=2 into x+y+z=6x + y + z = 6x+y+z=6.
x+(−1)+2=6 ⇒ x=5x + (-1) + 2 = 6 \;\Rightarrow\; x = 5x+(−1)+2=6⇒x=5
Multiply R2=[04−8]R_2 = \left[\begin{array}{cc|c}0 & 4 & -8\end{array}\right]R2=[04−8] by 14\tfrac{1}{4}41. The new row is:
Multiply every entry by 14\tfrac{1}{4}41.
(0, 44, −84)=(0, 1, −2)\left(0,\ \tfrac{4}{4},\ \tfrac{-8}{4}\right) = (0,\ 1,\ -2)(0, 44, 4−8)=(0, 1, −2)
So the row becomes [01−2]\left[\begin{array}{cc|c}0 & 1 & -2\end{array}\right][01−2].
A consistent system whose reduction produces a full row of zeros is called:
A zero row means one equation was a combination of the others, so it adds nothing. A consistent system like this has infinitely many solutions and is called dependent.
0=0⇒no new constraint0 = 0 \Rightarrow \text{no new constraint}0=0⇒no new constraint
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