MAT 242 Test 1 SOLUTIONS, FORM O 4. 1i5 pointel How many solutions does each of
ID: 3145215 • Letter: M
Question
MAT 242 Test 1 SOLUTIONS, FORM O 4. 1i5 pointel How many solutions does each of the following systens of inear equations haye rele the entries which led you to your conclusion. 1 1031-21 0 1 0 13 L00011-3J r13 3 31 2 0013-3 10 0 0 010 1 1 -3 1 21 1 00 0 00 -2 b. 00010 Solution: Correct answers are given above, and the relevant entzties are in boldface, Grading t+3 points for the correct answer, t2 points for indicating the entries (or for a brief explanation). Crading for common errors. 3 points for a Wrong answer fm points for not indi cating the relevant entries l points for many TOWExplanation / Answer
(a) since the matrix is in Reduced Row-Echelon Form (rref), and all leading entries are zero (which is as showh by bold), therefore, there will only be one solution as per the properties of rref.
We can also understand this by formulating the 4 equations from 4 rows of matrix:
from last row: (0)x1+(0)x2+(0)x3+(1)x4 = -3 which gives x4=-3
from 3rd row: (0)x1+(0)x2+(1)x3+(0)x4 = 0 which gives x3=0
and similarly going on , which in end will provide unique solution./
And remember, all the leading one's are the main clue here.
(b) since here the second column has only the top entry as non-zero, so it signifies that it will have many solutions since here, a time will come where there will be 1 equation and 2 variables remaining which means that whole line will be solution or infinite(many) solutions.
from last row: 0=0
from 3rd row: (0)x1+(0)x2+(0)x3+(1)x4 = 0 which gives x4=0
from 2nd row: (0)x1+(0)x2+(1)x3+(-3)x4 = -3 which gives x3=-3
from 1st row: (1)x1+(3)x2+(3)x3+(-3)x4 = -2 which gives x1 + 3x2 = 7 which is 1 equation and 2 variables that I have mentiones above.
(c) The last row of this matrix signifies (0)x1+(0)x2+(0)x3+(0)x4 = -2 which means 0=-2 which is not possible and hence this system of equation has no solution.
Remember that here last row is the main clue.
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