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# AUMAT120-Winter2014-Test1-Solutions.pdf

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University of Alberta

Mathematics

MATH102

Ian Blokland

Winter

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AUMAT 120 Term Test 1
Linear Algebra I February 5, 2014.
WINTER 2014 8:00–8:50 AM.
Instructor: J. Sylvestre — SOLUTIONS —
PART A — True or False
[1 mark each]
A1. If a matrix is row reduced to reduced row-echelon form by two different sequences of elementary
row operations, the resulting matrices might be different.
Solution. FALSE. RREF is unique. ▯
A2. If A is a matrix with rank 4, then the homogeneous system Ax˘0 requires 4 parameters.
Solution. FALSE. Rank tells you the number of variables that don’t need to become parameters.
▯
A3. If A and B are n£n matrices such that AB ˘0, then either A ˘0 or B ˘0.
Solution. FALSE. It is not too difﬁcult to come up with an example (size 2£2, say) of nonzero
matrices A and B so that AB evaluates to the zero matrix. ▯
A4. It is always true that A¯B ˘ B¯ A for every m£n matrices A and B.
Solution. TRUE. ▯
¡1 ¡1
A5. If A is an invertible square matrix, th)n ˘ A.
Solution. TRUE. ▯
A6. Every square matrix can be expressed as a product of elementary matrices.
Solution. FALSE. Only invertible square matrices can be expressed as a product of elementary
matrices. ▯
2 0 0 33
4 5
A7. The matrix 0 0 1 is symmetric.
¡3 ¡1 0
Solution. FALSE. This matrix is not equal to its transpose. ▯
PART B — Long Answer
[4 marks]B1. The augmented matrix below is in row-echelon form. Continue row-reducing until reduced row-
echelon form is reached. Then solve the corresponding system in the un1no2ns3x4,x5,x ,x ,x .
2 1 ¡1 1 ¡2 ¡2 3 3
6 0 1 2 1 0 1 7
4 0 0 0 1 1 ¡1 5
0 0 0 0 0 0
2 3
1 0 3 0 ¡1 3
6 0 1 2 0 ¡1 2 7
Solution. Three row operations will get you to 4R0F:0 0 1 1 ¡1 5.
0 0 0 0 0 0
The rank is 3 so we need two parameters. The third and ﬁfth columns do not contain a leading
one, so make these into parameters:5let x ˘3t and x ˘ s.
Page 1 of 3 Turning the rows back into equations, we get parametric solution (written in reverse order, as we
would solve it):
x5˘ t, x4˘¡1¡t, x3˘ s, x2˘2¡2s¯t, x1˘3¡3s¯t. ▯
2 3
0 1 0
[5 marks]B2. Let A ˘ 40 0 15 .
1 0 0
[1] (a) Is A an elementary matrix?
Solution. No, it would take more than one row swap to get it back to the identity▯matrix.
2 3
[2] (b) Compute A and A .
2 3 2 3
0 0 1 1 0 0
Solution. A ˘ 4 1 0 05, A ˘ 4 0 1 05. ▯
0 1 0 0 0 1
[2] (c) Recall that the inverse of A, if it exists, is the unique 3£3 matrix such˘ I3andA
¡1
A A ˘ I3, where3I is the 3£3 identity matrix.
Without further calculation, use your answers to part (b) to determine A and (A )1.
3 2 2
Solution. Since A ˘ I, we can write both AA ˘ I and A A ˘ I. By the uniqueness of the
inverse, this means that A˘ A .
¡ 1 T T
Then also (A ) ˘(A ¡1) ˘(A ) , which we notice is actually equal to A. ▯
[3 marks]B3. Use matrix algebra to isolate A in the following matrix equation. Assume that the matrices are
all square, are all of the same size, and are all invertible.
2 ¡1 ¡3 T ¡1
BC B AD C B ˘ BC
Answer. You need to “cancel” matrices off the side that contains A by multiplying both sides by an
appropriate inverse matrix. Take care about order of multiplication! Below we carry out two steps
in this process. (Though you don’t necessarily need to start with these two steps in this order.)
B¡1BC B ¡1AD ¡3C B ¡1˘ B¡1BC
IC B ¡1AD ¡3C B ¡1˘ IC

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