MATH 2065 Midterm: MATH 2065 LSU f18ex4a
Name: Solutions Exam 4
Instructions. Answer each of the questions on your own paper. Be sure to show your
work so that partial credit can be adequately assessed. Credit will not be given for answers
(even correct ones) without supporting work. Put your name on each page of your paper. A
table of Laplace transforms is appended to the exam.
1. [10 Points] Let A=
2 3
1−1
−3 1
and B=1 2
−3 1. Compute each of the following
matrices, if it exists. If it does not exist, explain why.
(a) AB (b) BA (c) A2(d) B2
◮Solution. (a) AB =
−7 7
4 1
−6−5
(b) BA does not exist since the number of columns of Bis 2, which is different from
the number of rows of A.
(c) A2does not exist since Ais not square.
(d) B2=−5 4
−6−5◭
2. [15 Points] Find all solutions to the linear system
x1−6x2−4x3=−5
2x1−10x2−9x3=−4
−x1+ 6x2+ 5x3= 3
Use Gauss-Jordan elimination.
◮Solution. The augmented matrix corresponding to the given system is
1−6−4−5
2−10 −9−4
−1 6 5 3
.
Apply Gauss-Jordan elimination to the augmented matrix:
1−6−4−5
2−10 −9−4
−1 6 5 3
−2R1+R2→R2
−→
R1+R−3→R3
1−6−4−5
0 2 −1 6
001−2
3R2+R1→R1
−→
R3+R2→R2
1 0 −7 13
0 2 0 4
0 0 1 −2
7R3+R1→R1
−→
(1/2)R2→R2
1 0 0 −1
0 1 0 2
0 0 1 −2
Math 2065 Section 2 November 26, 2018 1
Name: Solutions Exam 4
This matrix is in reduced row echelon form and the solution of the original system can
be read off as
x1
x2
x3
=
−1
2
−2
.
◭
3. [15 Points] Let A=
1−2 0
2−1 1
0 4 2
.
(a) Compute det A.
◮Solution. Use cofactor expansion along the first row:
det A= 1 ·det −1 1
4 2+ (−1)1+2(−2) det 2 1
0 2+ (−1)1+30·det 2−1
0 4
= ((−1)2 −1·4) + 2 ·(2 ·2−0·1) + 0 ·(2 ·4−0(−1))
=−6 + 8 = 2.
◭
(b) Compute the inverse of A.
◮Solution. Use row reduction of the augmented matrix [A I]:
A I=
1−20101
2−11010
0 4 2 0 0 1
(−2)R1+R2→R2
−→
1−2 0 1 0 0
0 3 1 −2 1 0
0 4 2 0 0 1
−R2+R3→R3
−→
1−2 0 1 0 0
0 3 1 −2 1 0
0 1 1 2 −1 1
−3R3+R2→R2
−→
2R3+R1→R1
1 0 2 5 −2 2
0 0 −2−8 4 −3
0 1 1 2 −1 1
R2↔R3
−→
1 0 2 5 −2 2
0 1 1 2 −1 1
0 0 −2−8 4 −3
(1/2)R3
−→
1 0 2 5 −2 2
0 1 1 2 −1 1
0 0 1 4 −23
2
−R3+R2→R2
−→
22R3+R1→R1
1 0 0 −3 2 −1
0 1 0 −2 1 −1
2
0 0 1 4 −23
2
Thus,
A−1=
−3 2 −1
−2 1 −1
2
4−23
2
.
◭
Math 2065 Section 2 November 26, 2018 2
Document Summary
Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. Put your name on each page of your paper. A table of laplace transforms is appended to the exam: [10 points] let a = . And b = (cid:20) 1 2 matrices, if it exists. If it does not exist, explain why. (a) ab (b) ba (c) a2 (d) b2. 6 5 (b) ba does not exist since the number of columns of b is 2, which is di erent from the number of rows of a. (c) a2 does not exist since a is not square. (d) b2 =(cid:20) 5. 4: [15 points] find all solutions to the linear system x1 6x2 4x3 = 5. The augmented matrix corresponding to the given system is.
