Page 334 - 35Linear Algebra
P. 334
334 Sample Second Midterm
Solutions
1. (a) Whenever detM = ad − bc 6= 0.
(b) Unit determinant bit matrices:
1 0 1 1 1 0 0 1 1 1 0 1
, , , , , .
0 1 0 1 1 1 1 0 1 0 1 1
(c) Bit matrices with vanishing determinant:
0 0 1 0 0 1 0 0 0 0
, , , , ,
0 0 0 0 0 0 1 0 0 1
1 1 0 0 1 0 0 1 1 1
, , , , .
0 0 1 1 1 0 0 1 1 1
As a check, count that the total number of 2×2 bit matrices is 2 (number of entries) =
4
2 = 16.
(d) To disprove this statement, we just need to find a single counterexam-
ple. All the unit determinant examples above are actually row equiva-
lent to the identity matrix, so focus on the bit matrices with vanishing
determinant. Then notice (for example), that
1 1 0 0
∼/ .
0 0 0 0
So we have found a pair of matrices that are not row equivalent but
do have the same determinant. It follows that the statement is false.
2.
detA = 1.(2.6 − 3.5) − 1.(2.6 − 3.4) + 1.(2.5 − 2.4) = −1 .
(i) Since detA 6= 0, the homogeneous system AX = 0 only has the solution
X = 0. (ii) It is efficient to compute the adjoint
T
−3 0 2 −3 −1 1
adj A = −1 2 −1 = 0 2 −1
1 −1 0 2 −1 0
Hence
3 1 −1
A −1 = 0 −2 1 .
−2 1 0
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