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MTH5112 Linear Algebra I 2012–
Coursework 4 — Solutions
Exercise 1. Recall that if A = (aij ) 2 × 2 , then
det(A) = a 11 a 22 − a 21 a 12.
(a) If B is obtained from A by swapping row 1 and row 2, then
B =
a 21 a 22
a 11 a 12
so det(B) = a 21 a 12 − a 11 a 22 = −(a 11 a 22 − a 21 a 12 ) = − det(A).
(b) If B is obtained from A by multiplying row 1 by α, then
B =
αa 11 αa 12
a 21 a 22
so det(B) = αa 11 a 22 − αa 21 a 12 = α(a 11 a 22 − a 21 a 12 ) = α det(A).
(c) If B is obtained from A by adding r times row 2 to row 1, then
B =
a 11 + ra 21 a 12 + ra 22
a 21 a 22
so det(B) = (a 11 + ra 21 )a 22 − a 21 (a 12 + ra 22 ) = a 11 a 22 + ra 21 a 22 − a 21 a 12 − ra 21 a 22 =
a 11 a 22 − a 21 a 12 = det(A).
For 2 × 2 matrices, the only assertions in Theorem 3.11 that have not been proved by the above are
(d) if B is obtained from A by multiplying row 2 by α then det(B) = α det(A);
(e) if B is obtained from A by adding a multiple of row 1 to row 2, then det(B) = det(A).
Now, (d) follows by observing that B can also be obtained from A by first swapping row 1 and row
2, then multiplying row 1 by α and then swapping row 1 and row 2 again. Thus, by (a) and (b)
above, det(B) = (−1) · α · (−1) det(A) = α det(A).
Finally, (e) follows by observing that B can also be obtained from A by first swapping row 1 and
row 2, then adding a multiple row 2 to row 1 and then swapping row 1 and row 2 again. Thus, by
(a) and (c) above, det(B) = (−1)
2 det(A) = det(A).
Exercise 2.
(a) Since A is invertible
det(A) det(A
− 1 ) = det(AA
− 1 ) = det(I) = 1 ,
so det(A
− 1 ) = (det(A))
− 1 .
(b) By (a), we know that det(S
− 1 ) = (det(S))
− 1 , thus
det(B) = det(S
− 1 AS) = det(S
− 1 ) det(A) det(S) =
det(S)
det(S)
det(A) = det(A).
Exercise* 3.
(a) We have
det(A) =
= R 2 − R 1
R 3 + R 1
so A is invertible.
(b) The (3, 2)-cofactor of A is
C 32 = (−1)
3+ det(A 32 ) = −
so the (2, 3)-entry of A
− 1 is
C 32
det(A)
(c) The system can be written in the form Ax = b, where
A =
(^) , x =
x
y
z
(^) , and b =
By Cramer’s rule
y =
det(A 2 (b))
det(A)
But
det(A 2 (b)) =
= R 2 − R 1
R 3 + R 1
and det(A) = − 3 as before, so
y =
Exercise 4.
(a) Recall that if A is a square matrix, and B is the matrix obtained from A by multiplying a row of
A by α, then det(B) = α det(A). Suppose now that A is an n×n matrix. Now, the matrix αA
is obtained from A by multiplying each of the n rows of A by α, thus det(αA) = α
n det(A).
(b) It follows from (a) that det(2In) = 2
n det(In) and hence that
det(In + In) = 2
n 6 = 2 = det(In) + det(In)
whenever n > 1. Thus, det(A + B) cannot equal det(A) + det(B) in general.
Exercise* 5.
(a) True. If A and B are any two square matrices of the same size, then
(A + B
T )
T = A
T
T )
T = A
T
But since the determinant of a matrix equals that of its transpose we have
det(A + B
T ) = det((A + B
T )
T ) = det(A
T
(b) False. For example, if A = O, then adj(A) = O, but A is not invertible. In fact, the adjugate
of a matrix always exists (irrespective of whether the matrix is invertible or not).
(c) True. Suppose that BA = CA. If det(A) 6 = 0, then A is invertible, and it follows that
B = BI = BAA
− 1 = CAA
− 1 = CI = C.
