Linear Algebra Exercises-n-Answers.pdf

160 Linear Algebra, by Hefferon
and conclude that the matrix is nonsingular if and only if either ah − bg = 0 or af − cd = 0. The
condition ‘ah−bg = 0 or af −cd = 0’ is equivalent to the condition ‘(ah−bg)(af −cd) = 0’. Multiplying
out and using the case assumption that ae − bd = 0 to substitute ae for bd gives this.
0 = ahaf − ahcd − bgaf + bgcd = ahaf − ahcd − bgaf + aegc = a(haf − hcd − bgf + egc)
Since a ≠ 0, we have that the matrix is nonsingular if and only if haf −hcd−bgf +egc = 0. Therefore,
in this a ≠ 0 and ae−bd = 0 case, the matrix is nonsingular when haf −hcd−bgf +egc−i(ae−bd) = 0.
The remaining cases are routine. Do the a = 0 but d ≠ 0 case and the a = 0 and d = 0 but g ≠ 0
case by first swapping rows and then going on as above. The a = 0, d = 0, and g = 0 case is easy —
that matrix is singular since the columns form a linearly dependent set, and the determinant comes
out to be zero.
Four.I.1.10
Figuring the determinant and doing some algebra gives this.
0 = y 1 x + x 2 y + x 1 y 2 − y 2 x − x 1 y − x 2 y 1
(x 2 − x 1 ) · y = (y 2 − y 1 ) · x + x 2 y 1 − x 1 y 2
y = y 2 − y 1
· x + x 2y 1 − x 1 y 2
x 2 − x 1 x 2 − x 1
Note that this is the equation of a line (in particular, in contains the familiar expression for the slope),
and note that (x 1 , y 1 ) and (x 2 , y 2 ) satisfy it.
Four.I.1.11 (a) The comparison with the formula given in the preamble to this section is easy.
(b) While it holds for 2×2 matrices
( )
h1,1 h 1,2 h 1,1
= h
h 2,1 h 2,2 h 1,1 h 2,2 + h 1,2 h 2,1
2,1
−h 2,1 h 1,2 − h 2,2 h 1,1
= h 1,1 h 2,2 − h 1,2 h 2,1
it does not hold for 4×4 matrices. An example is that this matrix is singular because the second
and third rows are equal
⎛
⎞
1 0 0 1
⎜ 0 1 1 0
⎟
⎝ 0 1 1 0⎠
−1 0 0 1
but following the scheme of the mnemonic does not give zero.
⎛
⎞
1 0 0 1 1 0 0
⎜ 0 1 1 0 0 1 1
⎟
⎝ 0 1 1 0 0 1 1⎠ = 1 + 0 + 0 + 0
−(−1) − 0 − 0 − 0
−1 0 0 1 −1 0 0
Four.I.1.12 The determinant is (x 2 y 3 − x 3 y 2 )⃗e 1 + (x 3 y 1 − x 1 y 3 )⃗e 2 + (x 1 y 2 − x 2 y 1 )⃗e 3 . To check perpendicularity,
we check that the dot product with the first vector is zero
⎛
⎝ x ⎞ ⎛
1
x 2
⎠ ⎝ x ⎞
2y 3 − x 3 y 2
x 3 y 1 − x 1 y 3
⎠ = x 1 x 2 y 3 − x 1 x 3 y 2 + x 2 x 3 y 1 − x 1 x 2 y 3 + x 1 x 3 y 2 − x 2 x 3 y 1 = 0
x 3 x 1 y 2 − x 2 y 1
and the dot product with the second vector is also zero.
⎛
⎝ y ⎞ ⎛
1
y 2
⎠ ⎝ x ⎞
2y 3 − x 3 y 2
x 3 y 1 − x 1 y 3
⎠ = x 2 y 1 y 3 − x 3 y 1 y 2 + x 3 y 1 y 2 − x 1 y 2 y 3 + x 1 y 2 y 3 − x 2 y 1 y 3 = 0
y 3 x 1 y 2 − x 2 y 1
Four.I.1.13
(a) Plug and chug: the determinant of the product is this
( ) ( ) ( )
a b w x aw + by ax + bz
det(
) = det(
)
c d y z
cw + dy cx + dz
= acwx + adwz + bcxy + bdyz
−acwx − bcwz − adxy − bdyz
while the product of the determinants is this.
( ) ( )
a b w x
det( ) · det( ) = (ad − bc) · (wz − xy)
c d y z
Verification that they are equal is easy.