Download Linear Algebra 1 Exam 2 Solutions 7/14/3

Linear Algebra 1 Exam 2 Solutions 7/14/3
Question 1
The line L has the symmetric equation:
x−2
y+3
z−4
=
=
.
1
2
−2
The line M has the parametric equation:
[x, y, z] = [−4, 10, 5] + s[10, −7, −2].
The line N is perpendicular to both the lines L and M and passes through
the point [2, −1, −3].
• Find the parametric and symmetric equations of the line N .
A direction vector for N is the cross product of the direction vectors
of L and M :
i
j
k
n = [1, 2, −2] × [10, −7, −2] = det 1 2 −2
10 −7 −2
= i(−4 − 14) − j(−2 + 20) + k(−7 − 20)
= −18i − 18j − 27k.
Dividing by −9, we may take the direction vector to be [2, 2, 3].
Then the parametric and symmetric equations of the line N are:
[x, y, z] = [2, −1, −3] + t[2, 2, 3] = [2 + 2t, −1 + 2t, −3 + 3t],
x−2
y+1
z+3
=
=
.
2
2
3
• Show that the lines L and N meet and find the intersection point P .
We put the parametric equation of N into the equation of L and solve
for t:
2 + 2t − 2
−1 + 2t + 3
−3 + 3t − 4
=
=
.
1
2
−2
The first equation gives: 2t = t + 1, so t = 1 and when t = 1, these
equations hold.
So L and N meet at the point with t = 1, so at the point P = [4, 1, 0].
• Show that the lines M and N meet and find the intersection point Q.
We put the parametric equation of M into the symmetric equation
of N and solve for s:
10 − 7s + 1
5 − 2s + 3
−4 + 10s − 2
=
=
.
2
2
3
The first equation gives: 10s − 6 = 11 − 7s, so s = 1 and when s = 1,
these equations hold.
So L and N meet at the point with s = 1, so at the point Q = [6, 3, 3].
• Calculate the distance P Q.
The distance P Q is :
p
√
√
P Q = (6 − 4)2 + (3 − 1)2 + (3 − 0)2 = 4 + 4 + 9 = 17.
Question 2
Let A = (2, 3, 5), B = (1, −1, 6), C = (2, 2, 4).
• Find the equation of the plane ABC.
Two vectors in the plane are :
AB = B − A = (1, −1, 6) − (2, 3, 5) = (−1, −4, 1),
AC = C − A = (2, 2, 4) − (2, 3, 5) = (0, −1, −1).
So a normal, n, to the plane is:
i
j
k
n = AB × AC = det −1 −4 1
0 −1 −1
= i(4 + 1) − j(1 − 0) + k(1 − 0)
= 5i − j + k
So the equation of the plane is:
[5, −1, 1].[x, y, z] = [5, −1, 1].[2, 3, 5] = 10 − 3 + 5 = 12,
5x − y + z = 12.
We may check that A, B and C all lie on this plane, as required.
• Find the area of the triangle ABC.
The triangle has area:
√
1√
3 3
1
25 + 1 + 1 =
.
|n| =
2
2
2
• How far is the point D = (2, −4, 7) from the plane ABC?
We need the component of AD in the direction of n:
√
1
1
([2, −4, 7] − [2, 3, 5]).[5, −1, 1] = √ (21 − 12) = 3.
|n|
27
√
So the point D is at a distance from the plane of 3 units of distance.
AD.n̂ =
• Find the parametric equation of the line L through D perpendicular to
the plane ABC.
The vector n is a direction vector for the line L, so the line L has the
parametric equation:
[x, y, z] = [2, −4, 7] + t[5, −1, 1] = [2 + 5t, −4 − t, 7 + t].
• Find the point where the line L meets the plane ABC.
We insert the formulas for x, y and z for the line L into the equation of the plane and solve for t:
5(2 + 5t) − (−4 − t) + 7 + t = 12,
21 + 27t = 12,
1
t=− .
3
So the required point is [2 − 35 , −4 + 13 , 7 − 13 ] = 13 [1, −11, 20].
It is easy to check that this point does indeed lie on the plane, as
required.
Question 3
Which of the following subsets of vector spaces are subspaces.
Explain your answers:
• The span S of the vectors [2, 3, −1] and [4, 6, 1] in R3 .
The span of any subset of a vector space is a subspace and this is
no exception!
• The intersection of S with the plane in R3 with equation:
x + 4y + 9z = 25.
A subspace must contain the zero vector, but x = y = z = t = 0 does
not satisfy the given equation.
So the given intersection does not contain the zero vector, so is not a
subspace.
• The set T of all polynomials p(x) in P (the vector space of all polynomials in the variable x) such that only even powers of x occur in p, i.e.
p(x) = p(−x).
If p(x) is a sum of multiples of even powers of x, then so is sp(x)
for any scalar s, since multiplying by a scalar cannot change an even
power to an odd power.
More formally, we have:
(sp)(x) − (sp)(−x) = sp(x) − sp(−x) = s(p(x) − p(−x)) = 0.
So the polynomial sp is even, if p is, for any scalar s.
So the set T is closed under scalar multiplication.
Next, if p(x) and q(x) are each sums of multiples of even powers of
x, then so is (p + q)(x) = p(x) + q(x) since addition of polynomials
cannot change an even power to an odd power.
More formally, we have:
(p + q)(x) − (p + q)(−x) = p(x) + q(x) − (p(−x) + q(−x))
= (p(x) − p(−x)) + (q(x) − q(−x)) = 0 + 0 = 0.
So the polynomial p + q is even, if p and q are.
So the set T is closed under addition.
So the set T is a subspace of P.
• The set U of all polynomials p(x) in P3 (the vector space of all polynomials in the variable x of degree no more than three), such that
p(0) = p(1).
If we write p(x) = ax3 +bx2 +cx+d, then p(0) = d and p(1) = a+b+c+d.
So the condition p(0) = p(1) gives the equation a + b + c = 0, so
a = −b − c.
Then p(x) = (−b − c)x3 + bx2 + cx + d = d(1) + b(x2 − x3 ) + c(x − x3 ),
where b, c and d are arbitrary.
So the space U is the span of the polynomials 1, x2 − x3 and x − x3 , so
is a subspace.
• The intersection T ∩ U.
The sets T and U may each be regarded as subspaces of P and the
intersection of subspaces of a vector space is itself a subspace, so T ∩ U
is a subspace of P and of P3 .
Question 4
Calculate the determinant of the following matrix B(t):
B(t) =
2−t
2
−1 5 − t
det(B(t)) = (2−t)(5−t)−(−1)(2) = t2 −7t+10+2 = t2 −7t+12 = (t−3)(t−4).
• Find the values of t for which the matrix B(t) has no inverse.
The matrix has no inverse iff it has zero determinant, iff t = 3 or
t = 4.
• For which values of t does the homogeneous system with coefficient
matrix B(t) have a non-trivial solution?
The homogeneous system with coefficient matrix B(t) has a non-trivial
solution iff B(t) has rank less than two, iff B(t) has no inverse, iff
det(B(t)) = 0, iff t = 3 or t = 4.
Let these values be t = t1 and t = t2 .
We put t1 = 3 and t2 = 4.
• Find a non-trivial solution X1 of the homogeneous system with coefficient matrix B(t1 ).
We have:
B(t1 ) = B(3) =
−1 2
−1 2
→
1 −2
.
0 0
The reduced system is x − 2y = 0.
So we may pick y = s and then x = 2s.
So the general solution is [x, y] = s[2, 1], with s arbitrary.
Taking s = 1, we may take X1 = [2, 1].
• Find a non-trivial solution X2 of the homogeneous system with coefficient matrix B(t2 ). We have:
B(t2 ) = B(4) =
−2 2
−1 1
→
1 −1
.
0 0
The reduced system is x − y = 0.
So we may pick y = t and then x = t.
So the general solution is [x, y] = t[1, 1], with t arbitrary.
Taking t = 1, we may take X2 = [1, 1].
• Show that X1 and X2 are linearly independent.
If pX1 + qX2 = 0, then we have:
p[2, 1] + q[1, 1] = [2p + q, p + q] = [0, 0].
So 2p + q = 0 and p + q = 0.
Subtracting these equations gives p = 0.
Back substitution into the second equation gives q = 0.
So p = q = 0 and the vectors X1 and X2 are linearly independent.
• Show that X1 and X2 span R2 .
We see that e1 = [1, 0] = X1 − X2 and e2 = [0, 1] = −X1 + 2X2 ,
so both e1 and e2 lie in the span of X1 and X2 .
So all vectors in R2 lie in the span of X1 and X2 , as required:
Explicitly, we have for any [x, y] in R2 :
[x, y] = xe1 +ye2 = x(X1 −X2 )+y(−X1 +2X2 ) = (x−y)X1 +(2y−x)X2 .
Check:
(x − y)X1 + (2y − x)X2 = (x − y)[2, 1] + (2y − x)[1, 1]
= [2x − 2y, x − y] + [2y − x, 2y − x] = [x, y]
Question 5
Let A, B and C be the following polynomials in P2 :
A = x2 − 3x + 2,
B = x2 − 6x + 8,
C = x2 − ax + 4.
Here a is a constant.
• Find the value of a such that A,B and C are linearly dependent.
We want, for scalars p, q and r:
0 = pA + qB + rC = p(x2 − 3x + 2) + q(x2 − 6x + 8) + r(x2 − ax + 4)
= x2 (p + q + r) + x(−3p − 6q − ar) + 2p + 8q + 4r.
So we obtain the linear system:
p + q + r = 0,
−3p − 6q − ar = 0,
2p + 8q + 4r = 0.
The first equation gives p = −q − r.
Substituting in the third equation, we get 0 = −2q − 2r + 8q + 4r =
6q + 2r.
So r = −3q and then p = −q − r = −q + 3q = 2q.
Substituting in the second equation of our system, we get the remaining
equation:
0 = −3(2q) − 6q − a(−3q) = −12q + 3aq = 3q(a − 4).
If a 6= 4, then q = 0 and then p = q = r = 0 and the polynomials are
linearly independent.
If a = 4, the last equation is satisfied and we have [p, q, r] = s[2, 1, −3],
with s arbitrary.
In particular, we have 2A + B − 3C = 0, when a = 4 and the polynomails are linearly dependent, as required.
So the required value of a is 4.
• For that value of a, find a non-trivial linear relation between the polynomials A, B and C.
As shown above such a relation is 2A + B − 3C = 0.
• Show that the polynomial D = x2 +7x−18 lies in the span of {A, B, C}.
The span of {A, B, C} equals the span of {A, B}, since C is a combination of A and B.
So it is sufficient to write D in terms of A and B.
So we need: x2 + 7x − 18 = p(x2 − 3x + 2) + q(x2 − 6x + 8).
Put x = 4 and B vanishes, so we get:
16 + 28 − 18 = 26 = p(16 − 12 + 2) = 6p.
Put x = 1 and A vanishes, so we get:
1 + 7 − 18 = −10 = q(1 − 6 + 8) = 3q.
So (p, q) = ( 13
, −10
).
3
3
Check:
13
10
1
A − B = (13x2 − 39x + 26 − 10x2 + 60x − 80)
3
3
3
1
= (3x2 + 21x − 54) = x2 + 7x − 18 = D.
3
• Do the polynomials A, B, C span P2 ?
Explain your answer.
When a = 4 all the polynomials A, B and C have the factor x − 2
so any polynomial in their span also has that factor.
But, for example, the polynomial x has no factor of x − 2, so cannot
lie in the span of {A, B, C}, so the given polynomials do not span P2 .
When a 6= 4, the three polynomials do span:
A = (x−2)(x−1),
B = (x−2)(x−4),
C = x2 −ax+4 = (x−2)2 +(4−a),
2A + B − 3C = 3(a − 4),
Since 3(a − 4) is non-zero, 1 lies in Span{A, B, C}.
Then D = x2 − 3x = A − 2(1) and E = x2 − 6x = B − 8(1) lie in
Span{A, B, C}.
Then 2D − E = x2 and 31 (D − E) lie in Span{A, B, C}.
Since 1, x and x2 lie in Span{A, B, C} and since Span{1, x, x2 } = P2 ,
we see that Span{A, B, C} = P2 , whenever a 6= 4.
Question 6
Let A, B, C and D be the following 2 × 2 matrices:
A =
1 0
,
2 1
−2 1
,
−3 −1
B =
C =
2 −3
,
4 −1
1 3
.
−1 4
D =
• Prove that A, B, C and D are linearly dependent and find a non-trivial
linear relation between them.
We solve the system:
xA + yB + zC + tD = 0,
0 0
0 0
x − 2y + 2z + t
y − 3z + 3t
.
2x − 3y + 4z − t x − y − z + 4t
=
We reduce as follows:
1 −2 2
1
0 1 −3 3
2 −3 4 −1
1 −1 −1 4
1 −2 2
1
0 0 −3 6
0 1
0 −3
0 0
0
0
1
0
0
0
→
1 −2 2
1
0 1 −3 3
0 1
0 −3
0 1 −3 3
→
1
0
0
0
0
0
1
0
0 −1
1 −2
0 −3
0 0
0 2 −5
0 −3 6
1 0 −3
0 0
0
→
1
0
0
0
1 −2 2
1
0 1 −3 3
0 1
0 −3
0 0
0
0
→
→
0
1
0
0
1
0
0
0
0
0
1
0
2 −5
1 −2
0 −3
0 0
→
0 −1
0 −3
.
1 −2
0 0
The reduced system is:
x − t = 0,
y − 3t = 0,
z − 2t = 0.
We pick t = s and then [x, y, z, t] = s[1, 3, 2, 1], with s arbitrary.
Picking s = 1, we have the non-trivial relation, showing that the set of
matrices {A, B, C, D} is a linearly dependent set:
A + 3B + 2C + D = 0.
Also note that D = −A − 3B − 2C.
So the span of {A, B, C, D} equals the span of {A, B, C}.
→
• Write down a set of four linearly independent 2 × 2 matrices that together span all of M22 .
A standard such set is {E, F, G, H}, where:
E =
1 0
,
0 0
F =
0 1
,
0 0
G =
xE + yF + zG + tH =
0 0
,
1 0
H =
0 0
,
0 1
x y
.
z t
The last equation shows that {E, F, G, H} spans M22 , since x, y, z and
t are arbitrary.
The same equation shows that if xE + yF + zG + tH = 0, then x =
y = z = t = 0.
So the set of matrices {E, F, G, H} is a linearly independent set that
spans M22 , as required.
• Hence, or otherwise, prove that the span of {A, B, C, D} cannot be all
of M22 .
If we suppose that {A, B, C, D} does span M22 , then each of E, F, G, H
is expressible as as a linear combination of {A, B, C, D}.
If we write the four linear systems for these linear combinations in one
augmented matrix, we get the following augmented matrix Z :
Z=
1 −2 2
1 1
0 1 −3 3 0
2 −3 4 −1 0
1 −1 −1 4 0
0
1
0
0
0
0
1
0
0
0
0
1
But this is also the augmented matrix for the linear system Y X = I,
with coefficient matrix Y the matrix given by the first four columns of
Z.
If a solution X exists, then det(Y ) det(X) = 1, so det(Y ) 6= 0 and Y is
invertible.
But we reduced the matrix Y above to a matrix with one zero row. So
Y has no inverse and the solution X cannot exist.
So at least one of the matrices E, F, G, H does not lie in Span{A, B, C, D},
so Span{A, B, C, D} =
6 M22 , as required.
Alternatively we may notice that for each of the matrices A, B, C
and D, the sum of the (11) entry and the (12) entry always equals the
(22) entry.
Then any matrix in their span must have the same property.
Since not every matrix has that property, they cannot span all of M22 .
For example the following matrix does not have the property, so does
not lie in Span{A, B, C, D}:
1 1
1 1