MATH1151 Algebra Test 2 2008 S1 v1AJanuary 28, 2015

These solutions were written and typed up by Gary Liang. Please be ethical with this resource.

It is for the use of MATHSOC members, so do not repost it on other forums or groups without

asking for permission. If you appreciate this resource, please consider supporting us by coming

to our events and buying our T-shirts! Also, happy studying :)

We cannot guarantee that our working is correct, or that it would obtain full marks - please

notify us of any errors or typos at unswmathsoc@gmail.com, or on our Facebook page. There

are sometimes multiple methods of solving the same question. Remember that in the real class

test, you will be expected to explain your steps and working out.

1. A matrix is invertible if and only if the determinant is not zero. So let’s find the determi-

nant.

det(A) = 1× det

(−2 −1

3 0

)− 1×

(2 −1

a 0

)+ a×

(2 −2

a 3

)= (−2× 0− (−1)× 3)− (2× 0− (−1)× a) + (a× (2× 3− (−2)× a)

= 3− a+ a(6 + 2a)

= 3 + 5a+ 2a2

1

Hence, the matrix is invertible if:

3 + 5a+ 2a2 6= 0

(2a+ 3)(a+ 1) 6= 0

a 6= −1,−3

2

2. (i) We should know that the cross product of two vectors is perpendicular to the two

vectors themselves. Hence, we want to find a× b.

To make it easy for ourselves, we should remember that

a× b =

∣∣∣∣∣∣∣e1 e2 e3

a1 a2 a3

b1 b2 b3

∣∣∣∣∣∣∣So

a× b =

∣∣∣∣∣∣∣e1 e2 e3

−1 4 2

−2 −1 3

∣∣∣∣∣∣∣= e1(4× 3− 2× (−1))− e2((−1)× 3− 2× (−2)) + e3((−1)× (−1)− 4× (−2))

=

14

0

0

−0

1

0

+

0

0

9

=

14

−1

9

(ii) We know that the projection of c on to a is given by

projac =

(a · c|a|2

)a

.

2

Hence,

projac =

(−142

)·(

8−5−7

)12 + 42 + 22

×

−1

4

2

= −2

−1

4

2

3. If the points were all on the line, we would have:

α+ β = 3

α+ 2β = −1

α+ 4β = 2

α+ 5β = 0

or: 1 1

1 2

1 4

1 5

(α

β

)=

3

−1

2

0

Let A =

1 1

1 2

1 4

1 5

, x =

(α

β

)and y =

3

−1

2

0

.

We should know that the least squares approximate solution x0 to Ax = y is the solution

to the normal equations

ATAx = ATy

3

Hence we want to solve

(1 1 1 1

1 2 4 5

)1 1

1 2

1 4

1 5

x =

(1 1 1 1

1 2 4 5

)3

−1

2

0

(

4 12

12 46

)x =

(4

9

)

If we solve this by row reduction and back substitution (hopefully you know how to do

this!), you should get α = 1910 and β = − 3

10 .

4. By reading the coefficients, we know that a vector normal to the plane is(

1−21

). From

this, we can tell that the line going through the point A perpendicular to the plane is

represented by

x =

2

7

−5

+ λ

1

−2

1

=

2 + λ

7− 2λ

−5 + λ

, λ ∈ R

Hence, point P lies on this line, and it also lies on the plane, so we can substitute the

equation of this line into the equation of the plane and solve for λ in order to find the

intersection between the line and plane. Substituting it in:

(2 + λ)− 2× (7− 2λ) + (−5 + λ) = 4

2 + λ− 14 + 4λ− 5 + λ = 4

6λ = 21

λ =7

2

Hence, point P is represented by 2 + 72

7− 2(72)

−5 + 72

=1

2

11

0

−3

4

MATH1151 Algebra Test 2 2009 v1bApril 17, 2019

These solutions were written and typed by Rui Tong using past resources by Brendan Trinh.

Please be ethical with this resource. It is for the use of MathSoc members, so do not repost it

on other forums or groups without asking for permission. If you appreciate this resource, please

consider supporting us by coming to our events! Also, happy studying :)

We cannot guarantee that our answers are correct - please notify us of any errors or typos at

unswmathsoc@gmail.com, or on our Facebook page. There are sometimes multiple methods of

solving the same question. Remember that in the real class test, you will be expected to explain

your steps and working out.

Question 1

Since we do not know if there will be points of intersection, and we are asked to find them if they

exist, we may as well attempt Gaussian elimination. Equating the two expressions gives 1

2

−5

+ λ

1

8

−1

=

3

−1

0

+ µ

−1

2

3

+ ν

2

1

−5

=⇒

−2

3

5

= µ

−1

2

3

+ ν

2

1

−5

+ λ

−1

−8

1

.

Hence consider −1 2 −1 −2

2 1 −8 3

3 −5 1 5

R2←R2+2R1−−−−−−−−→R3←R3+3R1

−1 2 −1 −2

0 5 −10 −1

0 1 −2 −1

R3←R3− 1

5R2−−−−−−−−→

−1 2 −1 −2

0 5 −10 −1

0 0 0 −45

.

We see that in a row-echelon form of the augmented matrix, the right-hand column is leading.

Hence this system has no solutions, so there are no points of intersection between the plane and

the line.

Question 2

1

Part i)

Note that you can jump straight to the final answer if you wish. In the test, however, the

non-calculator restriction can get annoying, so I prefer to do the computations slowly.

A2 =

(2 4

−3 −1

)(2 4

−3 −1

)

=

(4− 12 8− 4

−6 + 3 −12 + 1

)

=

(−8 4

−3 −11

).

So we require constants u and v that satisfy(−8 4

−3 −11

)= u

(2 4

−3 −1

)+ v

(1 0

0 1

).

If you can see by inspection that u = 1 and v = −10, well done. Otherwise, the most instructive

approach is just to equate components. It may be worth noting that the equality can be

expressed as (−8 4

−3 −11

)=

(2u+ v 4u

−3u −u+ v

)to make life easier, but this step is optional. We can equate the a21 or a12 components to get

u = 1 , and then either a11 or a22 components to get v = −10 .

Part ii)

Essentially what we do now is recursively use the fact that A2 = A− 10I . This allows us to

gradually build our powers up. To reduce the mess, we first consider

A4 = A2A2

= (A− 10I)(A− 10I) (*)

= A2 − 10A− 10A+ 100I2

= A2 − 20A+ 100I (because I2 = I)

= A− 10I − 20A+ 100I (*)

= −19A+ 90I.

2

Then

A5 = AA4

= A(−19A+ 90I) (*)

= −19A2 + 90A

= −19(A− 10I) + 90A

= 71A+ 190I

and hence x = 71, y = 190. Note that at every instance of (*), I used the boxed identity above.

Part iii) Usually, if nothing stands obvious, row reductions are a good way of commencing

determinant problems.

det

−1 3 2

3 −2 −7

−5 1 8

= det

−1 3 2

0 7 −1

0 −14 −2

(R2 ← R2 + 3R1 & R3 ← R3 − 5R1)

= det

−1 3 2

0 7 −1

0 0 −4

(R3 ← R3 + 2R2)

The nice thing about these is that ultimately we will end up with an upper triangular matrix,

at which point we can use the shortcut of multiplying the elements on the main diagonal to

obtain

det

−1 3 2

3 −2 −7

−5 1 8

= 28.

Note: Later problems will see alterations. Row reductions are more or less a go-to option when

there’s nothing else that’s convenient. Sometimes there are though!

Question 4

We have ∣∣∣−−→AB∣∣∣ =

∣∣∣∣∣∣∣ 1

−3

1

− 5

1

−2

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣−4

−4

3

∣∣∣∣∣∣∣ =√

41

and ∣∣∣−→AC∣∣∣ =

∣∣∣∣∣∣∣ 2

−4

1

− 5

1

−2

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣−3

−5

1

∣∣∣∣∣∣∣ =√

35

3

Also

−−→AB ·

−→AC =

−4

−4

3

·−3

−5

1

= 35

and hence cos θ = 35√41√35

=√

3541 . Note that

−→CA was used in favour of

−→AC because we required

the angle at A. Because we already found−−→AB, which starts at A and ends at B, we choose the

similar one for C.

4

MATH1151 Algebra Test 2 2018 v3aApril 17, 2019

These solutions were written and typed by Rui Tong. Please be ethical with this resource. It is

for the use of MathSoc members, so do not repost it on other forums or groups without asking

for permission. If you appreciate this resource, please consider supporting us by coming to our

events! Also, happy studying :)

We cannot guarantee that our answers are correct - please notify us of any errors or typos at

unswmathsoc@gmail.com, or on our Facebook page. There are sometimes multiple methods of

solving the same question. Remember that in the real class test, you will be expected to explain

your steps and working out.

Question 1

We present two valid approaches.

Method 1

This method relies on Gaussian elimination. If a is parallel to the plane, then a should be a

linear combination of the vectors that span the plane. That is, the equation

λ

1

−3

1

+ µ

2

−1

7

=

3

−2

5

should have a solution for λ and µ. This is now a system of equations, so we consider the row

reductions 1 2 3

−3 −1 −2

1 7 5

R2←R2+3Ri−−−−−−−−→R3←R3−R1

1 2 3

0 5 7

0 5 2

R3←R3−R2−−−−−−−→

1 2 3

0 5 7

0 0 −5

.

So we see that in the row-echelon form of the augmented matrix, the right column is leading.

Hence there is no solution, and thus a is not parallel to the plane.

Method 2

1

This method relies on vector geometry techniques. We start by finding a vector normal (i.e.

perpendicular) to the plane, which can of course be done so with the cross product.

n =

1

−3

1

× 2

−1

7

=

−20

−5

5

.

The idea is, a can only be parallel to the plane, if it is also perpendicular to n. This would

require a · n = 0, but in fact we see that

a · n =

3

−2

5

·−20

−5

5

= −25 6= 0.

So by contradiction, a is not parallel to the plane.

Remark: Any scalar multiple of n would have worked as well. So we could’ve considered dotting

against

−4

−1

1

to make the numbers smaller, and hence our lives easier.

Question 2

We take the augmented matrix (A | I) and find its reduced row-echelon form, which will be of

the form (I | B). Then B is our required matrix, i.e. B = A−1. Note that this process is fairly

2

tedious. −1 −2 2 1 0 0

−1 1 3 0 1 0

2 1 −4 0 0 1

R2←R2−R1−−−−−−−−→R3←R3+2R1

−1 −2 2 1 0 0

0 3 1 −1 1 0

0 −3 0 2 0 1

R3←R3+R2−−−−−−−→

−1 −2 2 1 0 0

0 3 1 −1 1 0

0 0 1 1 1 1

R1←R1−2R3−−−−−−−−→R2←R2−R3

−1 −2 0 −1 −2 −2

0 3 0 −2 0 −1

0 0 1 1 1 1

R2← 1

3R2−−−−−−−−→

R1←R1+2R2

−1 0 0 −73 −2 −8

3

0 1 0 −23 0 −1

3

0 0 1 1 1 1

R1←−R1−−−−−−→

1 0 0 73 −2 8

3

0 1 0 −23 0 −1

3

0 0 1 1 1 1

.

Therefore A−1 =

73 2 8

3

−23 0 −1

3

1 1 1

.

Question 3

Since the equation of the plane is given in Cartesian form, we can read off the coefficients to

get a normal vector

n =

2

−1

−4

.

Note that this is how the Cartesian and point-normal equations of a plane are linked.

For a point on the plane, I believe the easiest choice to consider is (0, 5, 0). A vector join-

ing this to (8,−5, 4) is 8

−5

4

−0

5

0

=

8

−10

4

.

A vector with the shortest distance between (8,−5, 4) and the plane can then be found by

3

projecting the above vector onto n. Note that since we just require the length

∣∣∣∣∣∣∣projn

8

−10

4

∣∣∣∣∣∣∣,

it suffices to consider

∣∣∣∣∣∣∣projn

8

−10

4

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣ 8

−10

4

· 2

−1

−4

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

2

−1

−4

∣∣∣∣∣∣∣

=|16 + 10 + 16|√

4 + 1 + 16

=42√21

= 2√

21.

Question 4

Recall that the area of the triangle is half that of the parallelogram, spanned by the vectors

joining the vertices. For two possible vectors, take 5

−2

0

− 3

1

−2

=

2

−3

2

and 6

0

−1

− 3

1

−2

=

3

−1

1

.

Recall that the area of the parallelogram spanned is found by taking the magnitude of the cross

products. Here, the cross product is 2

−3

2

× 3

−1

3

=

−1

4

7

.

So the area of the triangle is1

2

√1 + 16 + 49 =

√66

2.

4

MATH1151 Algebra Test 2 2018 v3bApril 17, 2019

These solutions were written by James Gao and typed by Rui Tong. Please be ethical with this

resource. It is for the use of MathSoc members, so do not repost it on other forums or groups

without asking for permission. If you appreciate this resource, please consider supporting us by

coming to our events! Also, happy studying :)

We cannot guarantee that our answers are correct - please notify us of any errors or typos at

unswmathsoc@gmail.com, or on our Facebook page. There are sometimes multiple methods of

solving the same question. Remember that in the real class test, you will be expected to explain

your steps and working out.

Question 1

For any points of intersection, upon equating, 3

−6

7

+ λ

1

1

−5

=

1

8

5

+ µ

−1

3

2

=⇒ µ

−1

3

2

+ λ

−1

−1

5

=

2

−14

2

If you can see by inspection that µ = −4 and λ = 2 , well done - you may skip the Gaussian

elimination. Otherwise, consider−1 −1 2

3 −1 −14

2 5 2

R2←R2+3R1−−−−−−−−→R3←R3+2R1

−1 −1 2

0 −4 −8

0 3 6

R2←R3−3R2−−−−−−−−→R2←− 1

4R2

−1 −1 2

0 1 2

0 0 0

.

So back subbing, R2 immediately gives us λ = 2. So subbing back, the point corresponds

to 3

−6

7

+ 2λ

1

1

−5

=

5

−4

−3

.

1

Note: If you have time, you probably don’t want to stop here. Back-sub further into R1 to get

µ = −4, and check that if you sub this value into the equation of the second line, you also get

the point (5,−4,−3).

Question 2

Once again, we find the required reduced row-echelon form. 1 1 −2 1 0 0

−3 1 −3 0 1 0

1 5 −8 0 0 1

R2←R2+3R1−−−−−−−−→R3←R3−R1

1 −1 2 1 0 0

0 −2 3 3 1 0

0 6 −10 −1 0 1

R3←R3+3R2−−−−−−−−→

1 −1 2 1 0 0

0 −2 3 3 1 0

0 0 −1 8 3 1

R1←R1+2R3−−−−−−−−→R2←R2+3R3

1 −1 0 17 6 2

0 −2 0 27 10 3

0 0 −1 8 3 1

R2←− 1

2R2−−−−−−−→

R3←−R3

1 −1 0 17 6 2

0 1 0 −272 −5 −3

2

0 0 1 −8 −3 −1

R1←R1+R2−−−−−−−→

1 0 0 72 1 1

2

0 1 0 −272 −5 −3

2

0 0 1 −8 −3 −1

.

So A−1 =

72 1 1

2

−272 −5 −3

2

−8 −3 −1

.

Question 3

We approach this similar to the other version. A normal vector is n =

1

−3

4

and by inspection

(11, 0, 0) lies on the plane. Hence for a vector from (−7, 8,−9) to the plane, take11

0

0

−−7

8

−9

=

18

−8

9

.

2

The shortest distance is

∣∣∣∣∣∣∣projn

18

−8

9

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣18

−8

9

· 1

−3

4

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1

−3

4

∣∣∣∣∣∣∣

=78√26

= 3√

26.

Question 4

We approach this similar to the other version. For two vectors that form the sides of the triangle,

consider 3

−3

8

− 1

−4

7

=

2

1

1

and 3

−3

8

−0

1

5

=

3

−4

3

.

The area of the triangle is then

1

2

∣∣∣∣∣∣∣2

1

1

× 3

−4

3

∣∣∣∣∣∣∣ =

1

2

∣∣∣∣∣∣∣ 7

−3

−11

∣∣∣∣∣∣∣ =

√179

2.

3

MATH1151 Algebra Test 2 2018 v4aApril 17, 2019

These solutions were written by James Gao and typed by Rui Tong. Please be ethical with this

resource. It is for the use of MathSoc members, so do not repost it on other forums or groups

without asking for permission. If you appreciate this resource, please consider supporting us by

coming to our events! Also, happy studying :)

We cannot guarantee that our answers are correct - please notify us of any errors or typos at

unswmathsoc@gmail.com, or on our Facebook page. There are sometimes multiple methods of

solving the same question. Remember that in the real class test, you will be expected to explain

your steps and working out.

Question 1

For the matrix to be invertible, we require det(A) 6= 0. So we first find

det(A) =

∣∣∣∣∣∣∣a 0 3

−2 a 2

−1 1 2

∣∣∣∣∣∣∣= a

∣∣∣∣∣a 2

1 2

∣∣∣∣∣+ 3

∣∣∣∣∣−2 a

−1 1

∣∣∣∣∣ (across first row)

= a(2a− 2) + 3(−2 + a) (2× 2 formula)

= 2a2 + a− 6

= (2a− 3)(a+ 2).

This expression is non-zero for all real a excluding a =3

2and a = −2, so those are the values

for which it is invertible.

Note: The decision to evaluate across the first row was more or less because of the 0 in the

first row. We could’ve also evaluated down the second column instead to take time, but then you

have more signs to keep track of.

Question 2

We’re essentially interested in the formula a ·b = |a||b| cos θ. Take note that since we’re dealing

1

with lines, we’re interested in using their respective direction vectors. We have−1

2

−1

·4

1

5

= −7,

∣∣∣∣∣∣∣−1

2

−1

∣∣∣∣∣∣∣ =√

6,

∣∣∣∣∣∣∣4

1

5

∣∣∣∣∣∣∣ =√

42.

So

cos θ =−7√

6×√

42

= − 7√6√

6√

7

= −√

7

6.

Note therefore that the acute angle is arccos

(√7

6

). If we took inverse cosine of the negative

angle, we would’ve arrived at the obtuse angle instead. (This more or less ties back to the

identity arccos(−x) = π − arccosx.)

Question 3

Part a)

In general, a Cartesian form will fall out immediately by starting with the point normal form,

and then expanding the dot product. We’ve been given a point it passes through and the normal

vector, so we just consider 2

0

−4

·x1x2x3

=

2

0

−4

·4

2

3

=⇒ 2x1 − 4x3 = 8− 12

=⇒ x1 − 2x3 = −2.

Part b)

Now that we have the Cartesian form, we can set some parameters. Since x2 is missing, set

2

λ = x2 and take µ = x3 to avoid fractions. Subbing in gives

x1 − 2µ−−2 =⇒ x1 = −2 + 2µ.

So the Cartesian form is

x =

−2 + 2µ

λ

µ

=

−2

0

0

+ λ

0

1

0

+ µ

2

0

1

.

where λ, µ ∈ R. (Note that the expression on the RHS is optional - you should be safe to jump

to the answer.)

Question 4

The following method you’ve learnt is essentially designed to work well in all cases (i.e. if the

lines intersect, are parallel or are skew). Here, we spend some time re-discussing how it works.

We start by observing that a vector normal (perpendicular) to both lines is given by

n =

1

−1

−2

×−2

3

3

=

3

1

1

.

We then pluck out any vector that represents a line segment, such that it joins any pair of

points (one on each line) together. Since the lines are given in Cartesian form, we may as well

use the points they’ve provided to find that line segment: 8

1

−4

− 1

2

−6

=

7

−1

2

.

The idea is, since we have a vector that is perpendicular to both lines, and the shortest distance

is chosen to be perpendicular to both, the vector projn

7

−1

2

will ultimately have length equal

to the distance we require! (You may need to draw a picture to visualise this.) Our answer is

3

hence

∣∣∣∣∣∣∣projn

7

−1

2

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣ 7

−1

2

·3

1

1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

3

1

1

∣∣∣∣∣∣∣

=22√11

= 2√

11.

4

MATH1151 Algebra Test 2 2018 v4bApril 17, 2019

These solutions were written and typed by Rui Tong. Please be ethical with this resource. It is

for the use of MathSoc members, so do not repost it on other forums or groups without asking

for permission. If you appreciate this resource, please consider supporting us by coming to our

events! Also, happy studying :)

We cannot guarantee that our answers are correct - please notify us of any errors or typos at

unswmathsoc@gmail.com, or on our Facebook page. There are sometimes multiple methods of

solving the same question. Remember that in the real class test, you will be expected to explain

your steps and working out.

Question 1

This question demonstrates how evaluating determinants is sometimes an art. Finding the

right technique at the right time makes life significantly easier, surprisingly. Recall that the row

operation Ri ← Ri+αRj does not change the determinant, so we do a row reduction first.∣∣∣∣∣∣∣∣∣∣1 2 −3 1

−1 1 4 −2

−3 −6 9 2

0 6 4 −5

∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣∣∣1 2 −3 1

0 3 1 −1

0 0 0 5

0 6 4 −5

∣∣∣∣∣∣∣∣∣∣(R2 ← R2 +R1 & R3 ← R3 + 3R1)

=

∣∣∣∣∣∣∣3 1 −1

0 0 5

6 4 −5

∣∣∣∣∣∣∣ (first column, note all 0’s vanish)

= −5

∣∣∣∣∣3 1

6 4

∣∣∣∣∣ (second row!)

= −5(12− 6)

= −30.

Note: When evaluating down rows/columns that aren’t the first one, make sure to keep track of

the alternating signs: ∣∣∣∣∣∣∣+ − +

− + −+ − +

∣∣∣∣∣∣∣Question 2

1

If we wish to use vectors, to find the length AB we first compute

−−→AB =

−2

7

3

−−3

5

−1

=

1

2

4

and then compute its length ∣∣∣−−→AB∣∣∣ =

√1 + 4 + 16 =

√21.

Take note that when using the dot-product-cosine-rule formula, since we require the angle at

B, we need both vectors to point towards B or both away from B! Since I’ve found−−→AB, I now

require−−→CB.

−−→CB =

−2

7

3

−−4

6

1

=

2

1

2

=⇒

∣∣∣−−→CB∣∣∣ =√

4 + 1 + 4 =√

17.

Then,

−−→AB ·

−−→CB =

∣∣∣−−→AB∣∣∣ ∣∣∣−−→CB∣∣∣ cos θ

=⇒ 2 + 2 + 8 =√

21√

17 cos θ

=⇒ cos θ =12√357

.

Question 3

We follow the same method as in the similar question in the other version.

Part a)

Considering the point-normal form, 6

1

2

·x1x2x3

=

6

1

2

·7

1

1

=⇒ 6x1 + x2 + x3 = 45.

Part b)

To avoid fractions, one possible choice of parameters is x1 = λ and x2 = µ. Then

6λ+ µ+ x3 = 45 =⇒ x3 = −6λ− µ+ 45.

2

Hence a parametric vector form is

x =

0

0

45

+ λ

1

0

−6

+ µ

0

1

−1

.

Question 4

We follow the same method as in the similar question in the other version. For a vector normal

to both lines,

n =

−1

1

1

×−3

1

2

=

1

−1

2

.

For one particular vector that represents an arbitrary line segment joining the lines, con-

sider −1

2

8

−−2

5

1

=

1

−3

7

.

Then the shortest distance is

∣∣∣∣∣∣∣projn

1

−3

7

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣ 1

−3

7

·−3

1

2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1

−1

2

∣∣∣∣∣∣∣

=18√

6

= 3√

6.

3