Question

Linear algebra

Answer 1

Hia! ^.^ Lets see here.

Answer 2

Here's my working

Answer 3

First one looks good! :)

Answer 4

You got good hand writting! ^.^

Answer 5

^ nice

Answer 6

Wait ill attach pics! :)

Answer 7

@Qwertty123 lol thanks :D.

Answer 8

lol You welcome! :)

Answer 9

@zzr0ck3r Come and help! ^.^

Answer 10

what is the question?

Answer 11

Do you just want to know if you found the inverse?

Answer 12

In the first part of the question, is it legal to perform column operations instead of row?

Answer 13

Yes it is! But stick to one!

Answer 14

Either column or row!

Answer 15

Part B is easy, matrix A will have a row which is a multiple of another so the determinant should be zero

Answer 16

Okay thanks @mecharv

Answer 17

lol why was I tagged?

Answer 18

One of my friends sent me his solution and I couldn't convince him why was he wrong

Answer 19

Just tell him his handwriting is bad! ;) Wait, Ill see

Answer 20

Honestly, I cant understand anything! :O

Answer 21

Hahaha sure I will do that, x1a1+x2b1+x3xc1=0
x1a2+x2b2+x3c2
x1a3+x2b3+x3c3
x1(a1 a2 a3)+x2(b1b2 b3)+x3(c1c2c3)=(0)
He equated x1=2 x2=1 and x3=-4 to the given

Answer 22

Are you there? @Abmon98

Answer 23

Yes

Answer 24

Did you see if those values satisfy?

Answer 25

I am in doubt because the question is 'How' many solutions....

Answer 26

I am not sure too, I tried x1=1 x=2 and x3=1

Answer 27

I also tried the x-values of my friend but this will imply all of matrix A entries to be zero.

Answer 28

The first line defines the matrix, and the second line shows clearly that the third column (c) is a linear combination of the first two (a, b).
So without doing any calculations, and using the properties of a matrix where one column is a linear combination of the other two, you should be able to tell
(1) the number of slack variables in the solution, and hence the number of solutions.
(2) if the matrix is invertible.

If you are not sure, read, for example:
http://mathworld.wolfram.com/InvertibleMatrixTheorem.html
and note item 4.

Answer 29

To me, for a) the answer is infinitely many solution. From row 1, we have \(b_1= 4c_1-2a_1\)
Hence, with infinitely many \(a_1, c_1\) satisfy this equation. Each pair of them define \(b_1\), Hence there are many matrices A formed by that reason. That gives us so many solution for Ax =0

Answer 30

for b) A is invertible if \(det A\neq 0\). That is we need to find det (A) and consider the condition of the entries make it =0.
I might be wrong when overestimating the problem, but that is what I thought.

Answer 31

If the are infinite number of possibilities for a and c
a1(b2c3-b3c2)-b1(a2c3-a3c2+c1(a2b3-a3c2)
a1((4c2-2a2)c3-((4c3-2a3)c2)-(4c1-2a1)(a2c3-a3c2)+c1(a2((4c3-2a3)-a3c2)=
a1(4c2c3-2a2c3-4c2c3+2a3c2)-(4a2c1c3-4a3c1c2-2a1a2c3+2a1a3c2)+c1(4a2c3-2a22a3-a3c2)=0

Answer 32

det(A)=0

Answer 33

@IrishBoy123 Can you please help?

Answer 34

agree with loser

they're the same equations re-packaged

so infinitely many

the mark scheme worries me though. for 20 marks you ought to have to do some work!!

experts on this include @Empty and @ikram002p and @ganeshie8

and a cast of many others.

Answer 35

if \[2\left(\begin{array}{c} a_1 \\a_2 \\a_3 \\\end{array}\right)+\left(\begin{array}{c} b_1 \\b_2 \\b_3 \\\end{array}\right)-4\left(\begin{array}{c} c_1 \\c_2 \\c_3 \\\end{array}\right)=\left(\begin{array}{c} 0 \\0 \\0 \\\end{array}\right)\]

then for any number \(t\) we have

\[2t\left(\begin{array}{c} a_1 \\a_2 \\a_3 \\\end{array}\right)+t\left(\begin{array}{c} b_1 \\b_2 \\b_3 \\\end{array}\right)-4t\left(\begin{array}{c} c_1 \\c_2 \\c_3 \\\end{array}\right)\]

\[=t\left[2\left(\begin{array}{c} a_1 \\a_2 \\a_3 \\\end{array}\right)+\left(\begin{array}{c} b_1 \\b_2 \\b_3 \\\end{array}\right)-4\left(\begin{array}{c} c_1 \\c_2 \\c_3 \\\end{array}\right)\right]=\left(\begin{array}{c} 0 \\0 \\0 \\\end{array}\right)\]

Answer 36

@Zale101

Answer 37

I did not fully get what looser is trying to emphasize yes I did notice that b1=4c1-2a1 but how does this help me answering the question.

Answer 38

Familiar with the properties of homogeneous linear equations:
https://math.dartmouth.edu/archive/m8s00/public_html/handouts/matrices2/node8.html
and linear dependence:
https://en.wikipedia.org/wiki/Linear_independence
would help you answer the questions.

Answer 39

Now that I know homogeneous equations, one column is a linear combination of the other two and that its either a unique or an infinite solution. How can I figure whether its infinite or unique?

Answer 40

"Every homogeneous system has at least one solution, known as the zero solution (or trivial solution), which is obtained by assigning the value of zero to each of the variables. If the system has a non-singular matrix (det(A) ≠ 0) then it is also the only solution. If the system has a singular matrix then there is a solution set with an infinite number of solutions. "
from Wiki, URL follows:
https://en.wikipedia.org/wiki/System_of_linear_equations

Also, read what @Zarcon wrote:
for any number t:
\(t\left[2\left(\begin{array}{c} a_1 \\a_2 \\a_3 \\\end{array}\right)+\left(\begin{array}{c} b_1 \\b_2 \\b_3 \\\end{array}\right)-4\left(\begin{array}{c} c_1 \\c_2 \\c_3 \\\end{array}\right)\right]=\left(\begin{array}{c} 0 \\0 \\0 \\\end{array}\right)\)

Since the right hand side is zero, we can multiply the left hand side by any number t, and the right-hand side remains zero. This implies that if the system has a non-trivial solution (i.e. not a zero vector), there is an infinite number of solutions. However, if the matrix is not invertible, then the only solution is the trivial solution.

So you can summarize the above and decide which case the given question falls into.

Answer 41

Rephrasing what everyone said. Take\[\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}2\\1\\-4\end{pmatrix}\]It is trivial to verify that \[\mathbf{x}=t\begin{pmatrix}2\\1\\-4\end{pmatrix}\] is also a solution to \(A\mathbf{x}=\mathbf{0}\) for any t.