Conditions on Matrix Imply Conditions on Elements

[SiddharthThakur]

Question 1
The reduced row echelon form of
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3
4 17 22
1 2 5
row
a
b
A is equal to
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0 0 0 0
0 0 1 2
1 2 0 3
R .
(a) What can you say about row 3 of A? Give an example of a possible third row for A.
(b) Determine the values of a and b.
(c) Determine the solution of the homogeneous system of equations Rx = 0 in parametric vector
form.
(d) What is the dimension of the column space of A? Do the columns of A span R3 ?

 

[Jameson]

Hi SiddharthThakur. Welcome to MHB. :) I can't really read your matrices well so until you fix them I can't comment on your specific problem but I can make a couple of comments in general about how to approach the problem.

(a) If there are any rows with entries of all 0, those rows should be below any non-zero rows.

(c) Make an augmented matrix and place a column of 0s in the last column. Now reduce to reduced row echelon form. From that point you should be able to write the general solution $R \hspace{1 mm} \vec{x}=\vec{0}$ in terms of your free variables.

(d) To find $\text{dim } \text{Col } A$, or simply $\text{rank } A$, find the pivot positions of the RREF matrix or just the echelon form and count the columns which have a pivot position in them. So if some matrix has a pivot position in columns 1,4,5,9 then $\text{dim } \text{Col }$ of that matrix is 4.

 

[SiddharthThakur]

The reduced row echelon form of A=[1,2,5,b_ 4,a,17,-22_ row3]
is equal to R=[1,2,0,3_ 0,0,1,-2_ 0,0,0,0]
(a) What can you say about row 3 of A? Give an example of a possible third row for A.
(b) Determine the values of a and b.
(c) Determine the solution of the homogeneous system of equations Rx = 0 in parametric vector
form.
(d) What is the dimension of the column space of A? Do the columns of A span

(its 3x4 matrices 3rows 4 columns)

 

[Jameson]

The reduced row echelon form of A=[1,2,5,b_ 4,a,17,-22_ row3]
is equal to R=[1,2,0,3_ 0,0,1,-2_ 0,0,0,0]

(a) What can you say about row 3 of A? Give an example of a possible third row for A.
(b) Determine the values of a and b.
(c) Determine the solution of the homogeneous system of equations Rx = 0 in parametric vector
form.
(d) What is the dimension of the column space of A? Do the columns of A span

(its 3x4 matrices 3rows 4 columns)

So \(\displaystyle A = \left( \begin{array}{cccc} 1 & 2 & 5 & b \\ 4 & a & 17 & -22 \\ i_{31} & i_{32} & i_{33} & i_{34} \end{array}\right)\)

and the RREF of \(\displaystyle A =\left( \begin{array}{cccc} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right)\)?

(a) When a row cancels out to all 0s that means it carries the same information as an above row. So if you have a row that is $x+y+z=2$ and another row that is $2x+2y+2z=4$, then the last row can be reduced to all 0s because it's just the same thing. How can you use that fact to give a possible row 3?

(b) What have you tried? Basically start trying to reduce A into RREF A and you should find some equalities to use. If you post what you've done I can help you continue. :)

(c), (d) We need to complete (b) to get a full answer for these.

 

[SiddharthThakur]

I'm not able to do this question.How can I reduced it into RREF without knowing the elements of row3...?
Its confusing me a lot.

 

[Jameson]

I'll use another matrix as an example.

\(\displaystyle A =\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 2 & 0 & 3 & 5 \\ 2 & 2 & 2 & 2 \end{array}\right)\).

The third row becomes all 0s very quickly when trying to find the RREF of this matrix. Why?

 

[Jameson]

Since the deadline for the assignment in question has passed, I will make some more comments on solving the problem so others can use the information in the future.

We are given a matrix $A$:

$\displaystyle A = \left( \begin{array}{cccc} 1 & 2 & 5 & b \\ 4 & a & 17 & -22 \\ i_{31} & i_{32} & i_{33} & i_{34} \end{array}\right)$ and we know that it is row-equivalent to: $\displaystyle A =\left( \begin{array}{cccc} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right)$.

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(a) Since the third row contains entries which are all 0, that means the third row is a linear combination of the first two. There are infinite possible third rows for matrix $A$ that reduce to all 0's, but here is one:

$\displaystyle A = \left( \begin{array}{cccc} 1 & 2 & 5 & b \\ 4 & a & 17 & -22 \\ 2 & 4 & 10 & 2b \end{array}\right)$

We can easily see that $R_3=2R_1$ in this example and we can easily reduce $R_3$ to all 0's.

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(b) Let's start row-reducing $A$, letting the third row start with all 0's because we know it is a linear combination of the first two rows.

$\displaystyle A = \left( \begin{array}{cccc} 1 & 2 & 5 & b \\ 4 & a & 17 & -22 \\ 0 & 0 & 0 & 0 \end{array}\right) \sim \left( \begin{array}{cccc} 1 & 2 & 5 & b \\ 0 & a-8 & -3 & -22-4b \\ 0 & 0 & 0 & 0 \end{array}\right) \sim \left( \begin{array}{cccc} 1 & 2 & 5 & b \\ 0 & \frac{a-8}{-3} & 1 & \frac{-22-4b}{-3} \\ 0 & 0 & 0 & 0 \end{array}\right)$

Now we want to eliminate the term $i_{13}$ by multiplying using the replacement equation $R_1=-5R_2+R_1$. However we want the term $i_{12}$ to remain as 2, so the only way that's possible is if:

\(\displaystyle \frac{a-8}{-3}=0 \implies a=8\)

Using the same replacement of $R_1=-5R_2+R_1$ we get the relation that:

\(\displaystyle -5 \left( \frac{-22-4b}{-3} \right)+b=3 \implies b=-7\)

We can check this on Wolfram by reducing the matrix: $\left( \begin{array}{cccc} 1 & 2 & 5 & -7 \\ 4 & 8 & 17 & -22 \\ 0 & 0 & 0 & 0 \end{array}\right)$ and we see that we have correctly found $a$ and $b$.

I will answer the other parts in a later post.