Conditions on Matrix Imply Conditions on Elements

SiddharthThakur

New member
Question 1
The reduced row echelon form of

 
3
4 17 22
1 2 5
row
a
b
A is equal to

 
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

Staff member
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

New member
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

Staff member
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

New member
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

Staff member
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

Staff member
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: $\displaystyle A =\left( \begin{array}{cccc} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right)$. [HR][/HR]

(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. [HR][/HR]

(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.