Go Classes LA for GATE DA | Homework: 2 | question: 6

Question

Q: The given matrix has solution for:

$\begin{bmatrix} 1 & 1 & 3\\ 1 & 2 & 5\\ 2 & 3 & 8 \end{bmatrix}$

a. All vectors b in $\mathbb{R}^{3}$

b. No vector b in $\mathbb{R}^{3}$

c. Some vectors b in $\mathbb{R}^{3}$

d.  Some vectors b in $\mathbb{R}^{2}$

Given Ans: C

Doubt: if we have $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$ as b ?

the colums of co-efficient matrix $\begin{bmatrix} 1 & 1 & 3\\ 1 & 2 & 5\\ 2 & 3 & 8 \end{bmatrix}$ are linearly dependent.

so, a solution like $\begin{bmatrix} 1\\ 2\\ -1 \end{bmatrix}$ will generate $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$ as b.

Now, my question are:

1) where does $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$ belongs ?

in $\mathbb{R}^{2}, \mathbb{R}^{3}, \mathbb{R}^{4},...$ ?

2) why option D is incorrect ?

Answer 1

It belongs to R3 because it has 3 entries

Comments

When b = $\begin{bmatrix}0\\0\\0\end{bmatrix}$, the matrix equation becomes:

$\begin{bmatrix}1 & 1 & 3\\1 & 2 & 5\\2 & 3 & 8\end{bmatrix} x = \begin{bmatrix}0\\0\\0\end{bmatrix}$

This system of linear equations has a trivial solution, where x = $\begin{bmatrix}0\\0\\0\end{bmatrix}$.

Now, If b is $\begin{bmatrix}0\\0\\0\end{bmatrix}$, then option D also seems like a valid choice for the given matrix equation, because $\begin{bmatrix}0\\0\\0\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$

(it's a point in space, not a vector because of zero magnitude. so, it should be equal in any dimention.)

I am thinking like this, is this wrong ?

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even 0 is also a vector in the given space.Since the columns of A are LD, therefore they do not fill the space, hence, a possibility arises that there might  be a vector in R3 which might not be a linear combination of the columns of A, hence, b may or may noot have a solution. Thus, option C is correct.

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I understood why option C is correct, but I didn't understood why option D is also not correct ?

if we see this way:
2 colums of matrix are llinearly independent, then their span will be a plane, which has 2 dimentions, so from this logic we can have some vectors of the form $\begin{bmatrix}\alpha\\\beta\\0\end{bmatrix}$ as b, where the 3rd dimention is zero, so can't we say that the given matrix has solution for some vectors in $\mathbb{R}^{2}$ because these vectors are geometrically same in both $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$, there is just the difference of representation.