GO Classes DA 2025 | Weekly Quiz 6 | Change of Basis & Linear Transformation | Question: 8

Question

Consider a linear system $A \mathbf{x}=\mathbf{b}$, where $A$ is a $3 \times 4$ matrix with $\operatorname{Rank}(A)=2$.
How many solutions does this system have if $(i) \operatorname{Rank}([A \mid \mathbf{b}])=2$, (ii) $\operatorname{Rank}([A \mid \mathbf{b}])=3$ respectively?

• A. (i) no solution, (ii) infinitely many solutions.
• B. (i) no solution, (ii) unique solution.
• C. (i) infinitely many solutions, (ii) no solution.
• D. (i) infinitely many solutions, (ii) unique solution.

Answer 1

A matrix is 3x4 and has rank = 2. So, nullity = 4 - 2 = 2

Therfore A matrix will be like this:
$\begin{bmatrix} * & * & * & * \\ 0 & * & * & * \\ 0 & 0 & 0 & 0 \end{bmatrix}$

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(i) Rank [A | b] = 2

It's in this form, where * is non-zero value
$\begin{bmatrix} * & * & * & *| &* \\ 0 & * & * & * |& * \\ 0 & 0 & 0 & 0 |& 0 \end{bmatrix}$

It's obvious from the matrix form is that Ax = b will have infinite solution since it has 2 nullity and no solution does not exist

Since it has 2 free variables, therefore Infinite solution exist

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(ii) Rank [A | b] = 3

Matrix will be like this:

$\begin{bmatrix} * & * & * & *| &* \\ 0 & * & * & * |& * \\ 0 & 0 & 0 & 0 |& * \end{bmatrix}$

We can see from the form, that this pattern [0 0 0 | non-zero] exist which does not have any solution

Therefore no solution is there for this case