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Let $A$ be a $3$ x $3$ matrix with rank $2$. Then, $AX=0$ has

1. The trivial solution $X=0$.
2. One independent solution.
3. Two independent solution.
4. Three independent solution.

### 1 comment

As per my understanding, rank of $A < 3$, so $A$ is not invertible. Hence, it will either have no solution or infinitely many solutions. Which of them falls under this condition?

The answer will be option A,B.

Since rank of matrix A is 2, so |A| = 0.This says that there is a non-trivial and non-zero solution. Non-trivial solution means along with a zero solution(0,0,0) there are many points of intersection in the plane.

Number of independent solutions or number of free variables = Total number of variables – rank of A

= 3 – 2 = 1

For eg:

x1 + x2 + x3 = 0

[1 1 1]$\begin{bmatrix} x1\\ x2\\ x3 \end{bmatrix}$ = [0]

Rank = 1 and total variables are 3.

Let x2 =k, x3=t , then x1= – k – t

Total independent variables are 2 (x2,x3) and dependent variable is 1 (x1).

Thanks, gotcha.

Which part i have told wrong...what you have told..i have told that only.

@samarpita sorry my bad .. i have wrongly interpret the comment as there is only  trivial solution possible .

A is a $3*3$ matrix .

Rank of A is $2$ which implies it has $2$ linearly independent vectors .

Now $AX=0$  has only two cases either it will have unique solution / trivial solution which is all the variable will contain 0 as solution .

Or it will have infinitely many solutions .

In case of $AX=0$ no solution case is not possible as (0,0,0) always will be solution for this kind of system of equation .

Now , $AX=0$ will have trivial solution only when $\left | A \right |\neq 0$ .

$AX=0$  will have infinitely many solution when $\left | A \right |=0$ . Now As in the problem $A$ is $3*3$ matrix and rank of               $A$ =2 which implies the determinant of $A$ is equal to $0$ .

So this conclude the system here  $AX=0$ has infinitely many solutions .

edited by

Answer given by them was (a) and (b).

Reason they stated –

If $r$ is the rank of matrix A and $n * n$ is the order of matrix then we shall have $(n– r)$ linearly independent non-trivial solutions. Any linear combination of these $(n– r)$ solutions will also be a solution of $AX = 0$.

I am not being able to understand the second sentence of their statement.

the solution(x) will be linear combinations of the linearly independent vectors.
No of linearly independent vectors in the solution(x)= no of free variables=n-r.
So
“Any linear combination of these (n–r) solutions will also be a solution of AX=0.”

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