GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 23

Answer 1

Option A:

We know that

1. row rank (A) = column rank (A) = rank(A)
2. row rank (A) <= m
3. column rank (A) <= n

Given $n < m$ . So, $rank (A) <= n$.

Note that Augmented matrix (A | b) has m rows and n+1 column.

So, $rank(A|b) <= n+1$. We already know that $n<m \equiv n+1 <= m$, hence we can say $rank(A|b) <= n+1 <=m$.

Therefore, option A is True.

 

Option B:

(Copied Suraj’s comment)

Note: For a matrix $A_{m \times n}$, when the rank equals the smallest dimension (i.e., min(m, n)), it is said to have full rank.

Consider $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} x_1 \end{bmatrix} $ = $\begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix}$.

This matrix has full rank but does not have a solution.

Therefore, option B is False.

 

Option C:

Consider $A = \begin{bmatrix} 3 & 6 \\ 2 & 4 \\ 1 & 2 \end{bmatrix}, X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $ and b = $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$.

We see that $AX=b$ has a solution ($x_1=1, x_2=0$) but $AX=0$ has infinitely many solutions.

Therefore, option C is False.

 

Option D:

(Copied Suraj’s comment)

Consider $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} x_1 \end{bmatrix} $ = $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$.

It has unique solution which is $x_1 =1$ but is not invertible.

Therefore, option D is False.

 

Comments

OPTION B:- Can we do it also as. Let  A(5*3) and with full rank(A) = 3. and augmented matrix is A|B (5*4). Now for instance we get rank(A|B) = 4.So rank(A) != rank(A|B) means we have NO SOLUTION.Hence B is INCORRECT.But still our matrix is with full rank.Correct me if i get wrong.

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@Rajender gill you're correct but it's easy to show a counter example than (dis)proving it.

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Yes it is correct approach. In exam you may not find counter example but if you have intuition then you can easily solve.

Answer 2

UNDERSTAND THE INTUITION BEHIND PART B AND D:

IMPORTANT POINT: Full rank matrix means rank = min(m, n) for m * n matrix.

The problem arises in these two parts when m > n. Consider the following scenarios for Part B and Part D:

PART B:

If m > n, in the row echelon form, it is possible to have a pivot in each column (this makes it full rank matrix) and still have [ 0 0 .. 0 | non-zero ] rows in the remaining rows (since m > n). Therefore, the system need not be consistent.

As a valid example for above scenario and counter example for PART B consider this system of linear equations:

$\begin{bmatrix} 4 & 0 \\ 8&2 \\ 4&2 \end{bmatrix} x = \begin{bmatrix} 4\\ 10\\ 7 \end{bmatrix}$.                           (AX = b)

In this system, matrix A is full rank, yet the system is inconsistent.

PART D:

The system AX = b, can have a unique solution, when m > n, if in the row echelon form, all columns must have a pivot and the remaining rows are [ 0 0 .. 0 | 0 ]. But since m > n, this is not even a square matrix and inverse is not defined. Therefore, A need not be invertible.

As a valid example for above scenario and counter example for PART D consider this system of linear equations:

$\begin{bmatrix} 4 & 0 \\ 8&2 \\ 4&2 \end{bmatrix} x = \begin{bmatrix} 4\\ 10\\ 6 \end{bmatrix}$.                           (AX = b)

In this system, matrix A has a unique solution, yet it is not invertible.