TIFR CSE 2023 | Part A | Question: 3

Question

$A$ is an $n \times n$ matrix with real-valued entries. Further, there exists a vector $x \neq 0$ such that $A x=0$. Now consider a given vector $b$ in $\mathbb{R}^{n}$. How many possible vectors $z$ exist, so that $A z=b?$ 

• A. $0$
• B. $1$
• C. $n-1$
• D. $n$
• E. Either $0$ or infinite

Answer 1

$A$ is a $n \times n$ matrix i.e. $A$ has $n$ column vectors in $\mathbb{R}^n$.

There exists a vector $x \neq 0$ such that $Ax=0$ this implies the column vectors of $A$ are Linearly Dependent ie $A$ has less than $n$ Linearly Independent column vectors.

Therefore, column vectors of $A$ doesn't span the entire $\mathbb{R}^n$ space. Therefore, there exists some $b$ for which $Az = b$ has no solution.

And for vectors $b$ which are some Linear Combination of column vectors of $A$, since column vectors of $A$ are Linearly Dependent, there are infinite Linear Combinations possible for each vector $b$. Therefore, there exists some $b$ for which $Az = b$ has infinite solutions.

Answer :- E

Comments

Awesome answer @shishir__roy.

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Just adding that since we Now know that A has linearly dependent vector too so echleon form will have one free variable.
but when we will write the augmented matrix for AX=b then [A|b] after transforming it into echelon form we may also get [000..0| non-zero ] for some b hence for some b there will be no solution and for some for which [000..0|non-zero] is npt possible there we know that [A|b] echelon form has free varaiable which means it will have infinite solutions hence ans is either 0 or infinite
ans ${E}$