I have the following question here:

A certain linear system $Ax=b$ consisting of $n$ unknowns has two distinct solutions $x_1$ and $x_2$ with $x_1 \neq x_2$. Which is the following statements is false?

$(A)$ The reduced row echelon form of $A$ is the $n \times n$ identity matrix.

$(B)$ The reduced row echelon form of $A$ has a row of zeros.

$(C)$ $det(A)=0$

$(D)$ There are an infinite number of distinct solutions of this system.

$(E)$ A is not invertible.

The correct answer is $(A)$ however I am not sure why. How are they justifying that this is the identity matrix after row reduction?

In any case, how are the other statements true?

It would be much appreciated if someone could explain each answer choice to me. Thanks!

  • 4
    $\begingroup$ Just an observation, all the options $B,C,D,E$ if true contradict $A$. $Ax_1=b=Ax_2\implies A(x_1-x_2)=0$. Since $x_1\neq x_2$, so $A$ behaves like the zero matrix in some sense. $\endgroup$ – Yadati Kiran Dec 17 '18 at 7:58
  • $\begingroup$ Ohh that makes sense! Thanks! @YadatiKiran $\endgroup$ – Future Math person Dec 17 '18 at 8:07

For the system $Ax=b$, we can either have a unique solution, no solution or infinitely many solutions. Since we already know two distinct solutions, the system must have infinitely many solutions $(D)$. This means the number of linearly independent rows (equations) is less than the number of unknowns.

If $A$ is a square matrix, this translates to the case when $\text{rank}(A)=\text{rank}([A|b])<n\iff$ there is at-least one zero row in the row echelon form of $A(B),\det A=0 (C),A$ is not invertible $(E)$.

Since the rest of the statements are true, $(A)$ must be false.

We know that $(A)$ is false because the rest are true, not because we understand why $(A)$ is false itself. Note that $A$ is invertible $\iff\text{rank}(A)=n\iff$ row echelon form of $A$ has no zero row.

Intuitively, to get the reduced row echelon form of $A$, we divide each row by its leading coefficient and make the rest of the entries in each column equal to $0$ through row transformations. This means the reduced row echelon form of $A$ has $n$ linearly independent rows with leading $1$s and rest of the entries $0$. Can you now see why the reduced row echelon form of any invertible matrix is the identity matrix and vice-versa?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.