GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 16

Answer 1

Solution.
$$
[A \mid b]=\left[\begin{array}{cc|c}
1 & 0 & 1 \\
-1 & 1 & 0 \\
k & 2 & 1
\end{array}\right] \underset{R 2 \rightarrow R 2+R 1, R 3 \rightarrow R 3-k R 1}{\rightarrow}\left[\begin{array}{cc|c}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 2 & 1-k
\end{array}\right] \underset{R 3 \rightarrow R 3-2 R 2}{\rightarrow}\left[\begin{array}{cc|c}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & -1-k
\end{array}\right]
$$
Therefore the system is consistent if and only if $-1-k=0.$

Comments

@Amit 003

actually, unique solution means matrix A is full rank for consistent system Ax=b

now full rank means all the columns of A must be LI regardless of m >= n (for m<n , columns cannot be LI)

​​​​​​Here, u are getting pivots in both the columns after reducing , so A is full rank (Ax=b cannot have infinite solutions)

in the question, after reducing (A|b) to echelon form, u are viewing -k-1 directly as a nonzero value but look carefully, the system will become inconsistent if -k-1 is not 0 

therefore, put -k-1 = 0 --> k = -1 

put this value and there will be no row like (00...0 | NZ)

alse note : (00...0|0) row in (A|b) does not implies infinite solutions, but if free columns exist in consistent system --> infinite solutions 

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 @Amit 003     we are getting a unique solution because we have two columns in matrix A and both have pivot elements.

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@Amit 003 Yes you are right we are getting [00|0] in the last row but that does not mean we have free variables as you can see here also there are two variables and we have two pivots also, so we have to make the last row last column in an augmented matrix to be zero, so that’s what we are doing here.