Question

The augmented matrix of a non-homogeneous system of equations \( AX = B \) is reduced to the following form after applying a series of elementary row transformations: \[ \begin{bmatrix}1&1&1&5\\ 0&0&-3&4\\ 0&0&\mu+1&\lambda^{2}-2\lambda+1\end{bmatrix} \] Then, which of the following is correct?

• A. Only for \( \mu \neq -1 \), \( AX=B \) has a unique solution
• B. Only for \( \mu = -1 \) and \( \lambda = 1 \), \( AX=B \) has an infinite number of solutions
• C. For any value of \( \mu \) and \( \lambda \), \( AX=B \) has an infinite number of solutions
• D. For all positive values of \( \mu \), \( AX=B \) has no solution

Hint:

When a column of the coefficient matrix contains no leading pivot entry (like the second column here), the system can never have a unique solution, regardless of the parameters chosen.

Answer 1

The Correct Option is B

Step 1: Recall the rank rules.
For a system, compare the rank of the coefficient part with the rank of the full augmented part. If they are equal but smaller than the number of unknowns, we get infinitely many solutions.
Step 2: Read the given matrix.
The last row is $[\,0\ \ 0\ \ \mu+1\ \mid\ (\lambda-1)^2\,]$, since $\lambda^2-2\lambda+1 = (\lambda-1)^2$.
Step 3: Make the last row vanish.
There are only three unknowns, and column two has no pivot. To get infinite solutions the whole last row must become zero. So we need \[ \mu+1 = 0 \quad\text{and}\quad (\lambda-1)^2 = 0 \]
Step 4: Solve these conditions.
From $\mu+1=0$ we get $\mu=-1$. From $(\lambda-1)^2=0$ we get $\lambda=1$.
Step 5: Check the ranks at these values.
With the last row gone, both ranks equal 2. Since 2 is less than the 3 unknowns, the system is consistent with infinitely many solutions.
Step 6: Pick the matching option.
So infinite solutions happen only when $\mu=-1$ and $\lambda=1$. \[ \boxed{\mu=-1,\ \lambda=1 \Rightarrow \text{infinite solutions}} \]