Find for which $\lambda$ the following system of equations has a solution

Question

I want to determine for which $\lambda\in\mathbb{R}$ the following linear system over $\mathbb{R}$ has a solution:

\begin{cases} x_1 + 2x_2 - 3x_3 = \lambda\\ 3x_1 + 5x_2 - 3x_3 = 2\lambda\\ -3x_1-4x_2-3x_3=1 \end{cases}

I would like to understand how to solve this exercise. Here's the augmented matrix to make it a little easier to solve:

\begin{bmatrix} 1&2&-3&\lambda\\ 3&5&-3&2\lambda\\ -3&-4&-3&1 \end{bmatrix}

I know that systems like these have a solution (one or infinite) if the rank of the coefficients matrix A matches the rank of the augmented matrix $A|b$. However, I don't know how I would go about evaluating the rank of $A|b$. Sure, I could try to calculate the determinant of all the submatrices of order 3 until I find that one has determinant $\neq 0$ but that would take a while.

Also, I've tried to reduce the matrix in row-echelon form but I can't see how to go from there.

\begin{bmatrix} 1&2&-3&\lambda\\ 0&-1&0&-\lambda\\ 0&1&-6&3\lambda \end{bmatrix}

Any hints?

Answer 1

In trying to reduce the augmented matrix to the reduced row-echelon form, we have

$$ \begin{align*} \left(\begin{array}{ccc|c} 1&2&-3&\lambda\\ 3&5&-3&2\lambda\\ -3&-4&-3&1 \end{array}\right)&\equiv \left(\begin{array}{ccc|c} 1&2&-3&\lambda\\ 0&-1&6&-\lambda\\ 0&2&-12&3\lambda+1 \end{array}\right)\tag{1}\\ &\equiv \left(\begin{array}{ccc|c} 1&0&9&-\lambda\\ 0&-1&6&-\lambda\\ 0&0&0&\lambda+1 \end{array}\right)\tag{2} \end{align*} $$ From the last augmented matrix, we see that we need to have $\lambda+1=0$ otherwise the system would be inconsistent. So to have solution, we need $\lambda=-1$.

The row operations that I did in 1 from top to bottom rows are $R_1\longrightarrow R_1$, $R_2\longrightarrow -3R_1+R_2$, and $R_3\longrightarrow 3R_1+R_3$, respectively.

The row operations that I did in 2 from top to bottom rows are $R_1\longrightarrow2R_2+R_1$, $R_2\longrightarrow R_2$, and $R_3\longrightarrow2R_2+R_3$.

Answer 2

Notice that the determinant of the first matrix is zero. So the equations are not linearly independent.

It is not hard to see that taking 4 times 2nd + 2 times 3rd minus 6 times 1st eliminates the lhs irrespective of the values of the $x_i$, and gives $\lambda=-1$.

So there are multiple solutions for $\lambda=-1$ and no solutions otherwise.