Solve simultaneous equations with matrix

Question

$\begin{cases} a-b+\lambda d=3 \\ -a+b-c=7 \\ -b+c-d=5 \\ \lambda a-c+d=4 \end{cases}$

With a certain value of $\lambda$ the simultaneous equations has infinitely many solutions. Define this $\lambda$-value and all solutions with that value of $\lambda$.

What I did:

$\begin{bmatrix} 1 & -1 & 0 & \lambda & 3 \\ -1 & 1 & -1 & 0 & 7 \\ 0 & -1 & 1 & -1 & 5 \\ \lambda & 0 & -1 & 1 & 4 \end{bmatrix}$

$\Rightarrow$ $R_1=R_1+R_2\,\,, \\ R_2=R_2+R_3\,\,, \\ R_3=R_3+R_4$

$\Rightarrow$ $\begin{bmatrix} 0 & 0 & -1 & \lambda & 10 \\ -1 & 0 & 0 & -1 & 12 \\ \lambda & -1 & 0 & 0 & 9 \\ \lambda & 0 & -1 & 1 & 4 \end{bmatrix}$

But I don't know how to go further from this. So how I get the value of $\lambda$ and solve all solutions with that? Thanks in advance.

Answer 1

Do the following row operations on the original matrix

1. $R_4 = R_4 + \lambda R_2$
2. $R_4 = R_4 + \lambda R_3$
3. $R_2 = R_2 + R_1$
4. $R_2 = R_2 + R_3$
5. $R_3 = R_3 - R_2$
6. $R_4 = R_4 + R_3$

After doing all these operations you will get last row as

[0 0 0 $1-2\lambda$ | $12\lambda-6$]

So for infinitely many solutions $\lambda$ should be 1/2 or 0.5 which will make whole last row as 0.

So 1/2 is the lambda value.

And then put the value of lambda to the matrix and then you can get infinite solutions for the system.

Hope it might help!