This is a question from a past exam paper.
The Question: Let $A \in R^{m\times{n}}$ with $m > n$ have full column rank and $B \in R^{k\times{n}}$. Find an expression for the constraint minimiser $x \in R^{k}$ in terms of $A$, $B$, $b \in R^{m}$, $c \in R^k$ and the Lagrange multiplier $\lambda$ of minimise $$\| b-Ax\|^2 : c^TBx=a $$
Then find the value of $\lambda$ which satisfies the constraint.
What I've done so far: First of all, I inspected for small values of $m$,$n$,$k$. Setting $m=3$, $n=2$, $k=2$, we get: $$A=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} B=\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} b=\begin{bmatrix} b_{1}\\ b_{2} \end{bmatrix} c=\begin{bmatrix} c_{1}\\ c_{2} \end{bmatrix}$$ By Lagrange's Multipliers, we have: $$L(x,\lambda)=\| b-Ax\|^2-\lambda(c^TBx-a)$$ Now we want to find some $x^*$, $\lambda^*$ such that we have $min\{L(x^*,\lambda^*)\}$. It then turns out that upon expanding $L(x,\lambda)$, we get: $$(*)\implies L(x,\lambda)=(b_1-a_{11}x_1-a_{12}x_2)^2+(b_2-a_{21}x_1-a_{22}x_2)^2+(b_3-a_{31}x_1-a_{32}x_2)^2-\lambda(x_1(c_1b_{11}+c_2b_{21})+x_2(c_1b_{12}+c_2b_{22}))$$
Observing the above, we can bring $L(x,\lambda)$ to a finite minimum by minimising the square terms. Also, $A$ has full column rank, implying linear independence, meaning that the following combination exists: $$(**)\implies \begin{bmatrix} a_{11}\\ a_{21}\\ a_{31} \end{bmatrix}x_1+\begin{bmatrix} a_{12}\\ a_{22}\\ a_{32} \end{bmatrix}x_2=0$$ That is, we can define our x to satisfy the above combination. With this in mind, we can substitute the system of equations from $(**)$ into each square term of $(*)$ which will result in the following $$(***)\implies b_{1}^2+b_{2}^2+b_{3}^2-\lambda(x_1(c_1b_{11}+c_2b_{21})+x_2(c_1b_{12}+c_2b_{22}))$$
We now find the partial derivatives of $(*)$: $$\frac{\partial L}{\partial x_1}=2(b_{1}-a_{11}x_{1}-a_{12}x_{2})(-a_{11})+2(b_{2}-a_{21}x_{1}-a_{22}x_{2})(-a_{21})+2(b_{3}-a_{31}x_{1}-a_{32}x_{2})(-a_{31})-\lambda(c_1b_{11}+c_2b_{21})=0$$ And substituting $(**)$ we get: $$(1)\implies -2a_{11}b_{1}-2a_{21}b_{2}-2a_{31}b_3-\lambda(c_1b_{11}+c_2b_{21})=0$$ And similarly when taking $\frac{\partial L}{\partial x_2}$ and by $(**)$ we get: $$(2)\implies -2a_{12}b_{1}-2a_{22}b_{2}-2a_{32}b_3-\lambda(c_1b_{12}+c_2b_{22})=0$$ Substituting $(1)$ & $(2)$ into $(***)$ yields: $$L(x,\lambda)=b_{1}^2+b_{2}^2+b_{3}^2+2(x_1(2a_{11}b_{1}+2a_{21}b_{2}+2a_{31}b_3)+x_2(2a_{12}b_{1}+2a_{22}b_{2}+2a_{32}b_3))$$ thus giving me my constraint minimiser $x$. It turns out that for any $m$, $n$, $k$, we can write: $$ L(x,\lambda)=\|b\|^2+2x^TAb \iff \begin{bmatrix} a_{11}\\ \vdots \\ a_{m1} \end{bmatrix}x_1+\dots+ \begin{bmatrix} a_{1n}\\ \vdots \\ a_{mn} \end{bmatrix}x_n$$
My Problem: The question asks to find the lagrange multiplier. The problem is that $\lambda$ is free in $L(x,\lambda)$, so something in my calculations must not make sense. I suspect this has got something to do when I take the partial derivatives. Help and solutions to this problem would be very much appreciated!