Simultaneous Equations Involving Matrices

Just to provide a more pedestrian look, here's what the equations boil down to when you evaluate them.

$2 x_1 + \lambda = 0 \\ 2 x_2 + 2 \lambda = 0 \\ 2 x_3 + 3 \lambda = 0 \\ x_1 + 2 x_2 + 3 x_3 = 4$

We have four linear equations to solve for four unknowns. Solving is standard procedure from there

Consider the system of equations: $Hx + A^T\lambda = \left[\begin{matrix}0&0&0\end{matrix}\right]^T$ $Ax = b$

Multiplying the first equation by $AH^{-1}$ gives:

$AH^{-1}Hx + AH^{-1}A^T\lambda = AH^{-1}\left[\begin{matrix}0&0&0\end{matrix}\right]^T$

Which gives:

$Ax + AH^{-1}A^T\lambda = 0$

Subtracting this equation by $Ax = b$ gives:

$AH^{-1}A^T\lambda = -b$

This gives:

$\lambda = -\left(AH^{-1}A^T\right)^{-1}b$

Finally, we have the equation:

$Hx + A^T\lambda = \left[\begin{matrix}0&0&0\end{matrix}\right]^T$

Replacing $\lambda$ gives:

$Hx -A^T\left(AH^{-1}A^T\right)^{-1}b = \left[\begin{matrix}0&0&0\end{matrix}\right]^T$

Finally,

$x = H^{-1}A^T\left(AH^{-1}A^T\right)^{-1}b$

At this point, one may substitute values and simplify. Evaluate the sizes of the resulting matrices at each stage as a sanity check .

Finally, we get:

$\boxed{x = \left[\begin{matrix}\frac{2}{7}&\frac{4}{7}&\frac{6}{7}\end{matrix}\right]^T}$

Through this easy question, I wanted to demonstrate that simultaneous equations involving matrices can be solved by effective manipulation.

An alternate approach to this problem is as such:

$Hx + A^T\lambda = \left[\begin{matrix}0&0&0\end{matrix}\right]^T$ $Ax = b$

This can be rearranged as:

$\left[\begin{matrix}H&A^T\\A&0\end{matrix}\right] \left[\begin{matrix}x\\ \lambda\end{matrix}\right] = \left[\begin{matrix}0\\0\\0\\b\end{matrix}\right]$

Therefore,

$\left[\begin{matrix}x\\ \lambda\end{matrix}\right]=\left[\begin{matrix}H&A^T\\A&0\end{matrix}\right]^{-1}\left[\begin{matrix}0\\0\\0\\b\end{matrix}\right]$

The last step can be evaluated by using a programmable calculator.