Transpose H

Given the measurement model $Z = HX$, we compute the transpose $H^T$ of the matrix $H = [[1, 1], [1, 0], [0, 1]]$.

$H^T = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$

Compute product $H^T$ H

Calculate the matrix product of $H^T$ and $H$.

$H^T H = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$

Compute inverse of $H^T$ H

The inverse of a $2 \times 2$ matrix $A = [[a, b], [c, d]]$ is $\frac{1}{ad-bc} [[d, -b], [-c, a]]$. Here $ad-bc = 2(2) - 1(1) = 3$.

$(H^T H)^{-1} = \frac{1}{3} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$

Compute product $H^T$ Z

Calculate the product of $H^T$ and the measurement vector $Z = [0.5, 0.45, 0.51]^T$.

$H^T Z = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0.5 \\ 0.45 \\ 0.51 \end{bmatrix} = \begin{bmatrix} 0.95 \\ 1.01 \end{bmatrix}$

Final Estimation

Multiply the inverse matrix by the vector $H^T Z$: $\hat{X} = \frac{1}{3} [[2, -1], [-1, 2]] [[0.95], [1.01]]$.

$\hat{X} = \frac{1}{3} \begin{bmatrix} 1.9 - 1.01 \\ -0.95 + 2.02 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 0.89 \\ 1.07 \end{bmatrix} \approx \begin{bmatrix} 0.296 \\ 0.356 \end{bmatrix} \approx \begin{bmatrix} 0.29 \\ 0.35 \end{bmatrix}$