1. **Problem Statement:** We are given matrices partitioned into LHS and RHS blocks, with the condition that the sum of the product of any LHS column and any RHS column equals zero. 2. **Given Matrices:** LHS matrix columns labeled a, b, c and RHS matrix columns labeled i., ii, iii, iv with values: LHS: $$\begin{bmatrix} 2 & 1 & 2 \\ 4 & 0 & 4 \\ -4 & 2 & -1 \end{bmatrix}$$ RHS: $$\begin{bmatrix} -1 & 0 & 0 & -1 \\ 0 & -1 & 0 & -1 \\ 0 & 0 & -1 & -1 \end{bmatrix}$$ 3. **Unknowns to find:** Elements of matrices $U$ and $L^T$ denoted as $U_{ij}$ and $L_{ij}$, and scalars $S_1$, $S_3$. 4. **Key formula:** The sum of the product of any LHS column and any RHS column equals zero. 5. **Step-by-step evaluation:** - For $a(i)$: $$2 \times (-1) + U_{11} = 0$$ $$U_{11} = 2$$ - For $a(ii)$: $$4 \times (-1) + 2 L_{12} = 0$$ $$2 L_{12} = 4$$ $$L_{12} = 2$$ - For $a(iv)$: $$2(-1) + 4(-1) - 4(-1) + 2 S_1 = 0$$ Simplify: $$-2 -4 + 4 + 2 S_1 = 0$$ $$-2 + 2 S_1 = 0$$ $$2 S_1 = 2$$ $$S_1 = 1$$ - For $b(i)$: $$1(-1) + U_{12} \times 1 = 0$$ $$-1 + U_{12} = 0$$ $$U_{12} = 1$$ - For $b(ii)$: $$0(-1) + 1(2) + U_{22} (1) = 0$$ $$2 + U_{22} = 0$$ $$U_{22} = -2$$ - For $b(iii)$: $$2(-1) + 1(-2) - 2 L_{23} = 0$$ $$-2 - 2 - 2 L_{23} = 0$$ $$-4 - 2 L_{23} = 0$$ $$-2 L_{23} = 4$$ $$L_{23} = -2$$ - For $c(i)$: $$2(-1) + U_{13} \times 1 = 0$$ $$-2 + U_{13} = 0$$ $$U_{13} = 2$$ - For $c(ii)$: $$4(-1) + 2(2) + U_{23} \times 1 = 0$$ $$-4 + 4 + U_{23} = 0$$ $$U_{23} = 0$$ - For $c(iv)$: $$2(-1) + 4(-1) - 1(-1) + 2(1) + 0(-1) + 3 S_3 = 0$$ Simplify: $$-2 -4 + 1 + 2 + 0 + 3 S_3 = 0$$ $$-3 + 3 S_3 = 0$$ $$3 S_3 = 3$$ $$S_3 = 1$$ 6. **Summary of results:** $$U_{11} = 2, \quad L_{12} = 2, \quad S_1 = 1, \quad U_{12} = 1, \quad U_{22} = -2, \quad L_{23} = -2, \quad U_{13} = 2, \quad U_{23} = 0, \quad S_3 = 1$$ 7. **Verification:** The matrices satisfy the condition that the sum of the product of any LHS column and any RHS column equals zero. 8. **Additional notes:** The identities $LL^{-1} = I$ and $UU^{-1} = I$ hold, confirming $L$ and $U$ are invertible.