1. Problem statement. Apply Crout LU to the matrix $A_{2b}$ and solve $A_{2b}x=b_{2b}$ via $L_{2b}y=b_{2b}$ and $U_{2b}x=y$, and state pivot checks. 2. Given data. The matrix and right-hand side are $$A_{2b}=\begin{pmatrix}4 & 2 & 0 & 0 & 1\\1 & 5 & 2 & 0 & 0\\0 & 1 & 4 & 2 & 0\\0 & 0 & 1 & 3 & 1\\1 & 0 & 0 & 2 & 6\end{pmatrix},\qquad b_{2b}=\begin{pmatrix}11\\14\\13\\9\\20\end{pmatrix}$$ 3. Crout formulas and rules. For Crout (unit diagonal in $U_{2b}$) use the formulas $\displaystyle l_{i,k}=a_{i,k}-\sum_{p=1}^{k-1} l_{i,p}u_{p,k}$ for $i\ge k$, $\displaystyle u_{k,j}=\frac{a_{k,j}-\sum_{p=1}^{k-1} l_{k,p}u_{p,j}}{l_{k,k}}$ for $j>k$, and $u_{k,k}=1$ for all $k$. Ensure each pivot $l_{k,k}\neq 0$ as you proceed. 4. Crout decomposition step-by-step (compute entries). Start with $k=1$. $l_{11}=a_{11}=4$. $l_{21}=a_{21}=1$, $l_{31}=0$, $l_{41}=0$, $l_{51}=1$. $u_{12}=\dfrac{a_{12}}{l_{11}}=\dfrac{2}{4}=\dfrac{1}{2}$, $u_{13}=0$, $u_{14}=0$, $u_{15}=\dfrac{1}{4}$. For $k=2$. $l_{22}=a_{22}-l_{21}u_{12}=5-1\cdot\tfrac{1}{2}=\tfrac{9}{2}$. $l_{32}=a_{32}-l_{31}u_{12}=1-0=1$. $l_{42}=0$, $l_{52}=a_{52}-l_{51}u_{12}=0-1\cdot\tfrac{1}{2}=-\tfrac{1}{2}$. $u_{23}=\dfrac{a_{23}-l_{21}u_{13}}{l_{22}}=\dfrac{2}{9/2}=\dfrac{4}{9}$, $u_{24}=0$, $u_{25}=\dfrac{-1/4}{9/2}=-\dfrac{1}{18}$. For $k=3$. $l_{33}=a_{33}-l_{31}u_{13}-l_{32}u_{23}=4-1\cdot\tfrac{4}{9}=\tfrac{32}{9}$. $l_{43}=a_{43}-l_{41}u_{13}-l_{42}u_{23}=1$. $l_{53}=a_{53}-l_{51}u_{13}-l_{52}u_{23}=0-(-\tfrac{1}{2})\cdot\tfrac{4}{9}=\tfrac{2}{9}$. $u_{34}=\dfrac{a_{34}-l_{31}u_{14}-l_{32}u_{24}}{l_{33}}=\dfrac{2}{32/9}=\dfrac{9}{16}$, $u_{35}=\dfrac{a_{35}-l_{31}u_{15}-l_{32}u_{25}}{l_{33}}=\dfrac{1/18}{32/9}=\dfrac{1}{64}$. For $k=4$. $l_{44}=a_{44}-l_{41}u_{14}-l_{42}u_{24}-l_{43}u_{34}=3-1\cdot\tfrac{9}{16}=\tfrac{39}{16}$. $l_{54}=a_{54}-l_{51}u_{14}-l_{52}u_{24}-l_{53}u_{34}=2-(\tfrac{2}{9})\cdot\tfrac{9}{16}=2-\tfrac{1}{8}=\tfrac{15}{8}$. $u_{45}=\dfrac{a_{45}-\sum_{p=1}^{3}l_{4,p}u_{p,5}}{l_{44}}=\dfrac{1-\tfrac{1}{64}}{39/16}=\dfrac{63/64}{39/16}=\dfrac{21}{52}$. For $k=5$. $l_{55}=a_{55}-\sum_{p=1}^{4} l_{5,p}u_{p,5}=6-\Big(1\cdot\tfrac{1}{4}+(-\tfrac{1}{2})(-\tfrac{1}{18})+(\tfrac{2}{9})(\tfrac{1}{64})+(\tfrac{15}{8})(\tfrac{21}{52})\Big)=6-\tfrac{27}{26}=\tfrac{129}{26}$. 5. Assembled factors. The Crout factors are $$L_{2b}=\begin{pmatrix}4 & 0 & 0 & 0 & 0\\1 & \tfrac{9}{2} & 0 & 0 & 0\\0 & 1 & \tfrac{32}{9} & 0 & 0\\0 & 0 & 1 & \tfrac{39}{16} & 0\\1 & -\tfrac{1}{2} & \tfrac{2}{9} & \tfrac{15}{8} & \tfrac{129}{26}\end{pmatrix},\qquad U_{2b}=\begin{pmatrix}1 & \tfrac{1}{2} & 0 & 0 & \tfrac{1}{4}\\0 & 1 & \tfrac{4}{9} & 0 & -\tfrac{1}{18}\\0 & 0 & 1 & \tfrac{9}{16} & \tfrac{1}{64}\\0 & 0 & 0 & 1 & \tfrac{21}{52}\\0 & 0 & 0 & 0 & 1\end{pmatrix}$$ 6. Forward substitution $L_{2b}y=b_{2b}$ (show intermediate values). From row 1: $4y_1=11\Rightarrow y_1=\tfrac{11}{4}$. Row 2: $1\cdot y_1+\tfrac{9}{2}y_2=14\Rightarrow y_2=\dfrac{14-y_1}{9/2}=\dfrac{5}{2}$. Row 3: $y_2+\tfrac{32}{9}y_3=13\Rightarrow y_3=\dfrac{13-y_2}{32/9}=\dfrac{189}{64}$. Row 4: $y_3+\tfrac{39}{16}y_4=9\Rightarrow y_4=\dfrac{9-y_3}{39/16}=\dfrac{129}{52}$. Row 5: $y_1-\tfrac{1}{2}y_2+\tfrac{2}{9}y_3+\tfrac{15}{8}y_4+\tfrac{129}{26}y_5=20\Rightarrow y_5=\dfrac{343}{129}$. So $$y=\begin{pmatrix}\tfrac{11}{4}\\\tfrac{5}{2}\\\tfrac{189}{64}\\\tfrac{129}{52}\\\tfrac{343}{129}\end{pmatrix}.$$ 7. Back substitution $U_{2b}x=y$ (compute each $x_i$). Row 5: $x_5=y_5=\dfrac{343}{129}\approx 2.6596899225$. Row 4: $x_4+\tfrac{21}{52}x_5=y_4\Rightarrow x_4=y_4-\tfrac{21}{52}x_5\approx 1.4067656513$. Row 3: $x_3+\tfrac{9}{16}x_4+\tfrac{1}{64}x_5=y_3\Rightarrow x_3\approx 2.1202610411$. Row 2: $x_2+\tfrac{4}{9}x_3-\tfrac{1}{18}x_5=y_2\Rightarrow x_2\approx 1.7054233997$. Row 1: $x_1+\tfrac{1}{2}x_2+\tfrac{1}{4}x_5=y_1\Rightarrow x_1\approx 1.2323658196$. Thus the solution (rounded) is $$x\approx\begin{pmatrix}1.2323658196\\1.7054233997\\2.1202610411\\1.4067656513\\2.6596899225\end{pmatrix}$$ 8. Pivot / permissibility checks for Crout without pivoting. While factoring we checked each pivot $l_{k,k}$ and found them nonzero: $l_{11}=4$, $l_{22}=9/2$, $l_{33}=32/9$, $l_{44}=39/16$, $l_{55}=129/26$, so no zero pivots occurred. Additionally $A_{2b}$ is strictly diagonally dominant by rows (each diagonal entry exceeds the sum of magnitudes of the other entries in its row), which guarantees nonsingularity and permits LU without pivoting. 9. Final answer. The Crout factors $L_{2b}$ and $U_{2b}$ are given above, the intermediate vector $y$ is given above, and the solution is approximately $x\approx\big(1.2323658196,\;1.7054233997,\;2.1202610411,\;1.4067656513,\;2.6596899225\big)^T$.