1 -31Forward elimination changes-1 -4 2 x = b to a triangular-52[11 -370 -3 -1 x = C:00-31x + 1y3z=-6- 1x - 4y + 2z = -5 ⇒-2x - 5y + 2z = -5Check that c =9.The equation - 3z = -6 in Ux= c comes from the original 1x + 1y - 3z = -6 inAx=b by subtacting 31 =times equation 1 and 32 =9. times the final equation 2. Reverse that to recover9.1in the last row of [A b] from the final [1 1 -3 -6] and [0 -3 -1 -11] and[0 03-6] in [U c].Row 3 of [A b] = (31 Row 1 + 32 Row 2 + 1 Row 3) of [U c].In matrix notation this is multiplication by L. So A = LU and b = Lc.What are the 3 by 3 triangular systems Lc = b and Ux = c?9.9.-28.1x + 1y3z-6-3y-12-11 →-3y - 4z = -171x + 1ly-3z=-6-3y - 12 = -11-32-69.C=X=-6-11 solves the first one. Which x solves the second one? x =-6

We have given matrix A and matrix b are, and system of equation is To find all solutions to using…

Forward elimination changes-1 [1 0 0 1 -31 -3 -1 x = C: 0 -3 1x + 1y3z = -6 - 1x - 4y + 2z = -5 ⇒ -2x - 5y + 2z = -5 Check that c= 1 1. 9. 1 -31 -4 2 x=b to a triangular -2 -5 2 1x + 1y3z-6 The equation -32= -6 in Ux= c comes from the original 1x + 1y - 3z = -6 in Ax=b by subtacting 31 = O times equation 1 and 32 = 9. times the final equation 2. Reverse that to recover [ ] in the last row of [A b] from the final [1 1 -3-6] and [0-3 -1 -11] and [00-3-6] in [U c]. Row 3 of [A b] = (31 Row 1 + 32 Row 2 + 1 Row 3) of [U c]. In matrix notation this is multiplication by L. So A = LU and b = Lc. What are the 3 by 3 triangular systems Lc = b and Ux = c? 41 -3y-12-11 → -3y4z=-17 1x + ly 3z = -6 -3y1z=-11 -32=-6 9. 8. C= X= -6 8 -11 solves the first one. Which x solves the second one? x = -6