restart;with( LinearAlgebra ):Define the consumption matrix, C. Note how this is formed as three column vectors; commas move to a new row and vertical struts move to a new column.C := < < .1, .3, .3> | <.6, .2, .1> | <.6, 0, .1> >;The 3x3 identity matrixI3 := IdentityMatrix(3);The final demand vectord := <0,18,0>;Intermediate demandd1 := C . d + d;and the next few intermediate demandsd2 := C . d1 + d;d3 := C . d2 + d;d4 := C . d3 + d;Or, in a loop:u := d;for k from 1 to 4 do u := C . u + d;end do;How many iterations are needed before these estimates settle down to 2 decimal places?The problem can also be solved by finding the inverse:Ainv := MatrixInverse( I3-C );Ainv . d;The same outline can be used for other problems. For extra credit, submit solutions to Exercises 13, 14, and 15 on p. 157 of the text by Friday, February 26.