# Octave commands for _Linear Algebra_ by Jim Hefferon,

- Topic: leontif.tex

a=[(25448-5395)/25448 -2664/30346;

-48/25448 (30346-9030)/30346];

b=[17589;

21243];

ans=a \ b;

printf("The answer to the first system is s=%0.0f and a=%0.0f\n",ans(1),ans(2));

b=[17489;

21243];

ans=a \ b;

printf("The answer to the second system is s=%0.0f and a=%0.0f\n",ans(1),ans(2));

- question 1

b=[17789;

21243];

ans=a \ b;

printf("The answer to question (1a) is s=%0.0f and a=%0.0f\n",ans(1),ans(2));

b=[17689;

21443];

ans=a \ b;

printf("The answer to question (1b) is s=%0.0f and a=%0.0f\n",ans(1),ans(2));

b=[17789;

21443];

ans=a \ b;

printf("The answer to question (1c) is s=%0.0f and a=%0.0f\n",ans(1),ans(2));

- question 2

printf("Current ratio for use of steel by auto is %0.4f\n",2664/30346);

a=[(25448-5395)/25448 -0.0500;

-48/25448 (30346-9030)/30346];

b=[17589;

21243];

ans=a \ b;

printf("The answer to 2(a) is s=%0.0f and a=%0.0f\n",ans(1),ans(2));

b=[17589;

21500];

ans=a \ b;

printf("The answer to 2(b) is s=%0.0f and a=%0.0f\n",ans(1),ans(2));

- question 3

printf("The value of steel used by others is %0.2f\n",18.69-(6.90+1.28));

printf("The value of autos used by others is %0.2f\n",14.27-(0+4.40));

a=[(18.69-6.90)/18.69 -1.28/14.27;

-0/18.69 (14.27-4.40)/14.27];

b=[1.10*(18.69-(6.90+1.28));

1.15*(14.27-(0+4.40))];

ans=a \ b;

printf("The answer to 3(a) is s=%0.2f and a=%0.2f\n",ans(1),ans(2));

printf("The 1947 ratio of steel used by steel is %0.2f\n",(18.69-6.90)/18.69);

printf("The 1947 ratio of steel used by autos is %0.2f\n",1.28/14.27);

printf("The 1947 ratio of autos used by steel is %0.2f\n",0/18.69);

printf("The 1947 ratio of autos used by autos is %0.2f\n",(14.27-4.40)/14.27);

printf("The 1958 ratio of steel used by steel is %0.2f\n",(25448-5395)/25448);

printf("The 1958 ratio of steel used by autos is %0.2f\n",2664/30346);

printf("The 1958 ratio of autos used by steel is %0.2f\n",48/25448);

printf("The 1958 ratio of autos used by autos is %0.2f\n",(30346-9030)/30346);

b=[17.598/1.30;

21.243/1.30];

ans=a \ b;

newans=1.30 * ans;

printf("The answer to 3(c) is (in billions of 1947 dollars) s=%0.2f and a=%0.2f\n and in billions of 1958 dollars it is s=%0.2f and a=%0.2f\n",ans(1),ans(2),newans(1),newans(2));