# Linear Algebra/Input-Output Analysis M File

`# 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));`