# Linear Algebra/Input-Output Analysis M File

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``` matlab
 # 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));
```

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