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2 in matrix form 2 4 1 1 1 b 1 0 g 0 1 3 5 2 4 y c g

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2
In matrix form:241±1±1±b10±g013524YCG35=24I0a±bT0035For both solution methods, we need the determinant of the coe¢ cient matrix:jAj=²²²²²²1±1±1±b10±g01²²²²²²= 1±b±g(a) The solution using matrix inversion is:24Y±C±G±35=241±1±1±b10±g0135°124I0a±bT0035The inverse is:A°1=adj(A)jAj=11±b±g24111b1±gbgg1±b35Thus, the solution is (after simplifying):24Y±C±G±35=11±b±g24111b1±gbgg1±b3524I0a±bT0035=264a°bT0+I01°b°g(1±g)a°bT0+I01°b°g±I0g³a°bT0+I01°b°g´375(b) Using Cramer±s rule:Y±=jA1jjAj=11±b±g²²²²²²I0±1±1a±bT010001²²²²²²=a±bT0+I01±b±gC±=jA2jjAj=11±b±g²²²²²²1I0±1±ba±bT00±g01²²²²²²= (1±g)a±bT0+I01±b±g±I0G±=jA3jjAj=11±b±g²²²²²²1±1I0±b1a±bT0±g00²²²²²²=gµa±bT0+I01±b±g

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Term
Spring
Professor
BBLECHA
Tags
Economics, Linear Algebra, Interest Rates, Invertible matrix, San Francisco State University

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