Explanation Given: The equations are 2 x − 3 y = 10 and 2 x + 2 y = 5 . Result used: “To solve linear system of equation using matrix inverse: Form the matrix equation A X = B from system of linear equations, where A is the matrix of the coefficients of the variables of system, X is the column matrix of the variables and B is the column matrix of the constants and X = A − 1 B . Inverse of a 2 × 2 matrix: If A = [ a b c d ] and det A ≠ 0 , then A − 1 = 1 det A [ d − b − c a ] . If det A = 0 , then A − 1 does not exist and A is a singular matrix.” Calculation: The given equations { 2 x − 3 y = 10 2 x + 2 y = 5 can be written in matrix equation form as shown below: [ 2 − 3 2 2 ] [ x y ] = [ 10 5 ] Hence, the coefficient matrix A = [ 2 − 3 2 2 ] . The determinant of the coefficient matrix is calculated to check whether A − 1 exists. det A = 2 ( 2 ) − 2 ( − 3 ) = 4 + 6 = 10 Since det A ≠ 0 , then the inverse of matrix A , which is A − 1 exists. Now, compare the 2 × 2 matrix A = [ 2 − 3 2 2 ] with the general form A = [ a b c d ] then the value of a = 2 , b = − 3 , c = 2 and d = 2

A Graphical Approach to College Algebra, Books a la Carte Edition (6th Edition)

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Chapter 6.6, Problem 38E

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To solve: The system of equations {2x−3y=102x+2y=5 using matrix inverse method.

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Chapter 6.6, Problem 37E

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A Graphical Approach to College Algebra, Books a la Carte Edition (6th Edition)

Ch. 6.1 - CHECKING ANALYTIC SKILLS Use each graph to...Ch. 6.1 - Prob. 8E Ch. 6.1 - CHECKING ANALYTIC SKILLS Use each graph to...Ch. 6.1 - Prob. 10E Ch. 6.1 - Prob. 11E Ch. 6.1 - Prob. 12E Ch. 6.1 - Prob. 13E Ch. 6.1 - Prob. 14E Ch. 6.1 - Prob. 15E Ch. 6.1 - Prob. 16E

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