Solve each system of equations by using A^-1. Note that the matrix of coefficients in each syste

Solve each system of equations by using A^-1. Note that the...

Key Concepts

Systems of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. The goal is to find the values of the variables that satisfy all the equations simultaneously. Understanding how these systems are structured and the conditions under which they have unique, infinite, or no solutions is fundamental in linear algebra.

Matrix Representation

Representing a system of linear equations using matrices involves creating a coefficient matrix for the variables, an unknown vector for the variables themselves, and a constant vector for the right-hand side of the equations. This representation simplifies the manipulation and solution of the system, particularly when using techniques such as matrix inversion.

Coefficient Matrix

The coefficient matrix is constructed by taking the coefficients of the variables from each equation in the system. This matrix is key to translating the system into a compact matrix equation form (Ax = b), which is essential for applying methods from linear algebra to solve the system.

Matrix Inverse

The inverse of a matrix, when it exists, is a matrix that, when multiplied by the original matrix, yields the identity matrix. The existence of an inverse is pivotal in solving systems of equations since it allows us to 'undo' the effect of the coefficient matrix. A matrix must be square and have a nonzero determinant to be invertible.

Solving Linear Systems Using Matrix Inversion

When a system of linear equations is written in the form Ax = b, and the coefficient matrix A is invertible, the solution for the variable vector x can be found by multiplying both sides of the equation by the inverse of A. This gives the formula x = A?¹b, which is a powerful method in linear algebra for finding the unique solution of the system.

Transcript

00:01 In this problem we have given the equations 2y is equal to 6, 3x plus 3y plus 2 z is equal to 16 and 2x plus 5y plus z is equal to 19.

00:21 Let's consider it equation 1, equation 2 and equation 3.

00:27 We can also write these linear equations as a is equal to 020 -332 -251 b is equal to 6, 16 and 19 and x is equals to x, y and z.

00:59 First we find a inverse for that we write augmented matrix a slash i now we use row operations to convert a slash i matrix into i slash a inverse we have a matrix that is 020332251 slash identity matrix of order 3 by 3 that is 1 000 010 by interchangeing the first and second row, r1 changes to r2 and by applying the operation r1 changes to 1 by 3 r1.

01:56 We get 112 by 3 020251 plus 0151 slash 1 by 31001010 now by using the operations r3 changes to minus 2 r1 plus r3 and r2 changes to 1 by 2 r2 r2 we get 1 1 2x3 010 -03 -0 -3 -0 -3 -1 -3 -1 -3 -1 by 0 -1 by 3 -1 by 2 -0 -0 minus 2 by 3 1.

03:01 Again by applying the operations, r3 changes to minus 3 r2 plus r3 and r1 changes to minus r2 plus r1.

03:20 We get 1 02 by 3 010 -00 -m -mine minus 1 by 3 slash minus 1 by 2 1 by 3 0 1 by 2 0 0 0 0 0 0 0 0 0 0 minus 2 by 3 by 3 1 1.

03:48 Since the 1 and second column are in the desired form we work on the third column by applying the operation r3 changes to minus 3 r1 and r1 changes to minus 2 by 3 r3 r3 plus r1.

04:09 We get 1 000 -00 -00 -1 slash minus 7 by 2 minus 1 2 minus 1 2 1 by 2 9 by 2, 2 minus 3.

04:32 Since the last matrix is in the form of i slash a inverse from here we get a inverse is equals to minus 7 by 2 minus 1 2 1 by 2 0 0 9 by 2 minus 3...

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