Solve the system of equations using the inverse of the coefficient matrix of the equivalent matr

Solve the system of equations using the inverse of the...

Key Concepts

Solving via Matrix Inversion

This method involves computing the inverse of the coefficient matrix (assuming it exists) and then multiplying it by the constant vector b to obtain the solution vector x. The solution is given by x = A?¹ * b, which is efficient especially for systems where the coefficient matrix is invertible.

System of Linear Equations

This refers to a set of equations that involve the same set of unknowns. The goal is to find values for these unknowns that satisfy all equations simultaneously, which is a central problem in algebra and linear algebra.

Matrix Representation

This concept involves expressing a system of linear equations in the form A*x = b, where A is the matrix containing the coefficients of the variables, x is the column vector of the unknowns, and b is the column vector of the constant terms. This form is particularly useful for applying matrix operations and computational techniques.

Coefficient Matrix

The coefficient matrix is the matrix formed by the coefficients of the variables in the system of equations. It encapsulates the structure of the linear system and is crucial for determining properties like consistency and invertibility of the system.

Matrix Inverse

The inverse of a matrix, when it exists, is the unique matrix that, when multiplied with the original matrix, produces the identity matrix. It plays a key role in solving systems of equations, as the inverse can be used to 'undo' the effect of the coefficient matrix.

Uniqueness and Invertibility Conditions

A unique solution to the system exists if and only if the coefficient matrix is invertible (non-singular). Invertibility is determined by the matrix having a non-zero determinant and a full rank, ensuring that the transformation represented by the matrix has a unique inverse.

Transcript

00:02 For the given equation, a into b should be equal to x to find b, the a inverse into x as the right.

00:09 Now, to find the given a is 5 minus 4, 3 minus 2, 1, 4 minus 2, 3, 2 minus 2, 3, 2 minus 9, 3 minus 5, 2 minus 4.

00:27 Now we have to find the a inverse that can be done by 5 minus 4 3 minus 2.

00:34 You have to make it as 1 0 0 0 0 then 1 4 minus 2 minus 2 minus 9 0 0 0 0 0 0 0 minus 5 2 minus 4 0.

00:50 Now first making this as 1 that can be done by swapping the first row in the second row.

00:58 So doing that we will get 1 .4 minus 2 3 0100 then 5 minus 4 3 minus 2 0 0 0 0 0 0 0.

01:13 2 minus 3 minus 6 minus 9 0 0 0 0 0.

01:25 Now eliminating this term, this term and this term that can.

01:31 Be done by 1 .4 minus 2 3 0100 0 minus 24 30 minus 17 1 minus 5 00 0 minus 11 10 minus 15 0 minus 15 0 minus 210 0 minus 17 0 minus 17 8 minus 13 0 minus 13 0 minus 3 0 1 1 now pivoting this point you are making this as 1.

02:10 That can be done by dividing it by 24.

02:14 So that is 1 .4 minus 2, 3.

02:18 0 .1 minus 13 by 24.

02:21 17 by 24.

02:23 0 minus 11, 10, minus 15.

02:26 0 minus 17, 8, minus 13.

02:30 And here it is 0 -1 -0.

02:32 Minus 1 by 24 5 by 24 0 0 minus 2 1 0 0 minus 3 0 0.

02:42 Now eliminating all these points may make it as 0 that can be done by doing row operations is 1 0 1 by 6 1 by 6 0 1 minus 13 by 24 17 by 24 0 097 by 24 minus 173 by 24 0 0 minus 29 by 24 minus 23 by 24 in this side it is 1 by 6 1 by 6 0 minus 1 by 24 5 by 24 0 minus 11 by 24 7 by 24 10 by 24 10 by 20 10 minus 17 by 24 10 10 minus 17 by 24 14 30 by 24 0 now pivoting this point by making it as 1 that is multiplying row 3 by 24 by 94...

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