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

Solve the system of equations using the inverse of the coefficient matrix of the equivalent matrix equation.
$$\begin{aligned}
2 w-3 x+4 y-5 z &=0 \\
3 w-2 x+7 y-3 z &=2 \\
w+x-y+z &=1 \\
-w-3 x-6 y+4 z &=6
\end{aligned}$$

Key Concepts

Matrix Multiplication

Matrix multiplication is used to combine matrices in the process of solving linear equations. In the inverse matrix method, once the inverse A?¹ is computed, it is multiplied by the constants vector b to yield the solution vector x in the equation Ax = b.

Determinant and Invertibility

The determinant of a matrix is a scalar value that indicates whether a matrix is invertible; a nonzero determinant implies that the inverse exists. Understanding this concept is fundamental when deciding which methods can be used to solve a system of linear equations.

Matrix Representation of Linear Systems

This concept involves expressing a set of linear equations in the compact form Ax = b, where A is the coefficient matrix containing the coefficients of the variables, x is the vector of unknowns, and b is the constants vector. This formulation is fundamental for applying matrix methods to solve systems of equations.

Coefficient Matrix

The coefficient matrix, denoted as A, is constructed by organizing the coefficients of the variables from the system of linear equations into a rectangular or square matrix. Its properties, such as invertibility, play a crucial role in determining whether the system has a unique solution.

Matrix Inverse

The inverse of a square matrix A, denoted as A?¹, is a matrix that satisfies the equation A?¹A = I, where I is the identity matrix. This concept is key in solving linear systems, as the unique solution can be found by computing x = A?¹b, provided that A is invertible.

Transcript

00:01 The given set of equation, the matrix is, can be written as 2 minus 3, 4, minus 5, 3 minus 2, 7, minus 3, 1, minus 1, minus 1, minus 1, minus 3, minus 6, 4.

00:25 And then we have wx, y, z, which will be equal to minus 6, which will be equal to 0 ,000.

00:41 216.

00:45 So solving this in terms of inverse of matrix we will get 2 minus 3, 3 minus 7 minus 3 1 1 minus 1 minus 3 6 4 in terms of 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 1.

01:11 And now we will swap r1 with r3 to get this as 1.

01:22 So doing that we will gain 1 -1 -1 -0 -1 -0 -0 -0 -1.

01:39 And this will be 3 minus 2, 7 minus 3, 2, 4 minus 5, minus 1, 3 minus 6.

01:49 Now eliminating these row elements.

01:53 How to eliminate its r2 is equal to r2 plus 3 r1 r3 is 2 r3 minus 2 r1 and r4 plus r1 so doing that we will get 1 1 1 1 1 0 0 minus 5 10 minus 6 0 1 minus 3 0 0 minus 5 10 minus 6 0 1 minus 3 0 0 minus 5 6 minus 7 minus 6 minus 6 minus 5 6 minus 7 0 0 0 minus 6 minus 7 minus 6 minus 7 0 0 minus 3 0 to 0 .0 minus 2, minus 7, 5, 0 ,0 1.

02:35 Now, we voting this point we will do 1 -1 -1 -0 -0 -0 -1 -0 -1 -1 -6 -5, 0 -1 -1 -6 -5, 0 -1 -1 -5, 0 -1 -5, 0 -1 -5 -5, 3 -5 -5 -0 -5, 0 -5 -6 -6 -7 -1 -1 -0 -0 -5 -6 -6 -7 -1 -1 -0 -0.

03:02 Minus two zero zero minus two minus seven five zero zero one one now eliminating this term this term and this term we will again one zero one by five zero one by five two by five zero one minus two by five six by five zero minus one by five three by five zero zero zero minus one by five three by five zero zero zero minus one minus five five five three by five zero zero zero minus 4 minus 1...

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