In Problems 41-60, use the inverses found in Problems 31-40...

In Problems $41-60$, use the inverses found in Problems $31-40$ to solve each system of equations.
$$\left\{\begin{array}{c}
x+y+z=2 \\
3 x+2 y-z=\frac{7}{3} \\
3 x+y+2 z=\frac{10}{3}
\end{array}\right.$$

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Key Concepts

Solving Systems Using the Inverse Matrix

When a system of linear equations is expressed in the matrix form Ax = b, and A is invertible, multiplying both sides of the equation by A?¹ leads to the solution x = A?¹b. This method simplifies the process of finding solutions, particularly for complex numerical systems.

Matrix Inverse

An inverse of a square matrix A, denoted A?¹, is a matrix that, when multiplied by A, yields the identity matrix (A?¹A = I). The existence of an inverse indicates that the matrix is non-singular (i.e., it has a non-zero determinant), and it is crucial for solving systems of linear equations uniquely.

System of Linear Equations

This concept involves multiple equations that are solved simultaneously for common variables. In such systems, each equation represents a constraint that the solution variables must satisfy, and the goal is to find a set of values that meet all these requirements at once.

Matrix Representation

A system of linear equations can be succinctly represented in matrix form as Ax = b, where A is the coefficient matrix, x is the vector of variables, and b is the constant vector. This representation is a cornerstone in linear algebra, enabling the utilization of powerful matrix techniques to solve the system.

Transcript

00:01 Okay, so from our given system of equations, we get that our matrix a is equal to this matrix here.

00:07 The matrix 111, 32, 2, negative 1, 312.

00:12 X is equal to the column vector, xyz, and b is equal to the column vector of our constants, 2, 7 thirds, and 10 thirds.

00:20 Now, from we get that x, right, or xyz is equal to a inverse times b.

00:30 So x is equal to a inverse times b.

00:36 So what is a inverse? well, we just take our matrix a.

00:39 We concatenated it.

00:40 They put a line and put the three by three identity.

00:43 Use elementary rule operations to move the identity over to the left -hand side.

00:46 And what we end up with on the right -hand side is our inverse.

00:49 We actually found the inverse already in a prior problem.

00:52 And we found that a inverse was equal to this matrix here, is equal to negative 5 seventh, one seventh, three seventh.

01:07 In our first row, and then we have nine seventh, one seventh, negative four seventh.

01:15 And then in our third row, we had three seventh, negative two seventh, and one seventh.

01:25 So there is a inverse.

01:30 There is a inverse.

01:32 Okay, so then to find x, right, to find x, y, and z, right, which is just the column vector, x, you know that x equal to a inverse times b...

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