In Problems 41-60, use the inverses found in Problems 31-40 to solve each system of equations. {

step 1 We're using matrices and inverses to solve them. So we've got our equation there on the right. S

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

Key Concepts

Systems of Linear Equations

Systems of linear equations consist of multiple equations that involve the same set of variables. The primary goal is to find values for these variables that simultaneously satisfy all the equations. This concept is foundational in algebra and is widely applicable in various fields including engineering, economics, and physics.

Matrix Representation of Systems

A system of linear equations can be expressed in matrix form as Ax = b, where A is the coefficient matrix, x is the vector of unknowns, and b is the constant vector. Representing the system in this way allows for the application of linear algebra techniques to simplify and solve the equations systematically.

Matrix Inversion

The matrix inversion concept involves finding a matrix A?¹ such that A?¹A = I, where I is the identity matrix. This property is crucial when the inverse exists because it provides a direct method for solving systems of equations by transforming the equation Ax = b into x = A?¹b.

Solving Systems Using Inverses

When the coefficient matrix A is invertible, the system of equations can be solved by computing the inverse of A and then multiplying it by the constant vector b to obtain x = A?¹b. This method streamlines the solution process, making it especially useful when dealing with more complex systems.

Transcript

00:01 We're using matrices and inverses to solve them.

00:04 So we've got our equation there on the right.

00:07 So matrix a is b, 3, b, and 2.

00:13 Matrix x is our variables, so x and y.

00:18 Matrix capital b is our solution, so 14 and 10.

00:24 And our inverse was found in our previous section.

00:28 So i have those written on my next page.

00:29 So b3, b2, our inverse is negative 2 over b, 3 over b, 1, and negative 1.

00:37 So negative 2 over b, 3 over b, 1, and negative 1, and negative 1.

00:44 And i'm just going to double check that.

00:46 That is what we have.

00:48 So now all i have to do is take my inverse.

00:52 2 over b, 3 over b, 1 and negative 1.

00:58 Multiply that by my solutions for my answers and that should get me my solutions for x and y so i'm going to end up with negative 2 over b times 14 plus 3 over b times 10 and 1 times 14 minus 1 times 10 so negative 2 times 14 is negative 28 over b and 3 times 10 is 30 over b and 1 times 14 is 14 minus 1 times 10 so negative 10 ease us with a 4 so we can go a little bit further on the top there negative 28 over b plus 30 over b is 2 over b and then we have 4 so we know that this is x and this is y.

02:04 So we can go a little bit further here, and we're going to try to solve for what the value of x is, by inputting the value of x and the value of y into one of our equations...

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