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

step 1 Alright, we're going to solve the following using inverses. So we have the formula written out t

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

Key Concepts

Condition for Invertibility

A key concept related to matrix inversion is the condition that a square matrix must have a nonzero determinant to be invertible. In the context of the given problem, the parameter 'a' must be nonzero (a ? 0) to ensure that the coefficient matrix is invertible. This requirement ensures that a unique solution exists for the system, making the problem well-posed and solvable using the inverse matrix method.

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. The goal is to find values for the variables that satisfy all equations simultaneously. This concept is fundamental in algebra and applied fields because it models many practical problems where relationships between variables need to be determined and solved concurrently.

Matrix Representation of a System

Representing a system of linear equations in matrix form involves writing the system as AX = B, where A is the matrix of coefficients, X is the vector of unknowns, and B is the constant vector. This form allows the use of powerful matrix techniques, including operations like inversion, to solve the system efficiently and provides a structured framework for understanding the relationships between the variables.

Matrix Inversion

The process of matrix inversion involves finding the inverse A?¹ of a square matrix A, such that A?¹A equals the identity matrix. When a system of linear equations is written in the form AX = B and the matrix A is invertible, the solution can be obtained easily using X = A?¹B. This method is particularly useful because it streamlines solving the system by reducing it to a single matrix multiplication operation.

Transcript

00:01 Alright, we're going to solve the following using inverses.

00:03 So we have the formula written out there on the right, so our matrix a is 2, 1, a, a, our matrix x would be our variables, in this case our x and y.

00:19 Our matrix b is our answers, so that's 7 over a and 5.

00:25 And our inverses we found in the previous section, so i have them all written on the next page, so we're looking for the 2 -1a -a matrix.

00:34 So we know then that our inverse is 1, negative 1 over a, negative 1 over a, and 2 over a.

00:44 So 1, negative 1 over a, negative 1 over a, and 2 over a.

00:53 So now i can take my inverse.

00:57 1, negative 1 over a, negative 1 over a, 2.

01:02 2 over a, i can multiply that by my answers, so 7 over a and 5, and that will help me solve for my matrix x.

01:14 So i have 1 times 7 over a plus negative 1 over a times 5, and negative 1 over a times 7 over a plus 2 over a times 5.

01:35 So 7 over a plus a negative 5 over a is a 2 over a.

01:45 And let's see on the bottom, negative 1 times 7 is negative 7.

01:49 A times a is an a squared.

01:55 So i'm going to end up with a negative 7a squared.

02:01 And 2 times 5 is 10.

02:12 So right now we know that x is equal to 2a and y is equal to this negative 7 over a squared plus 10 over a.

02:25 So we can throw that back into one of our equations and try to solve for a...

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