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

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

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Key Concepts

Matrix Equation Representation

This concept involves writing a system of linear equations in the compact form A · x = b, where A is the coefficient matrix, x is the column vector of variables, and b is the constant vector. This representation is fundamental because it allows the use of matrix operations to solve systems of equations, paving the way for efficient computational techniques.

Matrix Inversion

Matrix inversion is the process of finding a matrix A?¹ such that when multiplied with the original matrix A, it yields the identity matrix. In the context of solving systems of equations, if A is invertible, the solution can be directly obtained as x = A?¹ · b. This method is particularly useful for solving systems when an explicit formula for the inverse is known or can be computed.

Invertibility Conditions

For a matrix to have an inverse, it must be square and its determinant must be non-zero. This condition ensures that the matrix is full-rank, meaning its rows (or columns) are linearly independent. The invertibility condition is crucial when deciding whether the matrix inversion method can be applied to obtain the unique solution of a system of linear equations.

Transcript

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

00:04 So writing out our matrixes.

00:07 Our a matrix would be 3 .3, 1, 1, 2 ,1, 2, 1, 1.

00:15 Our x matrix would be our variables.

00:18 So x, y, z.

00:21 Our b matrix would be our answers.

00:26 So 8, 5, 4.

00:28 And our inverses are from our previous section.

00:31 So i have them all written down.

00:33 So we're looking for the one that matches to 3 .3 .1.

00:37 So 331 is this bottom one.

00:40 So 3 over 7, negative 4 over 7, 1 over 7, 1 over 7.

00:49 And then we've got 1 over 7, 1 over 7, negative 2 over 7.

01:01 And then for our last row, negative 5 over 7, 1 over 7, 3 over 7.

01:21 I'm going to do this a little differently than i've done some previous problems.

01:24 Because these are all over 7, i'm actually going to take that.

01:28 One sevenths out and that's going to get rid of all the bottoms and leave me with a little bit of a nicer matrix to deal with so three negative four one one one negative two negative five one three because that one seventh gets multiplied to everything in my matrix i can do that it just makes this step a little bit easier to put together so i'm gonna multiply these two matrixes first and then multiply my answer by one seventh...

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