Let A=[ 1 2 5 12 ], 𝐛1=[ -1 3 ], 𝐛2=[ 1 -5 ], 𝐛3=[ 2 6 ] and

Let A=[ 1 2 5 12 ], 𝐛1=[ -1 3 ], 𝐛2=[ ...

Key Concepts

Augmented Matrices

An augmented matrix is a compact representation that combines the coefficient matrix of a system of equations with its corresponding constant terms. This format is particularly useful when applying row operations to systematically solve the system by reducing the matrix to row echelon or reduced row echelon form.

Matrix Inversion

Matrix inversion is the process of finding a matrix, denoted as A?¹, which when multiplied with the original invertible matrix A yields the identity matrix. This concept is fundamental because it provides a direct method for solving systems of linear equations by using the relation A?¹A = I.

Systems of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. Solving these systems aims to find values for the variables that satisfy all the equations simultaneously. Techniques such as matrix inversion and row reduction facilitate efficient solution methods.

Row Reduction (Gaussian Elimination)

Row reduction, or Gaussian elimination, is a method used to simplify systems of linear equations by performing elementary row operations on an augmented matrix. This process systematically eliminates variables, leading to a form of the matrix that makes the solution for each variable straightforward, such as the reduced row echelon form.

Transcript

00:01 In this example, we have been provided with a matrix and four different vectors.

00:05 The first thing we'd like to do is calculate the inverse of this matrix a.

00:09 So to that end, let's start by calculating the determinant of a.

00:14 We know that the determinant of a is found by taking the product of the main diagonal first, so this will be 1 times 12, and then subtracting the product of the off diagonal, which is 5 and 2.

00:28 So we'll take 1 times 12 minus 5 times 2, and this results in a determinant of 12 minus 10, which is equal to 2.

00:39 So now that we have the determinant of this matrix, we can say that the inverse of a, since it's 2 by 2, is found by taking 1 1⁄2, times the following matrix where we first go to the main diagonal with 112 and interchange the order, so we have 12 and 1.

01:00 Then go to the off diagonal, which is 5 -2, and multiply the entries by negative 1.

01:05 This gives us a negative 5 and a negative 2, which goes here.

01:10 And ordinarily, what i'd like to do is say what this is equal to by scaling all entries by this factor of 1 half.

01:17 But it turns out this time, it's more convenient to leave the a inverse in this form with the scalar on the outside.

01:24 Let's explain why.

01:25 So our next step in this problem is to look at the matrix equation, a 2nd, and a 2nd.

01:30 Times x equals b1.

01:32 We can solve this by augmenting into a 2x3 matrix and then row reduction, but now that we have an inverse we can also say that the solution is of the form x equals a inverse times b1.

01:46 Now to solve for x we'll put in a inverse which is that half we mentioned, multiplying 12, negative 5, negative 2 1 times b1, which is negative 1 and 3.

02:01 So the reason we left the 1 half on the outside is that it's easier to deal with this multiplication here when we don't produce fractions from here and here.

02:12 So for the next step, we can say x is 1 half of the result of this matrix times this vector.

02:19 We have altogether the first entry is going to be equal to negative 18 and the second entry is positive 8 when we multiply the matrix and the vector.

02:28 Now scale that.

02:29 Result by a half and we find that x is going to be equal to negative 9 and 4 so that solves the first equation that we're interested in let's go to the next interesting equation a times x equals b2 copying our method from before we know that x is going to be first a inverse which is a half times the matrix 12 negative 5 negative 2 1 then multiply by b2 and b2 is given here.

03:05 So we multiply this by 1, negative 5.

03:09 So altogether, x is going to be equal to 1 half, multiplying the matrix times the vector, which results in 22, negative 10.

03:21 Then scale this vector by a half, and x is equal to 11 and negative 5.

03:27 Now let's go to the next matrix equation that we are interested in.

03:31 It's a x equals, if you have haven't guessed b3, then we'll also solve ax equals b4 next.

03:41 First, when ax equals b3, we find that x is equal to one -half, copying our strategy from before, put in the matrix 12 -5, negative -5, negative 2 -1, and b -3 is equal to 2 -6.

04:01 So in this case, we obtain one -half times the vector, which is going to be a 12 and negative 4.

04:10 So we have that x is equal to, when we are looking at a x equals b3, a 6, and negative 2, once we scale that last vector by 1 half.

04:20 Let me move some of these over so we can consider this last equation with a little more space.

04:26 So now that we have more space, once again, we can solve this matrix equation by writing x equals a inverse, which is a half times the matrix 12 -9 -5, negative 2 -1, and multiply by the vector b4, which is 3 .5.

04:45 In this case, we'll have x is equal to 1 half times the vector, which will be 13 and negative 5, when we multiply this matrix here with a vector.

04:57 Then scale by a half, and we see that x must be equal to, oh, i made a mistake here, should be of 13 negative 5 here, and so in this vector, let me delete some of these, after multiplying the matrix, with the vector, it was 26 and negative 10.

05:17 Then scaling by a half gives us 13 and negative 5.

05:21 So these are our two, or excuse me, four matrix equations that we have solved here, here, here, and here.

05:28 The respective solutions are negative 9 ,4, 11 negative 5, 6 negative 2, and 13 negative 5...

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