WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we are going to apply what we have learned with matrices so far,
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in order to solve systems of equations using matrices.
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OK, first: a system of equations can be written as a single matrix equation, and that is the first step to solving the system of equations.
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So, let's use an example here: now, recall that, in previous lessons, we talked about a lot of other methods to solve equations,
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such as substitution and elimination; and it depends on the situation which is the easiest method.
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But this gives you another option, and it can be the best method in certain situations.
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So, first, you are going to write one matrix using the coefficients of x and the coefficients of y; and I am going to call that matrix A.
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In the first column is the coefficients of x; in the second column, I am going to have coefficients of y.
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OK, that is the first matrix: the second matrix I will call B, and for this matrix,
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you are going to use the constants that are on the right side of the equation: 1 and 3.
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Finally, you are going to write a third matrix that I will call X; and that is going to consist of variables in the system of equations.
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OK, so now I have these three matrices; but I want to write them as an equation.
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And how I could write these is that I could say that matrix A times matrix X equals matrix B.
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And let's look at why: if I have matrix A, that is 5, 6, 4, 3 for the elements, times matrix X (that is my variables), it is going to give me my constants.
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Well, let's just look at what this is saying by using matrix multiplication.
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This is my matrix equation for this system of equations: I have written this matrix equation right here.
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All right, let's multiply this out and see what happens.
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Well, with matrix multiplication, I am going to multiply this first row times the column (I only have one column).
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And it is going to give me, for my product matrix, 5x + 4y; I can't go any farther.
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OK, now for this second row, I am going to also multiply it by the column; and then, that is going to give me 6x + 3y =...
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Recall the definition of equal matrices: equal matrices have corresponding elements that are equal.
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So, this element here (row 1, column 1) and this element here (row 2, column 2) correspond to row 1, column 1, row 2, column 1, here and here.
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OK, so that gives me 5x + 4y =...I look at the corresponding element in row 1, column 1.
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Here, I look at row 2, column 1, and my corresponding element here; and I get my original equations back.
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So, that shows that this system of equations is equivalent to this matrix equation.
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And that is what allows us to write this system of equations as a matrix equation.
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OK, now let's talk about solving systems of equations.
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We found this matrix equation AX = B: however, if I want to solve, what I really want to find is...
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I want to isolate X, because recall that the matrix X, we defined as containing the variables.
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And I want to isolate that, the same way as, if I am solving a regular system of equations, I want to isolate the variable.
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Well, we already talked about how this matrix equation, AX = B, can be written for a system of equations.
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So now, let's talk about isolating X.
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Recall that what we want to do is get rid of the A; so we can do that by multiplying it by its inverse.
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I am thinking for a second about why we are doing that: we are taking it, and we are multiplying it by its multiplicative inverse--matrix inverse--A^-1.
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Well, think about when we are working with just regular numbers.
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If I have 3x + 4, what I am going to do is multiply 3 by its multiplicative inverse, 1/3.
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If I have 3x = 4, and I want to isolate the x, I am going to divide by 3; or I could say I am going to multiply by 1/3.
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So, if I multiply 3 by its multiplicative inverse, 1/3, this is going to cancel out, and I am going to end up with 1.
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Well, what is going to happen here? Recall that, when we talked about identity matrices,
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we said they function somewhat like the number 1 in the world of matrices.
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So, if I am multiplying a matrix by its multiplicative inverse, then I am going to get the identity matrix.
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In the same, when I multiplied a number by its inverse, I got the number 1.
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So, A times A^-1, which we discussed in previous lessons on identity matrices, is the identity matrix.
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So, this is going to give me the identity matrix, times X, equals A^-1 times B.
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OK, now also recall that, if you multiply a matrix by its identity matrix, you get the original matrix back.
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What that tells me is that, if I multiply X by its identity matrix, that is the same as the matrix X.
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So, I can just write this as X = A^-1B.
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OK, so this is what is important to know; but this shows you how we got there.
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Now, in the last slide, we talked about writing the matrix equation AX = B.
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Here, we talked about isolating X; so what you really want to work with is this, where X equals A^-1 times B,
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because that will allow you to solve for this matrix, and therefore to solve for the variables in the system of equations.
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For example, write the system as a matrix equation.
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First, all this is asking is to write it as a matrix equation; we don't even need to solve.
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So, I have 2x - 3y = -7; this is 3x + 2y = 8; and we talked earlier about writing a matrix equation, AX = B.
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And that is going to consist of three matrices.
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The first matrix, A, is going to have, in the first column, the coefficients of x.
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In the second column, it is going to have the coefficients of y.
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Then, I am going to write a second matrix, B, which is going to have elements that consist of the constants, -7 and 8.
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Then, I am going to have a third matrix that contains the variables.
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The next thing to do is to put these together into an equation: that is going to give me A, times X, equals B.
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The matrix equation for this system of equations is shown here: and this is AX = B.
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Again, I wasn't asked to solve it, so I have done what I have been asked, which is to write the system as a matrix equation.
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And we can use that as the basis of solving systems of equations, just as we are going to do in this problem.
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The first thing to do is to write the matrix equation, so I need my three matrices: the first one is A;
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and the coefficient of x is 1; and the other is 2; the coefficients of y are -2 and 1.
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The second matrix, B, contains the constants for elements.
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And then, finally, we have X, with my variables x and y.
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Now, from this I could write AX = B; and we saw, earlier on, how, in order to solve it, I need to rearrange things
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and do some manipulation and end up with this, because I really want to solve for the matrix X, because it contains the variable.
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So, this is the equation I want; and I have A, B, and X, so what I need is A^-1.
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Recall from an earlier lesson that A^-1 is 1 over the determinant, times this matrix.
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OK, so all I am doing right now is finding A^-1, if it exists.
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This is ad, which is 1 times 1, minus -2 times 2.
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And just looking over at this, this is going to be -4, so it is 1 minus -4, so that is not 0.
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So, I am OK--the determinant does exist--so I am going to continue.
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I am going to switch these two, but they are the same number, so it ends up 1 and 1.
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Then, I am going to take -2 and reverse its sign; so that is going to be 2.
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Here, I am going to take 2 and reverse its sign (because that is c), and I am going to make it -2.
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This is going to give me 1 over 1 - -4, times this matrix.
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A^-1 is 1 over 1 + 4, which is going to give me 1/5.
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So, I don't even have to multiply this out, because I am going to be doing some multiplication over here.
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So, leaving this as it is: now I want to have X = A^-1 times B.
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So, X equals...here is A^-1, 1/5 times its matrix, 1, 2, -2, 1, times B; B is 3, 7.
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Now, I am going to go ahead and do my matrix multiplication to find the product matrix.
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So, the product matrix would be: for row 1, column 1, this is 1, 2, times 3, 7--so 1 times 3, plus 2 times 7, equals 3 + 14; that is 17.
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So, 17 goes right here; now, in row 1, column 2, right here, this is going to be 1...actually, it only has one row,
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so it is just row 2, column 1; that is all I need to find; row 2 right here is the next position, because I have no other column here.
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So, it is going to be row 2, column 1; that is going to give me -2 times 3, plus 1 times 7; that is -6 plus 7, which is going to give me 1.
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OK, I found that I have x = 1/5 times 17/1, which equals...I am going to multiply 1/5 by 17--that is 17/5.
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And multiplying 1/5 by 1, that is 1/5.
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Now, the matrix X equals this; recall that matrix X also equals xy; so these two are equal;
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therefore, the corresponding elements are equal; so I can say that x = 17/5 and y = 1/5.
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It is a little bit complicated, but you just have to take it one step at a time.
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I was asked to solve this using a matrix equation.
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I wanted to find this equation, because then I would have the matrix with the variables in it isolated.
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I started out by writing three matrices--one containing the x and y coefficients, one containing the constants, and one containing the variables.
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Then, I had A, but I needed to find A^-1; and I used my formula to find that.
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I went through and found that this is A^-1.
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With that, I could then insert it into this formula: A^-1, right here, times B.
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I went ahead and did matrix multiplication to find that I got this matrix as the product.
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And then, I multiplied it by 1/5, which I still had out here, to get the matrix X equals 17/5 and 1/5, which equals x and y.
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So, that is my solution to this system of equations.
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Example 3: again, solve using a matrix equation.
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So, I am going to go ahead and find my three matrices that I will need to have,
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because I eventually want to end up with X = A^-1B.
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So, A is a matrix with coefficients of x 3 and -1, and coefficients of y 7 and -2.
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Matrix B has elements that are the constants 4 and 3; and then, matrix x has the variables.
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The next thing is: I actually don't need A; I need A^-1.
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And I am going to recall my formula, that this equals 1 over the determinant, ad - bc, times this matrix.
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So, A^-1 is 1 over the determinant, which is 3 times -2, minus 7 times -1,
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times this matrix: a and d switch positions, so it is going to be -2, 3; b reverses its sign; and c reverses its sign.
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Proceeding to find the determinant: 3 times -2 is -6...that is minus -7, times this matrix...
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that is going to give me A^-1 is 1 over...well, this is -6 minus -7, or -6 + 7, so that is just 1.
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OK, I have A^-1: I found A, B, and C matrices; now I have A^-1.
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And this is what I want: I want that X = A^-1B.
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So, I am going to write A^-1 up here; and I am just going to write this as 1/1, times B; B is right here.
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I am doing some matrix multiplication: row 1, column 1 is going to be -2 times 4, plus 7 times 3.
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That is going to be -8; let me see, this should actually be a -7.
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This is going to be -2 times 4; and this is going to be 3, -7, times 3.
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So, row 1, column 1: -2 times 4 and -7 times 3.
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That is going to give me -8, minus 21, which is going to be -29 for row 1, column 1.
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Row 2, column 1: row 2 here, column 1 here: that is going to be 1 times 4, plus 3 times 3, equals 4 + 9, equals 13.
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It is actually -29; OK, so row 1, column 1 gave me -29; row 2, column 1 gave me 13.
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So, x equals 1 times this; well, since this is just 1, I am going to end up with x equals -29, 13 as elements.
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And I also know that x is this; corresponding elements in equal matrices are equal, so -29 equals x, and 13 equals y.
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So, this is my solution to this system of equations, using matrices.
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Again, solve using a matrix equation: and I am going to keep in mind that I want to find x = A^-1B,
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where A is a matrix with the coefficients of x and the coefficients of y;
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B is a matrix containing the constants 8 and 9; and x is a matrix containing variables--the variables in the system of equations.
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I need A^-1: A^-1 is 1 over the determinant of A, ad - bc, times the matrix d, -b, -c, a.
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Therefore, A^-1 is 4 times -6, minus -3 times 8, times this matrix.
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We are reversing a and d; so I take a, and I put it in the d position; d goes in the a position.
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For b, I am going to reverse the sign; I am going to make it 3.
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For c, I am going to reverse the sign; and that is -8.
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OK, here I have 4 times -6; that is -24, minus -3 times 8; that is -24; and you can probably already see what the issue is.
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That A^-1 would equal 1 over -24 minus -24, times its matrix; and we have a problem,
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because taking this one step further, this gives me 1/0.
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And it really doesn't matter what this matrix is over here, because this is undefined.
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A^-1 does not exist; I cannot use this method.
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So, the situation is that there is no unique solution; there may be no solution at all; there may be an infinite number of solutions.
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But this method only works if there is a unique solution--a unique solution for x, and a unique value for y, that satisfies the system of equations.
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So, I started out trying to use my method and writing this matrix equation.
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And I got my matrices OK; but once I got to this step, finding A^-1, then I discovered
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that the determinant for this matrix A is 0, and in that case, A^-1 does not exist; I can't use this method.
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OK, that concludes this lesson on solving systems of equations using matrices.
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