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 continue on talking about matrices, this time focusing on two special types of matrices, identity and inverse matrices.
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The **identity matrix** is a square nxn matrix, which has 1 for every element in the main diagonal, and 0 for every other element.
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So, an identity matrix is a square matrix...and let's look at an example--for example, a 2x2 matrix that is an identity matrix.
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It says that this has 1 for every element in the main diagonal; so that is 1 and 1; and 0 for every other element.
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Looking at another example of this for a 4x4 matrix: in the main diagonal (that would be here), I have all 1's.
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And I am going to fill in the other spots with 0's; so, down the main diagonal, I have 1's; everything else is a 0.
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And this is 1, 2, 3, 4; 1, 2, 3, 4; so it is a 4x4 square matrix.
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Now, let's talk about properties of these matrices.
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For any nxn matrix A, if you multiply that matrix times its identity matrix, then you will get the original matrix back.
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And you can also multiply these in either order.
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So, in a way, when you think about identity--the identity property of multiplication--think back to regular numbers;
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and when we talked about identity with multiplication, we would say that, if you multiplied the number n, any number,
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times 1, you got that original number back; and that was the identity property with multiplication.
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For example, if I took 3 times 1, I am going to get 3; or 10 times 1--I will get 10 back.
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This is the same idea, but with matrices; so, this functions the same way as 1 in this case,
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because if you multiply the matrix times its identity matrix, then you will get the original matrix back--
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the same way as, if you multiplied a number times 1, you get the number back.
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So, let's go ahead and try this--and remember that, unlike 1, though...it is just the number 1; but for identity matrices, there are multiple 1's.
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You see that, for a 2x2 matrix, the identity matrix will be different than for a 4x4 or a 3x3.
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OK, so let's use this 2x2, and let's say I had some matrix A, and it is going to be 3, -1, 2, 5.
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And I am going to multiply it by its identity matrix, which is going to be this one, because it's a 2x2; and let's see what I get.
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Well, recall that, for multiplication, if I am going to look for this position, row 1, column 1,
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I am going to multiply row 1 of this first matrix times this first column here; so that is going to give me 3 times 1.
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And then, I am going to find the sum of those products: 3 times 1, plus -1 times 0.
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That is going to give me 3 plus 0, or 3; OK, so I am going to get 3 up here.
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And let's look for row 1, column 2; it will be row 1, column 2; so 3 times 0, plus -1 times 1.
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This is going to give me 0 - 1; 0 - 1 is -1, so -1 goes right here.
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And continuing on now to the second row: row 2, column 1: row 2 of A, times column 1 of the identity matrix,
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is going to give me 2 times 1, plus 5 times 0, which is going to equal...2 + 0 is 2.
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OK, then finally, row 2, column 2: this is going to give me row 2, column 2--row 2 here, column 2 here.
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Row 2 is 2 times 0, plus 5 times 1; well, that is simply 0 + 5, or 5.
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So, you see that this property is shown here--that if I have a matrix A, and I multiply it by its identity matrix I, I get A back.
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In the same way, for numbers, with the identity property of multiplication: if I take a number and multiply it by 1, I get the original number back.
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OK, now talking about matrix inverses: if we have two nxn matrices (these are square matrices),
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and we say that they are inverses of each other...if I have two nxn matrices, A and B, they are **inverses** of each other
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if AB is equal to BA, and that product AB is equal to the identity matrix.
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So, if the product of A and B turns out to be the identity matrix, then those two matrices are inverses of each other.
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And the inverse of A, if it exists (and we are going to talk about that in a few minutes--that it may not always exist, and why),
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is written as A^-1: we pronounce this "A inverse."
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OK, again, let's say I was given two matrices, A and B, and then I was asked, "Are they inverses?"
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The way I would determine that is by multiplying those two out; and if the product is the identity matrix, then they are inverses.
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If it is not, then those two are not inverses of each other.
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Let's talk about finding the inverse of a 2x2 matrix.
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If we have some matrix, A, that looks like this; the inverse of A, if it exists, is given by this formula: A^-1 = 1/...
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and if you look at this, ad - bc, this is going to look familiar; and that is the determinant of this matrix.
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Remember that the determinant of a 2x2 matrix is given by the formula ad - bc; so it is 1 over the determinant, times this matrix.
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And it is pretty easy to remember how you form this matrix, if you just look at it this way.
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The way I got this is: I switched the positions of a and d, and then I took the opposite sign of b and the opposite sign of c.
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So, if I had numbers here...let's look at that: if I was told that A is the matrix 3, 4, 1, 2, and I was told to find A^-1:
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well, then I would take 1 over the determinant, which is 3(2) - 4(1), times...
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now, to find this matrix, I am going to switch these positions; I am going to put 2 here and 3 there.
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I am going to take this number; and I keep the number, but I am going to put the opposite sign.
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If this was a -4, I would have made it for; the same for the c position--I am going to keep it 1, but I am going to make it the opposite sign.
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Now, stopping here for a second: I mentioned "if the inverse exists."
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Looking at this, you can see that there could be a situation where the inverse does not exist.
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OK, here I have 1/(6 -4); so that is 1/2; well, that is allowable.
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But if this had been, say, 6 - 6, that would have been 0; that is not allowable--that is undefined.
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So, I wouldn't have been able to find A^-1.
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OK, therefore, there are situations (and it is when this denominator turns out to be 0) that you can't find the inverse.
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So, it is a good idea to look for this denominator right away, before you do any more work,
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to determine if we even can find the inverse, so you don't waste any more time looking for something that does not exist.
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Now that I have found that 1 over the determinant is actually 1/2,
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I can just multiply using scalar multiplication (this is functioning as a scalar) to figure out A¹.
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So, I am just going to multiply 1/2 times each element: 1/2 times 2 is 2/2, which is 1.
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And 1/2 times -4 is -4/2; that is -2; 1/2 times -1 is -1/2; and then, 1/2 times 3 is 3/2.
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Now, if you wanted to check your work, you could always check your work by taking A times A^-1,
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and seeing if it comes out to the identity matrix, which it should if you did things correctly.
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If I were to take 3, 4 times 1, 2, and multiply it by 1, -2, -1/2, 3/2, I would find that it does come out to the identity matrix for these 2x2 matrices.
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OK, so first we are asked, in Example 1, to determine if A and B are inverses of each other.
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So, are A and B inverses of each other? We are given these two matrices and asked if they are inverses.
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Well, recall that if two matrices are inverses of each other, then if I multiply them,
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I can actually multiply them in either order, and I am going to get the identity matrix.
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So, before I proceed, let's just think about what the identity matrix is for a 2x2.
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And I am going to have 1 along the main diagonal and 0 everywhere else, so this is the identity matrix in this situation.
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I am going to go ahead and multiply these and see what I end up with.
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OK, I am doing some matrix multiplication: first, row 1, column 1--first row, first column--that is 0 times 0, plus 1 times 1.
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And that is going to give me 1 in this position.
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Now, row 1, column 2: 0 times 0, plus...OK, then that would be 1, 1, times 2...row 1, column 2: I did 0 times 0, and now 1 times 2.
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That is 0 plus 2, so that is 2; so in this position, I am going to get 2.
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Now, row 2, column 1: 2 times 0, plus 1 times 1--that is going to give me 0 + 1 = 1 right here.
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Row 2, column 2: 2 times 0, plus 1 times 2: 0 plus 2 is 2.
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All right, so the question I was asked is if these are inverses of each other.
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Well, when I took their product (the product of A and B), I found that AB does not equal the identity matrix.
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This is not the identity matrix; this is; therefore, are A and B inverses of each other?
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No, A and B are not inverses of each other.
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And I was able to check that using this property.
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Find the inverse if it exists: so finding the inverse of a 2x2 matrix, recall, uses this formula: 1 over the determinant,
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which is ad - bc, times the matrix d, -b, -c, a.
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A^-1 (we are calling this matrix A) is 1 over the determinant.
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Here I have 1 times 4, minus 2 times 3.
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And just looking at this quickly, I see that this is 4 - 6, so that is -2; and therefore, that inverse is -1/2, so this inverse exists.
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I didn't get 0 down here (before I proceed any farther).
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That, times...da means I am going to switch these two: 4 will go in the a position; 1 will go in the d position.
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For b, I am just going to take the opposite sign; and for c, I am just going to take the opposite sign.
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Proceeding: this is going to give me 4 minus 6, as I said, which is going to give me -1/2, times 4, -2, -3, 1.
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Now, I need to multiply each element in here by -1/2; and that is going to give me -1/2, times 4, which is -2;
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-1/2 times -2 is -2/-2, which is 1; -1/2 times -3 is going to give me positive 3/2; and -1/2 times 1 is -1/2.
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Now, if I wanted to check this, I could check it by saying that A times A^-1 is the identity matrix.
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This is A^-1: let's go ahead and multiply this times A and see what happens.
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So, let's try A times A^-1, and just see what we get.
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Here, I had A; that is 1, 2, 3, 4; and A^-1, and that is -2, 1, 3/2, -1/2, equals...
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OK, working over here, row 1, column 1: that is 1, and I am just going to go ahead and do part of this mentally...
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1 times -2 is -2; 2 times 3/2...well, this cancels out; that just gives me 3.
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This is going to give me -2 + 3 = 1; so I get 1 right here.
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OK, row 1, column 2: that is 1 times 1 is 1; 2 times -1/2 is -2/2, which is -1.
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Since 1 - 1 is 0, 0 goes in the row 1, column 2 position.
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Let's go on to row 2, column 1: it is going to give me 3 times -2; that is -6.
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And then, this is 4 times 3/2 equals 12/2; that equals 6.
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So, -6 + 6 = 0; so I get 0 right here.
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Row 2, column 2--the last position--equals: 3 times 1 is 3; 4 times -1/2 would be -4/2, or -2; 3 minus 2 is 1.
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So you see, I actually did get the identity matrix; so I figured out A^-1, and I was checking
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that I was correct by saying, "Well, if this truly is the inverse, when I multiply these two together--
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if those two matrices are inverses of each other, I will get the identity matrix."
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And this is the identity matrix for a 2x2 matrix.
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So, the inverse--the answer to the question they are asking--is this right here.
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However, I checked my work right there.
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Example 3: Find the inverse, if it exists.
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I am using my formula, A^-1 = 1 over the determinant, ad - bc, times...switching d and a and changing the signs of b and c.
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So, A^-1 equals 1 over...this is ad; that is -1 times 0, minus bc (minus -3 times -1).
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Then, times this matrix...it was found by switching these two positions, 0 and -1.
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Now, for -3, which is in the b position, I am going to take the opposite sign; that is going to be 3.
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For -1, which is in the c position, I am going to take the opposite sign and make it 1.
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OK, let's make sure that this is going to work now.
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This is 0 minus -3 times -1; that is going to give me 3, so that is 1 over 0 minus 3; and there will be an inverse, because this is not 0.
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If this had come out to be 0, I couldn't have found the inverse.
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It is -1/3; 1 over 0 - 3 is -1/3...times this matrix.
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This is going to give me, if I multiply -1/3 by each element in here...I am going to get -1/3 times 0, and that is going to give me 0.
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Here, I have -1/3 times 3; that is going to give me -3/3, or -1, right here.
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1/3 times 1 is 1/3; and 1/3 times -1 is -1/3.
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So again, I was asked to find the inverse; I used this formula.
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I first found the determinant, and I determined that it was -1/3; therefore, I could find the inverse.
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And then, I multiplied it by the matrix, which consists of switching the positions of these two and reversing the signs of these two.
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This becomes a positive; this also becomes a positive--we are giving them the opposite signs from what they had.
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Then, I multiplied -1/3 by each element in this matrix to get this.
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And if I were to check it (which I could), it would be by multiplying A (the original matrix) times its inverse.
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And I would find that I get the identity matrix back.
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OK, find the inverse if it exists: again, I am recalling my formula: A^-1 is 1 over the determinant, ad - bc,
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times the matrix found by switching a and d, and reversing the sign on b, and reversing the sign on c.
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OK, so first the determinant: that is 1 times 6, minus 2 times 3.
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The matrix: switch these two positions, and reverse the sign on 3, and reverse the sign there.
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Now, you probably already saw that I didn't even need to go that far, because there is a problem.
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What I have here is 1 times 6 (which is 6), minus 6, times its matrix.
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Well, this turns out, obviously, to be 0; and since you can't have that--that is not allowed--it is undefined--
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we can stop right here, because this is undefined.
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In this case, the inverse does not exist; and the clue is: as soon as you see
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that you are having to divide by 0, you know that you are not allowed to do that.
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So, the inverse does not exist.
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That concludes this lesson on Educator.com on identity and inverse matrices, and I will see you again soon!