Inverse of 3x3 matrix

so now we're going to find the inverse of a 3x3 matrix so here we have a 3x3 matrix a and the first thing we're going to do is find the determinant of this matrix a and the way I like to find the determinant is just by copying the first two columns and placing them on the right side of the matrix so we'll start with this first column right here I'm going to copy it and place it to the right of the matrix so notice how I copy the numbers 1 0 5 and I placed them to the right side of the matrix and now I'm going to do the same exact thing with the column 2 I'm going to copy it and place it to the right side of the matrix now by copying these two columns notice how I'm able to create three diagonals I have one diagonal right here I have another diagonal right here and I have another diagonal right here what we need to do is we need to multiply the numbers in all three diagonals so that's what I'm going to do right now in this first diagonal we have 1 1 and 0 so I'm going to multiply all those numbers together 1 times 1 is just 1 1 times 0 is 0 and in the second diagonal we have 2 times 5 which is 10 and 10 times 5 which is 50 and in the third diagonal we have 3 times 0 which is 0 0 times 6 is also 0 and after we multiply these diagonals we need to do the exact same thing with the diagonals going the opposite way so notice how we have three diagonals going the opposite way we have one right here we have one right here and we have one right here and if we multiply these diagonals together starting with the first one we have 5 times 1 which is 5 5 times 3 is 15 in the second diagonally we have 6 times 5 which is 30 30 times 1 is also 30 and in the third diagonal we have 0 times 0 which is 0 0 times 2 is also 0 so now to find the determinant of matrix a the only thing that we need to do is take the red numbers and add them together and then subtract them from all the green numbers so with our red numbers we have a zero eight positive 50 and another zero and we need to subtract the red numbers from the green numbers and the green numbers are positive 15 positive 30 and zero so the determinant of matrix a is equal to zero plus 50 plus 0 which is just 50 minus 15 plus 30 which is 45 50 minus 45 is just equal to 5 so the determinant of matrix a is equal to 5 so after we find our determinant we have to find what I like to call a new matrix so let's just call our new matrix n so how do we know what goes inside of our new matrix n so let's take a look at our original matrix a let's take a look at the number in the top left position this number 1 if we cross out the row and the column of this number 1 notice how we're left with a 2 by 2 matrix 1 5 6 0 so in the top left position we need to copy that 2 by 2 matrix and put 1 5 6 0 so now we need to do the same exact thing for every number in the matrix a so now we'll move on to the top middle number this number 2 if we cross out the row and the column of this number 2 we're left with a 2 by 2 matrix 0 5 and 5 0 so I'm going to copy that 2 by 2 matrix and place it in the top middle position 0 5 5 0 so now let's do the same thing for the number in the top right position this number 3 if we cross out row and the column of the number three we're left with a 2 by 2 matrix 0 1 5 6 so I'm going to copy that 2 by 2 matrix in the top right position 0 1 5 6 so now let's do the same exact thing for the number in the middle left position this number 0 if we cross out the row and the column of this number 0 we're left with the 2 by 2 matrix 2 3 6 0 so I'm going to copy that 2 by 2 matrix in the middle left position 2 3 6 0 so now let's do the same as our thing for our middle number which is 1 I'm going to cross out the row and the column of this number 1 and we're left with the 2 by 2 matrix 1 3 5 0 so I'm going to copy that 2 by 2 matrix in the middle 1 3 5 0 so now let's do the same thing for the number in the middle right position this number 5 if we cross out the column and the row of this number 5 we're left with the 2x2 matrix 1 2 5 6 so I'm going to copy that 2x2 matrix into our middle right position 1 2 5 6 now if we do the same thing for the number in the bottom left position this number 5 we cross out the column and the row of this number 5 we're left with the 2x2 matrix 2 3 1 5 so I'm going to copy that 2x2 matrix in the bottom left position 2 3 1 5 and moving on to the number in the bottom middle position this number 6 if we cross out the column and the row of this number 6 we're left with a 2 by 2 matrix 1 3 0 5 so I'm going to copy that 2 by 2 matrix in the bottom middle position 1 3 0 5 and finally we can move on to our number in the bottom right position this number zero if we cross out the column and the row of this number zero we're left with a two by two matrix 1 2 0 1 so I'm going to copy that 2 by 2 matrix in the bottom right position 1 2 0 1 and now we have filled our new 3x3 matrix n with a bunch of smaller two-by-two matrices so now we can move on to our next step our next step is we need to replace our new matrix n with the determinants of all these two-by-two matrices so we'll start with the matrix in the top left corner to find the determinant of a 2 by 2 matrix you just need to multiply this diagonal and then multiply this diagonal and then subtract the both so that's what I'm going to do right now we have 1 times 0 which is 0 and we have 5 times 6 which is 30 0 minus 30 is equal to negative 30 now we'll move on to the top middle matrix we have 0 times 0 which is 0 and 5 times 5 which is 25 0 minus 25 is negative 25 now we'll move on to the top right corner we have 0 times 6 which is 0 and 5 times 1 which is 5 0 minus 5 is equal to negative 5 now moving on to the middle left we have 2 times 0 which is 0 and 3 times 6 which is 18 0 minus 18 is negative 18 moving on to the middle matrix we have 1 times 0 which is 0 3 times 5 which is 15 0 minus 15 is negative 15 moving on to the middle-right matrix we have 1 times 6 which is 6 and 2 times 5 which is 10 6 minus 10 is equal to negative 4 moving on to the bottom left matrix we have 2 times 5 which is equal to 10 we have 3 times 1 which is 3 10 minus 3 is equal to positive 7 when we go into the bottom middle matrix we have 1 times 5 which is 5 3 times 0 which is 0 5 minus 0 is equal to positive 5 and moving on to the bottom right we have 1 times 1 which is equal to 1 and 2 times 0 which is equal to 0 1 minus 0 is equal to positive 1 so now we have replaced our matrix n with the determinants of all these two-by-two matrices so moving on to our next step we need to alternate positive and negative signs in our matrix so if we start in the top left position we'll start with a positive sign then we gotta alternate to a negative positive negative positive negative positive negative and positive so now I'm going to scroll down just a bit just to give myself a little bit more space so now if we simplify our matrix n we have a positive and a negative 30 the positive and negative gives us a negative number we have a negative negative 25 two negatives gives us a positive 25 we have a positive negative 5 so that just stays negative 5 we have a negative negative 18 two negatives gives us a positive 18 we have a positive negative 15 so that just stays as a negative 15 we have a negative negative 4 two negatives gives us a positive 4 we have a plus 7 so that just stays positive 7 we have a negative 5 so that just stays negative 5 and we have a plus 1 so that just stays positive 1 so now we can move on to our next step our next step is we need to reflect all the numbers along this red diagonal which I'm drawing right now and if we reflect all the numbers across this red diagonal all the numbers that are actually laying on that diagonal are going to stay in the same position so the numbers that are on the diagonal negative 30 negative 50 and positive one they're gonna stay in the same position negative 30 negative 15 and positive 1 and if we reflect the number 18 and a number 25 along this red diagonal then they're going to switch places so this number 18 is going to go where the 25 used to be and this number 25 is going to go where the number 18 used to be and if we reflect this number 7 and this number negative 5 across the red diagonal then they're going to switch places so the negative 5 is going to go where the positive 7 used to be and the positive 7 is gonna go where the negative 5 used to be and I think you get the idea by now if we reflect these numbers across the red diagonal the negative 5 and the positive 4 we're going to switch places so the positive 4 is going to go where the negative 5 used to be and the negative 5 is going to go where the positive 4 used to be so now we have completely found our new matrix n and the nice thing about this matrix is that it allows us to find the inverse of our matrix a quite easily so I'm just going to scroll down a little bit just to give myself a little bit more space so the formula for the inverse of our matrix a is equal to 1 over the determinant of a all multiplied by this matrix n that we just found so we know from earlier that the determinant of a was 5 so 1 over the determinant of a is going to be 1 over 5 and we know that our matrix n is equal to negative 30 18 positive 7 so we have negative 30 18 positive 7 then we have 25 negative 15 negative 5 so we have 25 negative 15 negative 5 and we have negative 5 4 one negative five four and one and last but not least this one over five is a scalar so we need to multiply this by the entire matrix so I'm going to scroll down once again just to give myself a little bit more space the inverse matrix a is equal to one over five times negative 30 which is negative 30 over 5 or negative 6 1 over 5 times 18 is just 18 over 5 1 over 5 times 7 is equal to 7 over 5 1 over 5 times 25 is 25 over 5 or just 5 1 over 5 times negative 15 is negative 15 over 5 or negative 3 1 over 5 times negative 5 is negative 5 over 5 or just negative 1 1 over 5 times negative 5 is also negative 1 1 over 5 times 4 is 4 over 5 and 1 over 5 times 1 is just 1 over 5 so congratulations now we have found the inverse of matrix a we've probably spent over 10 minutes doing it which a computer could probably do in less than a second

Channel: Math Meeting

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Keywords: inverse 3x3 matrix, matrix inverse, determinant, matrices, linear algebra, mathmeeting

Id: pKZyszzmyeQ

Length: 14min 44sec (884 seconds)

Published: Thu Jul 19 2012