Find the inverse, if it exists, for each matrix. [3x3 matrix]

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Hello, everyone. We are asked to determine the inverse of the following matrix. If possible, we are given a three by three matrix with the first row of 303, 2nd row of zero, negative 30. And third row of 733, recall to find the inverse of a matrix, we need to multiply one divided by the determinant multiplied by the adjoint matrix. So we need to find the determinant and the adjoint matrix. So I'm gonna begin with the determinant. I'm gonna use the first row. So I have three multiplied by the minor matrix negative 30 is the first row 33 is the second row. Remembering that I need to alternate signs three was a positive. So now I do minus and my next element is a zero. So I'm not going to write the whole minor matrix because when I multiply by zero, it will be zero. So then I'm gonna add three multiplied by the minor matrix zero. Negative three is the first row 73 is the second row. We are going to find the determinant of these minor matrices. Recall you do that by multiplying from left to right diagonally and then subtracting the product from right to left. So this will give us three multiplied by negative nine minus zero plus three, multiplied by 21. So we have negative plus 63 and our determinant is 36. So we're gonna need that number in a little bit. We're gonna leave it there. It'll come back in the last couple of steps. Next to find the adjoint matrix. We first need to find the cofactor matrix. We are gonna find the Cofactor matrix by making minor matrices for each element. So for the first element of the first row, we are going to look at the three that is in the original matrix and eliminate the rows and columns it belongs to. So our minor matrix will be negative 30 is the first row 33 is the second row. My next element is a zero. So excluding the rows and columns that belongs to I get a minor matrix of first row 00, 2nd row 73. And my third element is a three. So excluding those rows and columns, I get a minor matrix of zero negative three for the first row 73, for the second row, recall how we did alternating positives and negatives. When we were finding the determinant, we need to do the same here. So if our first element is positive, I need to put a negative on my second element. My third element again is positive. This means I start my next row with a negative. So I'm gonna put that there. And then I'm finding the minor matrix that excludes the rows and columns that this first zero is in. So that would be a top row of 03, 2nd row of 33. Now, I need the minor matrix that excludes the negative three from the second row. So we have 33 is the first row. 73 is the second row of our minor matrix. That one was positive. So this next one is negative and we are going to exclude the rows and columns that have that third element zero from the second row. So we get a minor matrix top row 30, 2nd row 73 down to our third row. Um Our previous one was negative. So this one's a positive, we're gonna exclude the rows and columns that include that seven. So we get 03 is the first row in the minor matrix negative 30 is the second row. That matrix was positive. So we put a negative sign in front of the next element. So excluding the three, that's the second element of the third row. We get the minor matrix 33 is the first row 00 is the second row. Uh That one was negative. Our next one is positive and we're excluding the third element that is three in the third row. So we get the minor matrix 30 is the first row zero, negative three is the second row. We will now find the determinant of each of these minor matrices again multiply from left to right and then subtract the product from right to left. And when you do, so you get first row negative 90, 21 second row nine, negative 12, negative nine, third row 90, negative nine. So one more step before we have our full adjoint matrix to find the adjoint matrix, we need to transpose our columns and our rose. So when we do so our first row becomes our first column. So we're gonna write negative 90, 21 going down the first column and then nine negative 12, negative nine going down the second column, then 90, negative nine going down the third column. So now that we have our adjoint matrix, we have our determinant, we can solve the inverse. So we are going to do one divided by the determinant which is and that is multiplied by the adjoint matrix. We just found where the first row is negative 99, nine, uh second row is zero, negative 12, 0, third row is 21 negative nine, negative nine. We are going to multiply 1 36th as a scaler into this. So each element will be multiplied by 1 36th. And when we do, so we're gonna have a lot of fractions in this little space. So bear with me. We have negative 9 36th for the first element of the first row. 9 36 is the second element of the first row. 9 36 is again the third element of the first row for the second row, zero, divided by anything is zero. So I'm gonna just write that in as zero. Next I get negative 12 divided by and zero again, divided by 36 to 0 in the third row. I get 21 36th and then negative 36. And then again, negative 9 36. Since these fractions can be reduced, we will be reducing them. So when they are reduced, we get negative 1/4 1 4th 1/4 as the first row zero, negative one third zero is our second row. 7/12 negative 1/ negative 1/4 is the third row. So this is our inverse matrix. Have a nice day.

Find the inverse, if it exists, for each matrix. [3x3 matrix]

Key Concepts

Matrix Inversion

Determinant

Identity Matrix

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