Video: Finding an Unknown Matrix in an Equation Using the Inverse of a Matrix

Given that (1, −1, −1 and −2, 0, 0 and −8, −4, 0)( 𝑥 and 𝑦 and 𝑧) = (4 and 8 and 0), find the values of 𝑥, 𝑦, and 𝑧.

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Video Transcript

Given this system of matrices, find the values of 𝑥, 𝑦, and 𝑧.

Let’s recall the notation: 𝐴 multiplied by 𝑛 which is lowercase because it’s a vector is equal to 𝑏. To find the values of 𝑥, 𝑦, and 𝑧, which are the elements of the vector that we’ve called 𝑛, we need to find the inverse of 𝐴.

𝑛 is equal to the inverse of 𝐴 multiplied by 𝑏. Just a reminder though, the order does matter. And this doesn’t work if we work out 𝑏 times the inverse of 𝐴 because the number of columns in 𝑏 are not equal to the number of rows in 𝐴.

To find the inverse of 𝐴, we have a few convoluted steps to follow. Now, if you’re allowed to use a graphical calculator, you may do so at this point and skip onto the final step. Step one) Find the matrix of the cofactors, step two is to transpose the matrix of the cofactors to find its adjugate, and step three is to multiply this by one over the determinant of 𝐴.

For the given matrix, this is the formula we need to use to help us find the matrix of the cofactors. A good way to remember this is to cover up the row and column you’re looking at and find the determinant of the matrix that remains. You can then apply the relevant sign to this number alternating as you go. The top row is positive, negative, positive, the second row is negative, positive, negative, and the third is positive, negative, positive.

Remember if the formula tells you that the answer is negative, you must multiply by negative one, essentially changing the sign. First, we’ll find the determinant of the two-by-two matrix in the first row and first column. To do this, we multiply the top left element by the bottom right then subtract the product of the top right and bottom left elements.

The first element of the matrix is zero multiplied by zero minus zero multiplied by negative four, that’s zero. The element in the first row and second column is calculated by working out negative two times zero minus negative eight times zero which is once again zero. The element in the first row and third column is negative two multiplied by negative four minus zero multiplied by negative eight which is eight.

Let’s repeat this process for the second row. Negative one multiplied by zero minus negative one multiplied by negative four is negative four. When we substitute this into our matrix of cofactors though, we have to multiply by negative one. So our answer becomes positive four. One multiplied by zero minus negative one multiplied by negative eight is negative eight and one multiplied by negative four minus negative one multiplied by negative eight is negative 12.

When we substitute this into our table, the sign is negative. So we multiply by negative one to give us positive 12. And finally, our third row negative one multiplied by zero minus negative one multiplied by zero is zero. One multiplied by zero minus negative one multiplied by negative two is negative two. Then, when I pop that in the table, multiply it by negative one to go back to negative two. And finally, one multiplied by zero minus negative one multiplied by negative two is negative two. We have our matrix of cofactors.

Step two is to find the adjugate of the matrix of the cofactors. We do this by transposing the rows and columns, which looks like a reflection in the diagonal. The elements on the diagonal remain the same. We swap the zero with the four, the eight with the zero, and the 12 with the two.

Our final step is to multiply the adjugate by one over the determinant of the original matrix. To calculate the determinant, we can use this formula. We’re going to multiply each of the elements in the top row by their matching cofactors that we worked out at the start.

One multiplied by zero minus negative one multiplied by zero add negative one multiplied by eight is negative eight. We’re now going to multiply one over negative eight by the adjugate of the matrix of cofactors, which is as shown. We finally worked out the inverse of 𝐴.

We said that to solve this system of matrices, we multiply 𝑏 by the inverse of 𝐴. 𝑥 is given by zero multiplied by four add negative a half multiplied by eight add zero multiplied by zero which is negative four. 𝑦 is zero multiplied by four add one multiplied by eight add negative a quarter multiplied by zero which is eight. And 𝑧 is negative one multiplied by four add negative three-halves multiplied by eight and a quarter multiplied by zero which is negative 16.

𝑥 is negative four, 𝑦 is eight, and 𝑧 is negative 16.

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