# Question Video: Solving a Matrix Equation by Finding the Inverse Mathematics

Given that [−1, −6, −6 and 0, −2, −4 and 2, 4, 7] [𝑥, 𝑦, 𝑧] = [−8, 2, −9], find the values of 𝑥, 𝑦, and 𝑧.

07:50

### Video Transcript

Given that the product of a three-by-three matrix with elements negative one, negative six, negative six, zero, negative two, negative four, two, four, and seven and a vector containing elements 𝑥, 𝑦, and 𝑧 is equal to the vector containing elements negative eight, two, and negative nine, find the values of 𝑥, 𝑦, and 𝑧.

We know that if matrix 𝐴 is multiplied by vector 𝐗 and it equals vector 𝐁, that vector 𝐗 equals the inverse matrix of 𝐴 multiplied by vector 𝐁. We should also remember that order matters. We cannot multiply 𝐁 by the inverse matrix of 𝐴. We must multiply it in this order, the inverse of 𝐴 times 𝐁. This means that our first step is to find the inverse of matrix 𝐴.

You could use a graphing calculator to help you do this step. But we will take a look at how to do the steps by hand without a graphing calculator. This is a multistep process and will be the most complex part of solving this problem.

The first step in finding the inverse of matrix 𝐴, if the inverse exists, is to find the cofactor matrix, 𝐶, equal to the three-by-three matrix 𝐶 sub 𝑖𝑗, whose entries are the determinants of the corresponding matrix minors multiplied by the alternating sign, negative one, to the power of 𝑖 plus 𝑗. Then, we transpose the cofactor matrix to find the adjugate matrix, sometimes referred to as the classical adjoint matrix. Then, we will calculate the determinant of 𝐴 and multiply the adjugate matrix by the reciprocal of the determinant. Since 𝐴 is a three-by-three matrix, we must use the following formula to find the three-by-three cofactor matrix.

Notice the pattern of alternating positive and negative signs in front of each determinant. If we assign the wrong sign to one element, it can completely change our final result.

To begin, we find the determinant in the first row, first column. We look to matrix 𝐴 to find the four elements in the 𝑒-, 𝑓-, ℎ-, and 𝑖-positions. These are negative two, negative four, four, and seven. To calculate the determinant, we multiply the top-left element by the bottom-right element. And then we subtract the bottom-left element multiplied by the top-right element. And so the first element of our cofactor matrix is found by negative two times seven minus four times negative four, which is negative 14 minus negative 16, which equals positive two. So the top-left element in our new matrix is going to be positive two.

The determinant from the first row, second column would then look like this: zero times seven minus two times negative four, which is zero minus negative eight, which equals positive eight. Notice, however, that because of its position in the cofactor matrix, we need to apply a negative sign. So that element in the first row, second column is negative eight.

Now we’re looking for the element in the first row, third column: zero times four minus two times negative two. This element comes out to positive four. Let’s repeat this process for the second row of cofactors. The signs for this row are negative, positive, negative. The first determinant in this row comes out to negative 18. But we need to take the negative of negative 18. And so we have positive 18 in the second row, first column position.

In the second row, second column, we have positive five. And the determinant in the second row, third column comes out to positive eight. But we need to take the negative of that value. So this element will actually be negative eight.

And finally we’ll calculate the determinants in the third row using first a positive sign, then negative, then positive again. For the third row, first column, we get positive 12. In the third row, second column, we get four. But we need to multiply that by negative one, which gives us negative four. And in the final position, we get positive two.

We now have our complete cofactor matrix. Our next step is to find the adjugate matrix. We do this by transposing the rows and the columns. And that looks like a reflection across the diagonal. The elements along the diagonal remain the same. But we’ll switch the negative eight and positive 18, the four with the 12, and the negative eight with the negative four. So the adjugate of matrix 𝐴 is two, 18, 12, negative eight, five, negative four, four, negative eight, two.

Our final step is to then multiply the adjugate matrix by the reciprocal of the determinant of the original matrix 𝐴. To calculate the determinant, we can use this formula. We’ll multiply each element in the top row by the matching cofactors we found in the previous step. The determinant of the first row, first column was two. The first row, second column was eight. And the first row, third column was four. The expression simplifies to negative two plus 48 minus 24. So the determinant of 𝐴 is 22. So the reciprocal of the determinant of matrix 𝐴 is then one over 22. Altogether, the inverse of matrix 𝐴 is one over 22 times the adjugate.

After multiplying each element by one over 22, the product looks like this. As we originally said, to solve this system, we need to calculate the inverse of matrix 𝐴 times 𝐁. The resulting array will contain three elements: a value for 𝑥, a value for 𝑦, and a value for 𝑧.

To find 𝑥, we multiply negative eight by one over 11, two by nine over 11, and negative nine by six over 11, then find the sum of these products. This expression simplifies to negative 44 over 11. So the value of 𝑥 is negative four.

The value of 𝑦 is found by calculating negative eight times negative four over 11 plus two times five over 22 plus negative nine times negative two over 11, which simplifies to 55 over 11. So 𝑦 is equal to positive five.

And finally to solve for 𝑧, we must calculate negative eight times two over 11 plus two times negative four over 11 plus negative nine times one over 11. This expression simplifies to negative 33 over 11. So 𝑧 is equal to negative three.

Therefore, under these conditions, 𝑥 is equal to negative four, 𝑦 is equal to five, and 𝑧 is equal to negative three.