Question Video: Solving a Matrix Equation by Finding the Inverse of a Matrix Mathematics • 10th Grade

Video Transcript

Solve using the inverse of a matrix.

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 elements of the vector that we’ve called 𝑛, we first need to find the inverse of 𝐴. Remember the inverse of 𝐴 multiplied by 𝐴 is just the identity matrix. If we, therefore, multiplied both sides of our equation by the inverse of 𝐴, we get the solution to 𝑏 𝑛 equals the inverse of 𝐴 times 𝑏.

Just a quick reminder though, the order does matter. And this woudn’t work if we calculate 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 rather convoluted steps to follow.

Step one is to find the matrix of the cofactors. Step two is to transpose the elements in this matrix to find the adjugate. And step three is to multiply by one over the determinant of 𝐴.

Step one) Finding the matrix of the cofactors. 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. For all elements that have a negative sign, you’ll need to multiply by negative one, essentially changing the sign of this number.

The next step is a little more straightforward: we find the determinant of each of these elements. 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 in the matrix of cofactors is, therefore, one. It should remain a positive according to our formula.

The determinant of the second element in the first row is found by working out one multiplied by zero minus negative one multiplied by one. The determinant of the third element in the first row is found by working out one multiplied by one minus one multiplied by one, which is zero. The third element is zero.

Let’s follow this process for the remaining elements. The determinant of the matrix in the second row and the first column is calculated by working out negative one times zero minus negative one times one, which is one. In our matrix, it’s multiplied by negative one and becomes negative one. One multiplied by zero minus negative one multiplied by one is one and one multiplied by one minus negative one multiplied by one is two. The final element in our second row is, therefore, negative two.

The determinant of the two-by-two matrix in the third row and first column is negative one multiplied by negative one minus negative one multiplied by one, which is two. One multiplied by negative one minus negative one multiplied by one is equal to zero. And finally, one multiplied by one minus negative one multiplied by one is two. That’s step one completed. It’s by far the most time-consuming part.

The second step is to find the adjugate of the matrix of cofactors. This means to transpose the rows and columns or reflect them in this diagonal. The elements on this diagonal, therefore, remain the same. We swap negative one and negative one, which really doesn’t change a lot. We swap the zero and the two and finally we swap zero with negative two.

Our final step is to multiply this adjugate by one over the determinant of 𝑎, our original three-by-three matrix. To calculate the determinant, we multiply each of the elements in the top row by their matching cofactors. These are the determinants of their minor matrices that we worked out earlier. In this case, that’s one multiplied by one minus negative one multiplied by one minus negative one multiplied by zero, which is two.

Finally, we need to multiply our adjugate by one over the determinant or half. The inverse of 𝑎 is, therefore, as shown. Now, recall we said that to solve this system of linear equations, we multiply 𝑏 by the inverse of 𝐴. Now, all we need to do is multiply our matrices.

𝑥 is a half times nine add negative a half times negative 11 add one times six which is 16. 𝑦 is negative a half times nine add a half times negative 11 add zero times six which is negative 10. And 𝑧 is equal to zero times nine add negative one times negative 11 add one times six which is 17.

Our solution is 𝑥 equals 16, 𝑦 equals negative 10, and 𝑧 equals 17.