### 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.