Video: MATH-ALG+GEO-2018-S1-Q19

Using the inverse matrix, solve the system of linear equations: 2𝑧 βˆ’ 3𝑦 = 7, 𝑦 + 5π‘₯ = 4, π‘₯ βˆ’ 2𝑦 βˆ’ 𝑧 = 1.

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Video Transcript

Using the inverse matrix, solve the system of linear equations: two 𝑧 minus three 𝑦 equals seven, 𝑦 plus five π‘₯ equals four, and π‘₯ minus two 𝑦 minus 𝑧 equals one.

So we have a system of linear equations, three linear equations and the three unknowns π‘₯, 𝑦, and 𝑧. And we’re told to solve this system using the inverse matrix. So we’re going to want to write this system of linear equations as a single-matrix equation. To do this, we first rewrite each of our equations in the form π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 equals 𝑑 for some constants π‘Ž, 𝑏, 𝑐, and 𝑑.

So for example, we can rewrite the first equation two 𝑧 minus three 𝑦 equals seven. We can rewrite this as zero π‘₯ minus three 𝑦 plus two 𝑧 equals seven. You can see that this is in the form required. We had to swap the terms two 𝑧 and minus three 𝑦. And we had to write zero π‘₯ at the start of the expression. Of course, zero π‘₯ is just zero. And so, this doesn’t change the value of the expression on the left-hand side. We do the same with the second equation, swapping the terms five π‘₯ and 𝑦 and writing the coefficient of 𝑦 explicitly. We also have to write zero 𝑧 to get a 𝑧 term. The terms in the third equation are all there and they’re in the right order. All we have to do is write the coefficients of π‘₯ and 𝑧 explicitly.

And now that we have these three equations in the right form with the π‘₯, 𝑦, and 𝑧 terms lined up one above the other, we can write this system as a single-matrix equation. You can see that the layout of this matrix equation is very similar to the layout of the system of equations. You can check that this matrix equation represents the system of equations by multiplying out the left-hand side of the matrix equation using matrix multiplication. But we’re not going to do that. We’re just going to solve this matrix equation.

Now, how do we solve this matrix equation? Well, as the question suggest we use the inverse matrix left multiplying by it. On the left-hand side, the inverse matrix times the original matrix gives the identity matrix. And the identity matrix 𝐼 times π‘₯, 𝑦, 𝑧 is just π‘₯, 𝑦, 𝑧. That’s what the identity matrix does. So we get that π‘₯, 𝑦, 𝑧 is the inverse matrix of zero, negative three, two; five, one, zero; one, negative two, negative one times seven, four, one. So the main part of answering this question is to find this inverse matrix.

The inverse of a matrix 𝐴 is one over the determinant of that matrix 𝐴 multiplied by the adjugate of 𝐴. We can work out the determinant of our matrix by expanding along the first row. The first term of this expansion is the first entry of the first row, zero times its minor which is the determinant you get by removing the row and column containing zero. From this, we subtract the entry negative three times its minor, the determinant five, zero, one, negative one. And finally, we add the last entry of the first row two times its minor.

Of course, the first term being a zero times something is zero minus negative three times this determinant is the same as plus three times the determinant. And we can evaluate the determinant by multiplying along the leading diagonal five times negative one is negative five and then subtracting the product of the other two terms. Zero times one is zero. Negative five minus zero is negative five. So the value of the determinant is negative five. And the term becomes three times negative five. And similarly, the value of the remaining determinant is five times negative two which is negative 10 minus one times one which is one. Negative 10 minus one is negative 11. Three times negative five is negative 15 and two times negative 11 is negative 22. And so, the value of our three- by-three determinant is negative 15 plus negative 22 which is negative 37.

So that’s the determinant of our matrix. But to find the inverse of our matrix, the other ingredient we need is the adjugate. So let’s clear some room and find this. So what is the adjugate of our matrix? Well, it’s another three-by-three matrix which is most conveniently thought of as the transpose of another matrix. Now, what are the entries of this matrix that we take the transpose of to find our adjugate?

In our original matrix, the entry in the first row and column is this entry zero. The entry in the first row and column of our new matrix is the determinant you get by deleting the row and column containing the entry zero. This is called the minor of this entry. And you might remember that it’s turned up when we were finding the determinant of this matrix by expanding along the first row. Similarly, to find the entry in the first row and second column, we delete the first row and second column from our original matrix and form the determinant from the remaining entries.

This two-by-two determinant five, zero, one, negative one also appeared in the expansion for the three-by-three determinant. And we need to give it a minus sign because it’s in the first row and second column. And one plus two is three which is an odd number. When we give a sign to a minor in this way, that’s turned it into a cofactor. The next cofactor is the two-by-two determinant five, one, one, negative two. We don’t give this a minus sign as it’s in the first row and third column. And one plus three is four, an even number. But the next entry does have a minus sign because it’s in the second row and the first column.

Now, we continue to fill in our matrix of cofactors. Notice that the cofactors with minor signs form a diamond in our matrix. We now just need to evaluate all of our cofactors before taking the transpose. And you can think of taking the transpose as reflecting along the leading diagonal of the matrix. For example, the entry negative seven in the second row and first column goes to the second column and first row. This is therefore our adjugate, the matrix negative one, negative seven, negative two; five, negative two, 10; negative 11, negative three, 15.

Now, we have the two ingredients we need to find the inverse of our original matrix. Using our formula, we see that it is one over negative 37β€”that’s the determinant of our matrixβ€”times the three-by-three matrix negative one, negative seven, negative two; five, negative two, 10; negative 11, negative three, 15 which was the adjugate of our matrix. So let’s make this substitution. And now, we can just perform matrix multiplication on the right-hand side.

The matrix multiplication produces a three-by-one matrix, whose first entry, the entry in the first row, is negative one times seven plus negative seven times four plus negative two times one. That’s the dot product of the first row of the first matrix with the first and only column of our second matrix, that negative 37. Now, the dot product of the second row five, negative two, 10 with the column seven, four, one is 37. And the last entry is negative 11 times seven plus negative three times four plus 15 times one. This last entry is negative 74. And now, we can see that all the entries of our matrix are divisible by 37 and hence negative 37. We can simplify then. Negative 37 divided by negative 37 is one. 37 divided by negative 37 is negative one. And negative 74 divided by negative 37 is two.

Now, this is the value of the matrix of unknowns π‘₯, 𝑦, 𝑧. And we can see by comparing the entries of these matrices, so comparing these entries, we see that π‘₯ is one, 𝑦 is negative one, and 𝑧 is two. We should check that these values do indeed satisfy the system of linear equations that we started with. And hopefully, you can see that they do. And so, we’ve solved this system of linear equations by using inverse matrices.

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