### Video Transcript

The straight line whose equation is π¦ plus ππ₯ equals π passes through the two points negative five, five and three, 10. Using matrices, find π and π.

Before we can use matrices to solve this problem, weβre going to begin by attempting to write a system of linear equations. First, we know that the straight line with equation π¦ plus ππ₯ equals π passes through the points negative five, five and three, 10. In other words, when π₯ is equal to negative five, π¦ is equal to five. So we can substitute these values into the equation of our straight line π¦ plus ππ₯ equals π. Then weβll be able to repeat this with the coordinates three, 10 and we should then have a system of linear equations for π and π. Replacing π¦ with five and π₯ with negative five and our equation becomes five plus π times negative five equals π, which can alternatively be represented as five minus five π equals π.

Finally, weβre going to rearrange this equation to ensure that the variables are on the same side. And we do that by adding five π to both sides, giving us five π plus π equals five. Letβs now repeat this, but this time letting π₯ equal three and π¦ equal 10. That gives us 10 plus three π equals π. And then this time weβll subtract three π from both sides, giving us negative three π plus π equals 10. We now have a system of linear equations, so letβs clear some space and figure out where to go next.

We need to find a way to represent the system of linear equations using matrices. And so we recall that due to the way we multiply a two-by-two matrix by a column matrix, the system of linear equations of the form ππ₯ plus ππ¦ equals π and ππ₯ plus ππ¦ equals π can be represented in matrix notation as the two-by-two matrix π, π, π, π times the column matrix π₯, π¦ equals the column matrix π, π. In other words, we simply take the coefficients of π₯ and π¦ and represent that in a two-by-two matrix. Now, in this case, weβre working with the variables π and π, which shouldnβt be confused with the constants π and π in our general formula.

The coefficient of π in our first equation is five, and the coefficient of π in our first equation is one. Then the coefficient of π in our second equation is negative three, and once again the coefficient of π is one. So the left-hand side of our system of linear equations can be represented as the matrix five, one, negative three, one times the column matrix π, π. Then we take the constants five and 10 and we create a second column matrix as shown. Our job will be to solve this matrix equation by essentially making the column matrix π, π the subject. And so letβs imagine weβre dealing with some general matrix equation π΄π₯ equals π΅, where π΄ is a two-by-two matrix, π₯ is the variable column matrix, and π΅ is another column matrix.

We solve this equation by multiplying both sides by the inverse of π΄. And this works because the inverse of π΄ times π΄ just gives us the identity matrix. And then the identity matrix when multiplying any other matrix just leaves that original matrix. So our matrix π΄ is the two-by-two matrix five, one, negative three, one. So it should be quite clear that the next step will be to find the inverse of this matrix. Letβs take the general two-by-two matrix π, π, π, π and define this to be equal to π΄.

To find its inverse, we multiply a two-by-two matrix which weβll define in a second by one over the determinant of π΄, where the determinant of π΄ is the product of the elements in the top left and bottom right minus the product of the elements in the top right and bottom left. The two-by-two matrix that we multiply this by is found by switching the elements in the top left and bottom right and then changing the sign of the remaining two.

So letβs begin by finding the determinant of our matrix. Itβs five multiplied by one minus one multiplied by negative three. Thatβs five minus negative three, which is of course equal to eight. So to find the inverse of our matrix π΄, weβre going to multiply one over eight by some two-by-two matrix. This matrix is found by switching the elements in the top left and bottom right and then simply changing the sign of the remaining two. So the inverse of π΄ is an eighth times the matrix one, negative one, three, five. And whilst we could distribute one-eighth across our two-by-two matrix, that would create a matrix thatβs very fraction heavy. So weβre going to do that at the end.

Remember to find the variable matrix π₯ in the matrix equation π΄π₯ equals π΅, we multiply both sides by the inverse of π΄. And on the left-hand side, that just leaves us with the variable matrix π₯. Now, of course, our variable matrix is the column matrix π, π. And this is equal to the inverse of π΄ times the constant matrix five, 10. So how do we multiply this pair of matrices? We begin by finding the dot product of the elements in the first row of our first matrix and the elements in our column matrix. Thatβs one times five plus negative one times 10, which is equal to negative five.

Then we find the dot product of the elements in the second row and the column of our column matrix. This time, thatβs three times five plus five times 10, which is equal to 65. So our matrix π, π is equal to one-eighth times the matrix with these elements. And now we can distribute the eighth across this matrix, and we do so simply by multiplying negative five by one-eighth and 65 by one-eighth. And we may as well leave that in fractional form. And when we do, we find the matrix π, π is equal to the matrix negative five-eighths, sixty-five eighths.

Now, of course, weβre actually trying to find the values of π and π, so we will extract this from its matrix form. If two matrices are equal, it follows that their individual elements must be equal. So our element π is equal to the element negative five-eighths and π must be equal to 65 over eight. And so weβre done. We found the values of π and π. They are negative five-eighths and sixty-five eighths, respectively.