### Video Transcript

Consider the following system of linear equations in matrix form. The matrix nine, one, five, three multiplied by the matrix π₯, π¦ is equal to the matrix negative one, negative three. Write the determinant Ξ sub π₯. Write the determinant Ξ sub π¦. Write the determinant Ξ.

In this question, weβre given a linear system of equations written in matrix form. And we can see this is a system of two equations in two unknowns. This is because our matrix of unknowns contains only two unknowns π₯ and π¦. We need to use the system to write the determinants Ξ sub π₯, Ξ sub π¦, and Ξ. And to do this, we need to recall what is meant by these three determinants.

And we can do this by first recalling the two-by-two matrix written at the start of this expression is called the matrix of coefficients. And this is because if we evaluate the matrix multiplication on the left-hand side of this equation, the components of this matrix are the coefficients of our variables π₯ and π¦. And this is useful to note because Ξ is the determinant of the matrix of coefficients. So Ξ is the determinant of the two-by-two matrix nine, one, five, three. And we can use the matrix of determinants to find expressions for the determinants Ξ sub π₯ and Ξ sub π¦.

And we can do this by recalling the determinant Ξ sub π₯ is the determinant of the matrix of coefficients where we replace the column for the coefficients of π₯ with the entries of the answer matrix. Since the first column is the coefficients of π₯, this means we need the first column to be the entries of the answer matrix. The first column will be negative one, negative three. Then the remainder of this matrix will be the same as the coefficients matrix. This is just the extra column one, three, which gives us our final answer. Ξ sub π₯ is the determinant of the two-by-two matrix negative one, one, negative three, three.

Letβs now apply this process one more time to find the determinant Ξ sub π¦. This time, we want to change the column for the coefficients of π¦. And in particular, this means the column for the coefficients of π₯ will remain unchanged. So the first column of the matrix in this expression will be nine, five. Then the second column for the matrix in this expression will be the entries in the solution matrix. And this gives us the determinant Ξ sub π¦ is the determinant of the two-by-two matrix nine, negative one, five, negative three.

And although itβs not necessary to answer this question, itβs worth noting that now we found expressions for these determinants, we could apply Cramerβs rule, which tells us if the value of Ξ is not equal to zero, then π₯ is equal to Ξ sub π₯ divided by Ξ and π¦ is equal to Ξ sub π¦ divided by Ξ is the unique solution to the system of equations. And since the value of Ξ is not zero in this case, we could use this to check our answers. We could find the values of π₯ and π¦ and check they are valid solutions to the system of equations. However, this is not necessary to answer the question.