Video Transcript
Use determinants to solve the system 𝑥 minus five 𝑦 plus three 𝑧 is equal to five, three 𝑥 minus four 𝑦 plus two 𝑧 is equal to negative five, and negative 𝑥 plus three 𝑦 minus two 𝑧 is equal to negative five.
In this question, we’re given a system of three linear equations in three unknowns. And we’re asked to solve this system by using determinants. And we can recall to use determinants to solve a system of equations, we need to use Cramer’s rule. So let’s start by recalling the shortened version of Cramer’s rule for a system of three equations in three unknowns. This tells us if the determinant of the matrix of coefficients △ is nonzero, then 𝑥 is equal to △ sub 𝑥 over △, 𝑦 is equal to △ sub 𝑦 over △, and 𝑧 is equal to △ sub 𝑧 over △ is the unique solution to the system of equations. So we can use Cramer’s rule to solve a system of equations by using determinants. We just need to find △, △ sub 𝑥, △ sub 𝑦, and △ sub 𝑧. And we need our value of △ to be nonzero.
So let’s start by finding △. Remember, △ is the determinant of the matrix of coefficients. And that’s the three-by-three matrix we get by taking entries as the coefficients of our variables. And it’s very important when writing △ out to make sure we include the signs of all of the coefficients. We get that △ is equal to the determinant of the three-by-three matrix one, negative five, three, three, negative four, two, negative one, three, negative two.
We can now evaluate this determinant by expanding over the first row. We get one times the determinant of the two-by-two matrix negative four, two, three, negative two minus negative five multiplied by the determinant of the two-by-two matrix three, two, negative one, negative two plus three times the determinant of the two-by-two matrix three, negative four, negative one, three. Now, all we need to do is evaluate the determinant of these three two-by-two matrices. And we do this by taking the difference in the product of the diagonals. This then gives us eight minus six plus five times negative six minus negative two plus three times nine minus four.
And now we just need to evaluate this expression. We get two minus 20 plus 15, which is equal to negative three. Therefore, we’ve shown the value of △ is negative three. In particular, we’ve shown that △ is nonzero. So this system of equations has a unique solution, and we can find this by using Cramer’s rule. So let’s clear some space and keep note of the fact that △ is equal to negative three and determine the values of △ sub 𝑥, △ sub 𝑦, and △ sub 𝑧.
To do this, we first need to recall what we mean by △ sub 𝑥, 𝑦, and 𝑧. And let’s start with △ sub 𝑥. △ sub 𝑥 is the determinant of the matrix of coefficients where we replace the column for the coefficients of 𝑥 with the column given by the answers to our system of equations. And in particular, this means we can find an expression for △ sub 𝑥 by replacing the first column in our expression for △ with the column given by the answers to our system of equations. So we’ll replace the first column with five, negative five, negative five. And then we leave the other two columns unchanged. △ sub 𝑥 is the determinant of the three-by-three matrix five, negative five, three, negative five, negative four, two, negative five, three, negative two.
Now we can evaluate this determinant by expanding over the first row. We get five times the determinant of the two-by-two matrix negative four, two, three, negative two minus negative five multiplied by the determinant of the two-by-two matrix negative five, two, negative five, negative two plus three times the determinant of the two-by-two matrix negative five, negative four, negative five, three. And now we can evaluate the determinant of these three matrices by using the difference in the product of the diagonals. We get five times eight minus six plus five times 10 plus 10 plus three multiplied by negative 15 minus 20. And now we can just evaluate this expression. We get five times two plus five times 20 plus three multiplied by negative 35, which simplifies to give us 10 plus 100 minus 105, which is just equal to five.
So △ sub 𝑥 is equal to five. And this means we can determine the value of 𝑥. It’s △ sub 𝑥 over △, which is five divided by negative three. However, we also need to find the values of △ sub 𝑦 and △ sub 𝑧. So let’s clear some space and determine the value of △ sub 𝑦. And to do this, we need to recall that △ sub 𝑦 is the determinant of the matrix of coefficients where we replace the column for the coefficients of 𝑦 with the values of the answers to our system of equations. So, to write an expression for △ sub 𝑦, we’ll start with the second column. It’s the column five, negative five, negative five, the constants in our simultaneous equations. And then our first and third columns remain unchanged. They’re still just the coefficients of 𝑥 and 𝑧. So △ sub 𝑦 is the determinant of the three-by-three matrix one, five, three, three, negative five, two, negative one, negative five, negative two.
And once again we can evaluate this determinant by expanding over the first row. We get one times the determinant of the two-by-two matrix negative five, two, negative five, negative two minus five times the determinant of the two-by-two matrix three, two, negative one, negative two plus three multiplied by the determinant of the two-by-two matrix three, negative five, negative one, negative five. And now all that’s left to do is evaluate this expression. To evaluate the determinant of these matrices, we take the difference in the product of the diagonals. This gives us one times 10 plus 10 minus five multiplied by negative six plus two plus three times negative 15 minus five. And we can then evaluate this expression. It’s equal to negative 20.
So let’s keep note of the fact that △ sub 𝑦 is equal to negative 20 and apply this process one more time to determine the value of △ sub 𝑧. And let’s start with the fact that the first two columns in our expression for △ sub 𝑧 will remain unchanged. They’ll be the coefficients of 𝑥 and 𝑦 in the simultaneous equations, respectively. Then, the third column in △ sub 𝑧 should be the constants in our simultaneous equations. That’s five, negative five, negative five. So △ sub 𝑧 is the determinant of the three-by-three matrix one, negative five, five, three, negative four, negative five, negative one, three, negative five. And once again we can evaluate this determinant by expanding over the first row. We get one times the determinant of the two-by-two matrix negative four, negative five, three, negative five minus negative five times the determinant of the two-by-two matrix three, negative five, negative one, negative five plus five times the determinant of the two-by-two matrix three, negative four, negative one, three.
And now we just need to evaluate this expression. We get 20 plus 15 plus five times negative 15 minus five plus five multiplied by nine minus four. And we can evaluate this expression. It’s equal to negative 40. And now that we found the values of △ sub 𝑥, △ sub 𝑦, △ sub 𝑧, and △, we can substitute these into our equations to determine the solution to the system. And doing this and simplifying gives us our final answer. The solution to the system is 𝑥 is equal to negative five over three, 𝑦 is equal to 20 over three, and 𝑧 is equal to 40 over three.
And it’s worth noting we can check our answer by substituting these values of 𝑥, 𝑦, and 𝑧 into the system of equations to check that they are indeed solutions. Therefore, we were able to use determinants to solve the given system of equations. We got 𝑥 is equal to negative five over three, 𝑦 is equal to 20 over three, and 𝑧 is equal to 40 over three.
