### Video Transcript

Given that the matrix π plus π, π minus π, π plus π plus π, π minus seven π minus π is equal to the matrix negative three, negative 17, negative five, negative 64, determine the values of π, π, π, and π.

Well, the first thing we know is that the two matrices are equal to each other. So therefore, they must have the same number of rows, columns. And each corresponding term must be equal. So as we know this, we can set up for equations.

The first one is π plus π is equal to negative three. And thatβs cause theyβre the corresponding terms in each of the matrices. Then we have π minus π is equal to negative 17. Then we have π plus π plus π is equal to negative five. And then, finally, π minus seven π minus π is equal to negative 64. And what Iβve done is Iβve labeled each of the equations one, two, three, and four, because this will help us when we explain what weβre gonna do next.

So the first thing weβre gonna do is rearrange equation one to make a new equation one. This gives us π equals negative three minus π. And what Iβve done is subtracted π from each side of the equation. And Iβve done this because I want to make π the subject. We couldβve done this and made π the subject and used that as well. Iβve just chosen to make π the subject.

So now what weβre gonna do is use the substitution method because what weβre gonna do is substitute π equals negative three minus π β so this is now our equation one β into equation two. And when we do that, we get negative three minus π minus π is equal to negative 17. And thatβs because weβve got negative three minus π instead of our π. So this is gonna give us negative three minus two π equals negative 17.

So then what we can do is add three to both sides of the equation. So we get negative two π is equal to negative 14. So then if we divide each side of the equation by negative two to get π, we get π is equal to seven. And thatβs because if you divide negative 14 by negative two, you get seven, because a negative divided by a negative is a positive.

So now what we want to do is use this to find π. And the way weβre gonna do that is by substituting π is equal to seven into equation one. And weβre gonna substitute it into the equation one after we rearrange to make π the subject. And once we substitute it in, weβre gonna get π is equal to negative three minus seven, which is gonna give us an π-value of negative 10. So therefore, we found π and π. And what we want to do now is find π and π.

So next, what weβre gonna do is find π. And to find π, what weβre gonna do is substitute π equals negative 10 and π equals seven into π plus π plus π is equal to negative five. And when we do that, weβre gonna get negative 10 add seven plus π is equal to negative five. So then if we simplify, we get negative three plus π is equal to negative five. So now if we add three to each side of the equation, itβs gonna leave us with π on its own on the left-hand side. And when we do that, we get π is equal to negative two. So great, weβve now found π.

So finally, we need to find π. So therefore, in order to find π, what weβre gonna do is substitute π equals negative 10 and π equals seven. But this time, weβre gonna substitute it into equation four. And so when we do that, weβre gonna get negative 10 minus, then seven multiplied by seven minus π is equal to negative 64. So then weβre gonna have negative 10 minus 49 minus π is equal to negative 64.

So now what weβre gonna do is tidy this up. So then weβre gonna have negative 59 minus π is equal to negative 64. So then we can add 59 to each side of the equation. And when we do that, we get negative π is equal to negative five. So finally, we wanna find out what π is. So weβre gonna divide each side of the equation by negative one, which is gonna give us a π-value of π is equal to five.

So therefore, given that our matrices were equal to each other, we can therefore say that the values of π, π, π, and π are negative 10, seven, negative two, and five, respectively. So weβve finished the question and worked out the answer.

The last thing I wanted to do is quickly show you the elimination method because I said we used the substitution method to work out π and π. But we couldβve used the elimination method to get us started. Iβm just gonna quickly demonstrate how youβd do that.

So in order to use the elimination method, what we do is choose one of the variables with the same coefficient in each equation. So in our two equations, we couldβve chosen π or π. Iβve just chosen π. We then look at the sign that corresponds to our variable. So in this case, we can see that we have a positive and a negative. So we actually have different signs.

Itβs worth noting at this point that if weβd chosen π to be our variables, then these would in fact have the same sign. And a common mistake is students see different sign in the middle and assume thatβs a different sign. But itβs not. Itβs the sign of the coefficient. So both of our πs wouldβve been positive.

Okay, so like I said in the example Iβm doing, weβve got different signs. And we have a little memory aid to help us remember what to do. And that is, if you have same signs, then subtract. But if we have different signs, weβre going to add. So weβre gonna add equation one and equation two. And when we do that, we get two π. And thatβs because π plus π is two π. Then we get zero because positive π add negative π is zero. And then weβve got negative 20, and thatβs cause negative three add negative 17 is negative 20. So then what we do is divide both sides of the equation by two to find π. So we get π is equal to negative 10. So itβs the same as we got when we used the other method, so the substitution method. And then what we do from here is carry on using the same as before to find π, π, and π.