In Exercises 37–44, use Cramer's Rule to solve each system.

Question

In Exercises 37–44, use Cramer's Rule to solve each system. $\begin{cases}x + y + z = 0 \2x - y + z = -1 \-x + 3y - z = -8\end{cases}$

Accepted Answer

Hey everyone here we are asked to apply Cramer's rule to solve the given system of equations. So our given equations are x plus y minus Z is equal to negative +45, X plus two, Y minus Z is equal to negative one and negative two, X plus y minus three, Z is equal to negative 12. And now before we begin solving we first need to recall Cramer's rule which tells us that X is equal to D. Sub X divided by D. Y is equal to D sub Y divided by D. And Z is equal to D. Sub Z divided by D. And now we can begin solving by finding these variables in each equation. So we can start with finding variable D. And now we need to recall that D is a three by three matrix where its first column is the x coefficients from each equation and its 2nd and 3rd columns are the Y n z coefficients from each equation. So doing now we can find our first column, looking at the first equation for our x coefficient, we see that X has a coefficient of one, so we place one in the top left corner of the matrix. We then move on to our second equation and we see that we have a coefficient for X of five. And lastly with our third equation, we have a x coefficient of negative two. Now we can move on to our second column, looking at our y coefficient, our first equation has a Y coefficient of one. Our second equation has a Y coefficient of two and our third equation has a Y coefficient of one. Now we can move on to our final column looking at the the coefficient, our 1st and 2nd equations both have negative one as the the coefficient and our third equation has negative three as the coefficient. And from here we can start solving to find the determinant. So doing this, we first take the first term which is in the top left corner, here is one and then we cross out the row and the column where this value resides. And we're left with four numbers. We take the products of their diagonals and then subtract them. So we start with two times negative three which is negative six minus negative one times one which is negative one we then move on to the second term in the matrix the first row which is also one. But we subtract this second term. So we have minus one times the quantity of well we have to cross out the first row and the column, which this value resides. So we have five times negative three which is negative 15 minus negative one times negative two which is positive two. We then repeat this process for the last term in the first row which is negative one. So we have plus negative one or just minus one times the quantity of five times one which is five and minus two times negative two which is negative four. And now if we were to simplify we find that D. Is equal to three. And now we can move on to the rest of the d variables from the formulas. So we can go ahead and look at D sub X. Well D sub X is also a three by three matrix. But instead of having X coefficients as the first column, they are replaced by these values on the right hand side of each equation. So our first column becomes negative four negative one and negative 12. Now our 2nd and 3rd columns are simply just the Y and Z coefficient. So our second column is 1 to 1 and our third column is negative one negative one negative three. Now solving just like we did before we take the first value which is negative four times the quantity of two times negative three which is negative six minus negative one times one which is negative one. We then subtract the second term which is one times the quantity of negative one times negative three which is positive three minus one times negative 12 which is positive plus the last value which is negative one. So we have minus one times the quantity of negative one times one which is negative, one minus two times negative 12 which is negative 24. Now simplifying, we see that D sub X is equal to six and now we can move on to D sub y again D sub Y is a three by three matrix. The first column are the X coefficients. So we have 15 and negative two. For the first column, our second column instead of the y coefficients, they are replaced by this these values on the right hand side of each equation. So we have negative four negative one negative 12. And our last column the third column is just the Z coefficients which are negative one, negative one, negative three. Now we can solve, so we take the first value which is one times the quantity of negative one, times negative three, which is positive three minus times negative 12, which is positive 12 minus the second value, which is negative four. So we actually have plus four times the quantity of five times negative three, which is negative 15 minus negative one times negative two which is positive two. And finally we take plus the last term in the first row, which is negative one. So we have minus one times the quantity of five times negative 12, which is negative, 60 minus one times negative two, which is positive two, simplifying. We see that D sub y is equal to negative 15. And now we can move on to our final term which is D sub Z. Again the sub Z is a three by three matrix where the 1st and 2nd columns are the X and y coefficient respectively. So for our first column we have 15 and negative two and our second column we have 12 and one. Our third column, instead of having the z coefficients, they are replaced by the terms on the right hand side Of each equation. So we have negative 4 -1 and -12 as our third column. Now we can go ahead and solve. So we take the first turn which is one times the quantity of two times negative 12, which is negative 24 minus negative one times one which is negative one minus the second term in the first row, which is one times the quantity of five times negative 12, which is negative 60 minus one times negative two, which is positive two. Then we add the third term in the first row, which is actually negative four. So we have minus four times the quantity of five times one, which is five minus negative two times two which is negative four, simplifying. We find that D sub Z is equal to three and now we have found each of our variables. So we can go ahead and plug in these values into our formulas we identified above. So starting with X, we see that we have D sub X which is six divided by D which is three. So X is two. Moving on to why we have D sub Y which is negative divided by D which is three, so Y is negative five. Finally we have D. We have Z. So we have D sub Z which is three divided by D, which is three, so Z is one. Now, we just have to write our found values as a set. So we have a final solution of the set of two negative five and one which is our answer here for choice. D. Thanks for watching. I hope you found this video helpful, and I'll see you again next time.

In Exercises 37–44, use Cramer's Rule to solve each system. $\begin{cases}x + y + z = 0 \\2x - y + z = -1 \\-x + 3y - z = -8\end{cases}$

Key Concepts

Cramer's Rule

Determinants of Matrices

Systems of Linear Equations

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