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. $\begin{cases}x + 2z = 4 \2y - z = 5 \2x + 3y = 13\end{cases}$

Accepted Answer

Hey everyone here we are asked to apply Cramer's rule to solve the given system of equations. So the given equations are two, X minus D is equal to negative 28, Y minus D is equal to negative 46 X plus five Y is equal to negative 23. And now before we can begin solving this system we 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 from each formula. So we can start with variable D. Well, D is a three by three matrix where its first column are the x coefficients from each given equation and its 2nd and 3rd columns are the Y and Z coefficients. So beginning with the first column, looking at the x coefficient, we start at the first given equation where we have two X. So our x coefficient is two. Which we place on the top left corner of our matrix. We then move on to our second equation which we can see does not have any X terms. So the matrix gets a zero next with our third equation, we have just X. So we see that the coefficient is just one and now we can move on to our second column. Looking at the y coefficients. Our first equation does not have any Y Y terms. So the matrix gets a zero. Moving on to the second equation, we have eight Y. So we placed eight in the matrix. Then with our third equation we have five Y. So we have a coefficient of five. Now we can move on to our third column looking at the z coefficients. So again beginning at the first equation, we actually see that both the 1st and 2nd equations have negative one as the z coefficient. And our third equation does not have any easy terms. So the matrix gets a zero. Now we can begin solving and to solve this, we need to take the first term and write it out and then we need to cross out its row and the column where it resides. We then take the crux of the diagonals and subtract them from each other. So we have two times the quantity of eight times zero which is just zero minus negative one times five which is negative five. Next we move on to the second term but in this case tracked this second term. So we have minus zero times the quantity. Again crossing out the row and its column, we have zero times zero which is just zero minus negative one times one which is just negative one. Then we take the third term but we add this term but it is negative one. So we have minus one times the quantity while we need to cross out its row and its column and we're left with zero times five which is just zero minus eight times one which is eight. Now simplifying, we find that D equal to 18 and now we can move on to finding the rest of the parts of the formulas. So we can start with D sub x. D sub X is also a three by three matrix. But instead of having X coefficients in the first column, they are replaced by these terms on the right hand side of the equations. So our first column becomes negative to negative 46 negative 23. Now our 2nd and 3rd columns are just the Y and Z coefficients. So for our second column we have 085 and four. Our third column we have negative one negative 10. Now we can solve just like we did earlier taking the first term which is negative two times the quantity of eight times zero which is zero minus negative one times five which is negative five minus the second term, which is just zero. So we have negative 46 times zero which is just zero minus 23 times negative one is just positive plus the third term which is minus one. So we have minus one times the quantity of negative 46 times five which is negative minus eight times negative 23 which is negative 184. Now simplifying we find that D sub X is equal to 36. Now we can move on to finding d sub y D sub y again is a three by three matrix where its first column are the x coefficients. So we have 20, won its second column instead of being y coefficients is now replaced by the terms on the right hand side. So we have negative to negative 46 negative 23. And our third column are just the Z coefficient. So we have 1 -10. Now we can solve like we have been, so we take the first term which is two times the quantity of negative 46 times zero, which is just zero minus negative one times negative 23 which is positive 23 minus the second term, which is negative two. So we have plus two times the quantity of zero times zero, which is just zero minus negative one times one which is negative one plus the last term which is negative one. So we have minus one times the quantity of zero times negative 23 which is just zero minus negative 46 times one, which is just negative 46. Simplifying, we find that D sub y is equal to negative 90. And now we can find our last term which is D sub Z, which is also a three by three matrix where its 1st and 2nd columns are just the X and Y coefficients respectively. So our first column is simply 201 and our second column is 085. And now our third column, instead of being the coefficients are the terms on the right hand side of each equation. So for our third column we have negative to -46 and -23. And now we can just begin solving. So we begin with the first term which is two times the quantity of eight times negative 23 which is negative 184 minus negative 46 times five which is negative 230 minus the second term, which is just zero. So we have zero times negative 23 which is just zero and negative 46 times one, which is negative 46 plus the last term which is negative two. So we have minus two times the quantity of zero times five which is zero minus eight times one, which is just eight simplifying. We find that D subsidy is equal to 108 and now we have all of our unknown. So we can go ahead and plug in these values into our formulas from Cramer's rule. So starting with X we have D sub X which we found to be 36 divided by D which is 18. This simplifies to X. Being to now we can move on to Y. We have D sub Y which we found as negative 90 divided by D which is 18, simplifying. We see that Y is equal to negative five. Now we can move on to Z. We have D sub Z, which is 108 divided by D, which is 18 simplifying. We see that Z is equal to six. Now we have found our solution but we need to rewrite it as a set of points. So we see that our final solution is the set to negative 56, which is answer here for choice. A 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. $\begin{cases}x + 2z = 4 \\2y - z = 5 \\2x + 3y = 13\end{cases}$

Key Concepts

Cramer's Rule

Determinants of Matrices

Setting up the Coefficient Matrix and Constants Vector

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