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

x +...

Question

In Exercises 37–44, use Cramer's Rule to solve each system. x + y + z = 4 x - 2y + z = 7 x + 3y + 2z = 4

Accepted Answer

Hey everyone here we are asked to apply Cramer's rule to solve the given set of equations. So our given equations are five, X plus five, Y plus two. Z is equal to 22 three, X minus Y plus Z is equal to negative and two X plus three, Y plus two, Z is equal to 17. 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 from here we can begin solving by first finding the variables of each formula here. So we can start with variable D. D is a three by three matrix where its first column are the x coefficients from each given equation and the 2nd and 3rd columns are simply the Y. And the coefficients again from each equation. So beginning with our first column, looking at the x coefficients, we see that in the first equation we have five X. So we place five in the top left corner of the matrix. We then move on to our second equation which has three X. So we place three and then our third equation has a x coefficient of two. We can now move on to our second column, looking at the y coefficients, our first equation has a y coefficient of five. Our second equation has a y coefficient of negative one and our third equation has a coefficient of three. Now moving on to our final column, the z coefficients, Our first equation has a c coefficient of two. Our second equation has a coefficient of one and our third equation also has a coefficient of two and now we can begin solving. So if you recall to solve this matrix here we have to take the first term which is in the top left corner of the matrix. So we have five and then we multiply it by the quantity of the remaining value. So we need to cross out the row and the column which this value resides in. We then multiply take the product of the diagonals and then subtract them accordingly. So we have five times the quantity of negative one times two which is negative two minus one times three which is three we then can move on to the second term which we have to subtract. So we have minus five times the quantity of the remaining values which we cross out the row and the column we have three times two which is six minus one times two which is two. We then repeat this but add our final term here. So we have plus two times the quantity of three times three which is nine minus two times negative one which is negative two. And now we solve this and find that D. Is equal to negative 23. We can now move on to the rest of the variables from the formula. So looking to find D sub x. Well, again we have a three by three matrix, but instead of our first column being the x coefficient, they are replaced by these terms on the right hand side of the equation. So our first column is 22 negative 15 and 17 however, are 2nd and 3rd columns are just the Y and Z coefficients. So our second column is five negative and our third column is 212. Now we can solve just like we did before. So we take the first term which is 22 times the quantity of negative one times two which is negative two minus one times three, which is three minus the second term, which is five times the quantity of negative 15 times two which is negative, 30 minus one times 17, which is just 17 plus the third term, which is two times the quantity of negative 15 times three which is negative, 45 minus 91 times 17 which is negative 17, simplifying, we find that D sub x is equal to 69. Now we can move on to finding the supply. Again, this is a three by three matrix. Its first column are simply the X coefficients. So we have 53 to Its second column is instead of the y coefficients, they're replaced by again these terms on the right hand side of the equations. So our second column is 22 negative 17 and our third column is simply the z coefficients. So we have 212 solving we take the first term which is five times the quantity of negative 15 times two which is negative 30 minus one times 17, which is just 17 minus the second term, 22 times the quantity of three times two, which is six minus one times two which is two plus the third term, which is two times the quantity of three times 17, which is 51 -15 times two, which is negative 30 simplifying. We see that D sub y is equal to negative 161. Now we can move on to our final variable here, which is D sub Z D sub Z is again a three by three matrix where the 1st and 2nd columns are simply the X and Y coefficients. So for our first column we have and for our second column we have five negative and our third column instead of z coefficients, we replace them with the terms on the right hand side of each equation. So we have 22 -1517. Now we can begin solving, so we take the first term five times the quantity of negative one, times 17, which is negative 17 minus negative 15 times three, which is negative, 45 minus the second term five times the quantity of three times 17 which is minus negative 15 times two which is negative plus the third term which is 22 times the quantity of three times three which is nine minus negative one times two. Giving us negative two. Simplifying. We find that the sub is equal to negative 23. Now we have our necessary values all of our unknowns. So we can simply plug in these values into the formulas we addressed above. So starting with X. We have D. Sub X which is 69 Divided by D. Which is -23. This simplifies to x being negative three. Moving on to Y we have D. Sub Y. Which is negative 161 divided by D. Which is negative 23 simplifying to why being seven. And now finally with Z. We have D. Sub Z. Which is negative 23 divided by D. Which is also negative 23 simplifying to Z being one. Now we just have to write down these final terms as a set for our final answer. So we see that our final answer is the set of numbers -371 which is our answer here for choice. C. Thanks for watching. I hope you found this video helpful and I'll see you again next time

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In Exercises 37–44, use Cramer's Rule to solve each system. x + y + z = 4 x - 2y + z = 7 x + 3y + 2z = 4