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. $\begin{cases}4x - 5y - 6z = -1 \x - 2y - 5z = -12 \2x - y = 7\end{cases}$

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 two x minus three, Y minus four. Z is equal to negative 16, x minus three, Y plus four, Z is equal to seven and three X plus two Y is equal to seven. 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 finally Z is equal to D. Sub Z divided by D. And now from here we can begin solving by finding the variables in each formula. So we can start with variable D. Well three by three matrix where its first column is the x coefficients from each equation and the 2nd and 3rd columns are simply the Y and Z coefficients respectively. So beginning with our first column, we look at the x coefficients, beginning at our first given equation, we have two X. So our first coefficient of x is two. Then we move on to the second equation, we see that X has a coefficient of one. And from our third equation we see that X has a coefficient of three. Now we can move on to our second column, looking at the y coefficients, we see that both the first and the second equations have negative three as a y coefficient and our third equation has two as it's y coefficient and now we can move on to our final column. The third column which looks at the z coefficient. So starting again at our first equation, we see that we have negative for Z. So our coefficient is negative four. Our second equation has positive four and our third equation does not have a Z term. Therefore the matrix gets a zero and now we can begin solving. So we first need to take the first term which is two and then we cross out the row and the column this value resides in and then we take the product of the diagonals and subtract them from each other. So we have two times the quantity of negative three times zero which is just zero minus four times two which is eight. We then move on to the second term. But with this term we need to subtract so we have minus negative three. So we have negative three times the quantity of the remaining terms. So we cross out the row and the column where it lies. So we're left with one times zero which is just zero minus four times three which is 12. Lastly we do the same thing but with our third term we add. So we have minus four times the quantity of the remaining terms which are one times two which is two minus negative three times three which is negative nine. Simplifying. We see that D is equal to negative 96. And now we can move on and start finding D sub x. D. Sub x is also a three by three matrix. But instead of having X terms in the first column, they are replaced by the terms on the right hand side of each equation. So our first column becomes negative 16 77. However, our second two columns are simply the Y and Z coefficients like they were four D. So our second column is negative three, negative 32. And our third column is negative 440, solving just how we did earlier. We have negative 16 times the quantity of negative three times zero which is zero minus the quantity of four times two which is eight minus the second term which is negative three times the quantity of seven times zero which is zero minus four times seven which is the last term which is negative four. So we have minus four times the quantity of seven times two which is 14 minus negative three times seven which is negative 21 simplifying, we see that D sub x is equal to negative 96. Now we can move on to D sub Y. D sub Y is also a three by three matrix. However, its first column are the X coefficient. So we have 213. But its second column which would be the y coefficients are replaced by the terms on the right hand side of each equation. So we have negative 16 and our third column are just the coefficients. So we have negative 440. Now solving we take the first turn which is two times the quantity of seven times zero which is zero minus seven times four which is minus the second term, which is negative 16 times the quantity of one times zero which is zero minus four times three which is 12 plus the last term, which is negative four times the quantity of one times seven which is seven minus seven times three which is 21. Simplifying, we find that D sub, y is equal to negative 192. Lastly we can look for D sub Z, D sub Z is a three by three matrix. However, its 1st and 2nd columns are simply the X and y coefficients respectively. So for the first column we have 213 and for our second column we have negative three negative 32. And our third column, instead of being z coefficient, they are replaced by the terms on the right hand side of each equation. So we have negative 1677. Now we can solve just how we have been. So we take the first term which is two times the quantity of negative three times seven which is negative 21 minus seven times two which is 14 minus the second term which is negative three times the quantity of one times seven which is seven minus seven times three which is 21 plus the last term which is negative 16. So we have minus 16 times the quantity of one times two which is two minus negative three times three which is negative nine. Simplifying. We get that D sub Z is equal to negative 288. Now we have found all of the necessary values so we can just plug them into our formulas above. So starting with X, we have D. Sub X which is negative 96 divided by D. Which is also negative 96. So we see that X is just one now for why we have D supply which is negative 192 divided by D. Which is negative 96. And we see that Y is equal to two. Finally with Z we have D sub Z which is negative 288 divided by D. Which is negative 96 giving us Z is equal to three. Now we just have to write these values as a set and we see that our final solution is the set of points 123 which is our answer here for choice B 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. $\begin{cases}4x - 5y - 6z = -1 \\x - 2y - 5z = -12 \\2x - y = 7\end{cases}$

Key Concepts

Cramer's Rule

Determinants of Matrices

Systems of Linear Equations

Watch next