In Exercises 45–48, explain why the system of equations cannot...

Question

In Exercises 45–48, explain why the system of equations cannot be solved using Cramer's Rule. Then use Gaussian elimination to solve the system.

Image showing a system of three linear equations with variables x, y, and z arranged in matrix form.

Accepted Answer

hey everyone here we are asked to use the Gaussian elimination to solve the system. But first we must explain why we cannot use Cramer's rule. Well first our given equations are X plus three y minus six Z is equal to negative 23 negative two, Y plus four, Z is equal to 14. And finally five X plus eight Y minus 16. Z is equal to negative 66. Now to first determine why we cannot use Cramer's rule, we must find the determinant. So if we recall the determinant, D is the three by three matrix where it's 1st, 2nd and 3rd columns are the X, Y and Z coefficients from the given equations. So beginning with our first column, the x coefficients in starting with our first given equation, we see that the ex coefficients are 105. The second column, the y coefficients are three negative 28. And the third column these d coefficients are negative 64 and negative 16. And now from here we can just begin solving like we normally would with this determinant here, so we take the first value one times the quantity of negative two times negative 16, which is 32 minus four times eight which is also minus the second value in the first row which is three times the quantity of zero times negative 16, which is zero minus four times five which gives us 20 plus the last term in the first row which is negative six, so minus six times the quantity of zero times eight which is zero minus negative two times five which gives us negative 10. And now simplifying, we see that D is equal to zero, which tells us that the system is inconsistent or that it contains dependent equations and we cannot use this to find our solution. So we must move on to using the Gaussian elimination. Now the first step for this elimination method is to set up the augment matrix. And so for this matrix, we must include this bar here to represent the equal sign and to the right of the equal sign lies each of the terms on the right hand side of the equations. So this column starting at the first equation is negative 23 14 and negative 66. And next the left hand side of the equation are the x, y and Z columns. The coefficients from each equation. So starting with the x coefficient column, we have 10 and five from the y coefficients we have three negative 28. And from the z coefficients we have negative 64 and negative 16. And now from here we want to start manipulating this matrix into reduced row echelon form as much as possible. So that means that we need this diagonal here, the one negative two and negative 16 to all be ones. And we want the rest of these numbers here to be zeros. So from this we can start by looking at the first column and we see that we have the appropriate one in zero in the first two rows. But this five here must be changed to a zero. So to do this, we can notice that this one in the first row can be multiplied by negative five and then added to this five to get zero. With that information we know that we can let row three be equal to the result of negative five times row one plus row three. And with this our reduced matrix becomes 13, negative six is equal to negative and zero negative 24 and 14. And finally the new row, the third row becomes zero, negative seven, negative or positive, sorry, 14 and 49. And now we can continue to reduce. And the first thing I notice here is that this middle term. This negative two can easily become a one by dividing by negative two. So with this we know that we can let row to be equal to negative row two divided by two. So with this our matrix is now 13 negative six is equal to negative 23 for the first row. And our second row becomes 01, negative two and negative seven. And our last row, our third row remains as zero negative 7, 14 and 49. Now another manipulation we can notice is that row two can be multiplied by negative three and added to the first row. And so we can get two zeros. So this three and this negative six can become zeros. So with this we can let roll one be equal to negative three times row two plus row one. With this our matrix becomes role one, changing too negative two. And then our second row remains as 01, negative two, negative seven. And our last row as zero negative 7, 14 and 49. And now I see one less one last manipulation that we can do and that we can multiply row two by seven and simply add it to row three to get a few more zeros. So we see that negative seven and 14 will become zeros. So if we let row three be equal to the results of seven times row two plus row three. With this, our final reduced matrix is for row one we have 100 -2. For row two we have 01, negative two and negative seven. And for our final roll it becomes 0000. And so now we can evaluate this matrix here and identify our solution. So from the first row we see that X is equal to negative two. Because if we recall from earlier these columns here are the X, Y and Z. And then from this we can also see from the second row that one y minus two, Z is equal to negative seven. And the last row simply tells us that zero is equal to zero which doesn't really give us any information but it proves that the system is true. So from this we have identified one value, the X value. So now we just need to manipulate this second equation here to find the relationship between Y and Z. So if we get y alone by transposing negative two Z, we have Y is equal to two Z minus seven. And now from here to express our solution in terms of T. We must let Z be equal to T. And now with this we have our equation here for Z, and our equation for Y now becomes to t minus seven. And so now we have each of our three solutions for the set. So our final solution becomes the set of negative two to t minus seven and T. 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 45–48, explain why the system of equations cannot be solved using Cramer's Rule. Then use Gaussian elimination to solve the system.

Key Concepts

Cramer's Rule and Determinants

Gaussian Elimination

Consistency and Dependence of Systems

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