Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations. 2x+y+
z=-3 3x-5y+3z=-4 5x-y+2z=-2
1 answer:
Answer:
(x, y, z) = (1, -1, -4)
Step-by-step explanation:
A suitable graphing or scientific calculator can find the reduced row-echelon form for you. There are on-line calculators that will do that, too.
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In general, if you want to do this by hand, you want to use row operations on the augmented matrix to make the diagonal elements 1 and the off-diagonal elements 0 as shown in the attached result.
If a[i,j] represents the element at row i, column j, you do that by dividing row i by a[i, i] (to make a[i, i] = 1), then subtracting the product of row i and a[k,i] from row k. (for all rows k ≠ i) For this 3-row matrix, repeat these steps for i = 1 to 3.
In the general case of an n by n+1 augmented matrix, you will be doing n^2 row operations, each one involving evaluation of n+1 expressions. The work rapidly grows with matrix size, so readily justifies use of a calculator.
As with many "elimination" problems, appropriate choice of sequence can reduce the work. The above algorithm always produces the reduced row-echelon form, but may result in messy arithmetic along the way.
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So, n = 3, is a 3rd degree polynomial, roots are -2 and 2i
well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too
so, the roots are -2, 2i, -2i
now...![\bf \begin{cases} x=-2\implies x+2=0\implies &(x+2)=0\\ x=2i\implies x-2i=0\implies &(x-2i)=0\\ x=-2i\implies x+2i=0\implies &(x+2i)=0 \end{cases} \\\\\\ (x+2)\underline{(x-2i)(x+2i)}=0\\\\ -----------------------------\\\\ \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -----------------------------\\\\ (x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0 \\\\\\ (x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0 \\\\\\ x^3+2x^2+4x+8=0](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ax%3D-2%5Cimplies%20x%2B2%3D0%5Cimplies%20%26%28x%2B2%29%3D0%5C%5C%0Ax%3D2i%5Cimplies%20x-2i%3D0%5Cimplies%20%26%28x-2i%29%3D0%5C%5C%0Ax%3D-2i%5Cimplies%20x%2B2i%3D0%5Cimplies%20%26%28x%2B2i%29%3D0%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x%2B2%29%5Cunderline%7B%28x-2i%29%28x%2B2i%29%7D%3D0%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bdifference%20of%20squares%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%28a-b%29%28a%2Bb%29%20%3D%20a%5E2-b%5E2%5Cqquad%20%5Cqquad%20%0Aa%5E2-b%5E2%20%3D%20%28a-b%29%28a%2Bb%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%28x%2B2%29%5Bx%5E2-%282i%29%5E2%5D%3D0%5Cimplies%20%28x%2B2%29%5Bx%5E2-%282%5E2i%5E2%29%5D%3D0%0A%5C%5C%5C%5C%5C%5C%0A%28x%2B2%29%5Bx%5E2-%284%5Ccdot%20-1%29%5D%3D0%5Cimplies%20%28x%2B2%29%28x%5E2%2B4%29%3D0%0A%5C%5C%5C%5C%5C%5C%0Ax%5E3%2B2x%5E2%2B4x%2B8%3D0)
now, if we check f(-1), we end up with 5, not 15
hmmm
so, how to turn our 5 to 15? well, 3*5, thus
usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15
well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too
so, the roots are -2, 2i, -2i
now...
now, if we check f(-1), we end up with 5, not 15
hmmm
so, how to turn our 5 to 15? well, 3*5, thus
usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15