Science:Math Exam Resources/Courses/MATH152/April 2012/Question 02 (b)

MATH152 April 2012

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[hide] Question 02 (b)
Consider the following system of linear equations ${\begin{aligned}2x-y+2z&=a\\x+3y-6z&=b\\3x+2y-4z&=c,\end{aligned}}$ where a,b, and c $\in \mathbb {R}$ . For what values of a, b, and c is the system homogeneous? Solve the resulting homogeneous system by calculating the reduced row echelon form of the augmented matrix.

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[show] Hint 1
A homogeneous problem is one in which the source vector is zero.

[show] Hint 2
The source vector is the vector $\mathbf {b}$ in $A\mathbf {x} =\mathbf {b}$

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[show] Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We obtain the homogeneous problem by setting a=b=c=0. We can then solve the system by Gaussian elimination. First we switch the second and first row so that we have a 1 as the first entry. Doing this we get, $\left[{\begin{array}{ccc\|c}1&3&-6&0\\2&-1&2&0\\3&2&-4&0\end{array}}\right]$ Next we want zeros in the rest of the entries of the first column. To do this we subtract 2 multiples of row 1 from row 2 and 3 multiples of row 1 from row 3 $\left[{\begin{array}{ccc\|c}1&3&-6&0\\0&-7&14&0\\0&-7&14&0\end{array}}\right]$ Now divide the second row by -7 to get a 1 in the second column of the second row $\left[{\begin{array}{ccc\|c}1&3&-6&0\\0&1&-2&0\\0&-7&14&0\end{array}}\right]$ Next subtract 3 multiples of row 2 from row 1 and add 7 multiples of row 2 to row 3 $\left[{\begin{array}{ccc\|c}1&0&0&0\\0&1&-2&0\\0&0&0&0\end{array}}\right]$ The system is now in reduced row echelon form and we can now solve the system. From the first two rows we get ${\begin{array}{cccc}x&&&=0\\&y&-2z&=0\end{array}}$ The third row is a little more interesting. No matter what we multiply the third row with, we will get the zero vector which implies that we have a free variable. This free variable can't be x since we know that x=0 and so it will be either y or z. Let's choose it to be the z variable and set $\displaystyle {}z=t$ to be a parameter. Using the relation from row 2 then we get $\displaystyle {}y=2z=2t$ We can therefore write that the solution to the homogeneous problem is $\mathbf {x} ={\begin{bmatrix}x\\y\\z\end{bmatrix}}=t{\begin{bmatrix}0\\2\\1\end{bmatrix}}$ for any number t.