Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 03 (b)

MATH152 April 2013

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Question B 03 (b)
Let A be a certain $2\times 3$ matrix. Suppose that using elementary row operations, the matrix A can be transformed into the matrix $U={\begin{bmatrix}1&0&-7\\0&1&4\end{bmatrix}}$ Can you find the solution of $Ax={\begin{bmatrix}1\\0\end{bmatrix}}$ with the given information? Justify your answer.

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Hint
If you were to solve a non-homogeneous system, would the right hand side change as you do elementary row operations? Would this affect the solution?

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. No, this is not possible. Elementary row operations change the right hand side and hence, in general, solutions of Ax = b are not related to solutions of Ux = b. Only in the special case of a homogeneous system, when b = 0, is this possible. An easy way to see this is by a counter example. We will look at two matrices A1 and A2 that can both be row reduced to U, but have different solutions to $A\mathbf {x} ={\begin{bmatrix}1\\0\end{bmatrix}}$ Let us choose A1 = U and $A_{2}={\begin{bmatrix}1&0&-7\\1&1&-3\end{bmatrix}}$ A2 was obtained by adding the first row of U to the second row of U. Hence, by subtracting row 1 from row 2 in A2 gives U, which shows that A2 can be row reduced to U. Solving $A_{1}\mathbf {x} _{1}={\begin{bmatrix}1\\0\end{bmatrix}}$ we get ${\begin{bmatrix}1&0&-7&|&1\\0&1&4&|&0\end{bmatrix}}$ with solution $\mathbf {x} _{1}=a{\begin{bmatrix}7\\-4\\1\end{bmatrix}}+{\begin{bmatrix}1\\0\\0\end{bmatrix}}$ While solving $A_{2}\mathbf {x} _{2}={\begin{bmatrix}1\\0\end{bmatrix}}$ we get ${\begin{bmatrix}1&0&-7&|&1\\1&1&-3&|&0\end{bmatrix}}\rightarrow {\begin{bmatrix}1&0&-7&|&1\\0&1&4&|&\color {red}-1\end{bmatrix}}$ with solution $\mathbf {x} _{2}=b{\begin{bmatrix}7\\-4\\1\end{bmatrix}}+{\begin{bmatrix}1\\\color {red}-1\\0\end{bmatrix}}$ As we see, regardless of a and b we always find $\mathbf {x} _{1}\not =\mathbf {x} _{2}$. Thus using the given information, we cannot tell which solution set is correct.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Another way to view this question is to notice that elementary row operations combine to give us that $\displaystyle A=EU$ where E is an invertible matrix. Thus solving $\displaystyle Ax={\begin{bmatrix}1\\0\end{bmatrix}}$ Reduces to solving $\displaystyle Ux=E^{-1}{\begin{bmatrix}1\\0\end{bmatrix}}$ This last solution varies with the matrix E. Notice that if the vector on the right were the zero vector, then this matrix would not matter (as it didn't in part (a)) but here it can make a huge difference. A concrete counter example is given in solution 1.

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