Science:Math Exam Resources/Courses/MATH152/April 2010/Question A 04

MATH152 April 2010

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[hide] Question A 04
Circle the one correct answer below. A linear system of four equations in three unknowns has (a)always a unique solution (b)either a unique solution or no solutions (c)either a unique solution or an infinite number of solutions (d)either no solutions or an infinite number of solutions (e)either no solutions, a unique solution, or an infinite number of solutions

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[show] Hint
Conditions on one, infinite, or no solutions depend on zero rows in the matrix, what kind of conditions create those?

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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. Recall the conditions for a unique solution, no solution and an infinite number of solutions. For a unique solution, the row reduced matrix must be the identity matrix so that each unknown has precisely one determined value (hence unique). For no solutions to occur we need either a contradiction (i.e. trying to determine that an unknown takes on two values) or we need an impossible situation (i.e. trying to determine that 0x=7). For an infinite number of solutions we need a free variable in the form 0z=0. We have 4 equations and 3 unknowns so our matrix is $4\times {3}$. We are trying to determine three variables with 4 conditions. Recall that to determine n unknowns we need exactly n equations (i.e. we only need 3 conditions here). Therefore, the only way we will get a unique solution is if one of the equations is redundant information (so that in actuality we only have 3 conditions). This would appear in the matrix as a row of zeros (since one row will be a multiple of another). For example, consider a system row reduced as follows $\left[{\begin{array}{ccc|c}1&0&0&7\\0&1&0&2\\0&0&1&4\\0&0&0&0\end{array}}\right]$. If we call our unknowns, x,y, and z then we see that x=7, y=2, z=4 and then the last row tells us 0x+0y+0z=0 which holds true and so we have a unique solution. If the 4th condition was not simply redundant information but rather a independent condition then we could row reduce to get something of the form $\left[{\begin{array}{ccc|c}1&0&0&7\\0&1&0&2\\0&0&1&4\\0&0&0&12\end{array}}\right]$. Here we still are determining that x=7, y=2, and z=4 but we also require that 0x+0y+0z=12 which doesn't hold true for these values (in fact it will never hold true) and therefore all 4 conditions can't be satisfied so this results in no solution. We have considered when there is no redundancy in the constraints and when one condition is redundant. What if multiple conditions are redundant? Let's say we have a model with the following constraints ${\begin{aligned}1x+2y+3z&=5\\1x+3y+6z&=2\\2x+4y+8z&=10\\3x+6y+12z&=15.\end{aligned}}$ Notice that equation 3 is just 2 times equation 1 and equation 4 is 3 times equation 1. Therefore, really 3 of the four equations are just saying that 1x+2y+3z=5. The row reduced matrix would be $\left[{\begin{array}{ccc|c}1&0&0&11\\0&1&2&-3\\0&0&0&0\\0&0&0&0\end{array}}\right]$. In this example we have a free variable because we only have 2 pivots. Therefore let z=t. Then we conclude that ${\begin{aligned}x&=11\\y&=-3-2z=-3-2t\\z&=t\end{aligned}}$ or ${\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}11\\-3\\0\end{bmatrix}}+t{\begin{bmatrix}0\\-2\\1\end{bmatrix}}.$ Here we see that there are actually an infinite number of solutions depending on the parameter t. Therefore we have constructed examples where we can have unique solutions, no solutions, or an infinite number of solutions and thus we see the answer is (e)