9.7+OFSA+Solutions

OFSA SOLUTIONS for Matrix Theory

This page contains peer generated solutions and error explanations to OFSA questions. As you read or view the solutions, be critical: check for accuracy, but also for more efficient solution strategies. If you have a better method or different idea/answer, post a discussion and monitor the responses.


 * Quick Directions**
 * Post answers, solutions and error explanations to each OFSA question below.
 * For each "distractor" or incorrect answer choice, explain the error that would lead to that incorrect answer choice.
 * You may either do the above in typed format or using a pencast.
 * Separate each question with a section bar.
 * After each solution, provide a hyperlink back to the corresponding OFSA page.
 * Follow example below.
 * Click here to refer to solution format in 7.7


 * Question 1**
 * __Solution 1__**

math \left[ {\begin{array}{*{20}{c}} {2x + y}&8 \\ 9&y \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {3x}&{x - y} \\ 4&x \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {17}&9 \\ {13}&5 \end{array}} \right] math

math x + y = 5 \to y = 5 - x math

math \begin{gathered} 2x + y + 3x = 17 \hfill \\ 2x + (5 - x) + 3x = 17 \hfill \\ 4x = 12 \hfill \\ x = 3 \hfill \\ \end{gathered} math

math \begin{gathered} y = 5 - 3 \hfill \\ y = 2 \hfill \\ \end{gathered} math


 * __Error Explanation 1__**

A) x=3, y=2 ...................Correct! B) x=22/6, y=-4/3.......... Error. Student solved the first equation as y=x-5 C) x=2, y=3................... Error. Stutdent scrambled x and y. D) x=1, y=12................. Error. Student extracted the incorrect equation x+y=13

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math \left[ {\begin{array}{*{20}{c}} {2a}&b \\ e&d \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} c&c \\ d&{ - 2e} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} b&d \\ 5&c \end{array}} \right] math
 * Question 2**
 * __Solution 2__**

math \begin{gathered} e = d + 5 \hfill \\ d = e - 5 \hfill \\ \end{gathered} math

math \begin{gathered} d + 2e = c \hfill \\ (e - 5) + 2e = c \hfill \\ c = 3e - 5 \hfill \\ \end{gathered} math

math \begin{gathered} b - c = d \hfill \\ b - (3e - 5) = (e - 5) \hfill \\ b = 4e + 10 \hfill \\ \end{gathered} math

math \begin{gathered} 2a - c = b \hfill \\ 2a - (3e - 5) = (4e - 10) \hfill \\ 2a = 7e - 15 \hfill \\ \boxed{a = \frac{2}} \hfill \\ \end{gathered} math


 * __Error Explanation 2__**

A) 7e-5...................Error. Student did not distribute the negative in step four, and forgot to divide by two. B) 7e-15.................Error. Student did not divide by two. C) (7e-5)/2..............Error. Student did not distribute the negative in step four. D) (7e-15)/2............Correct!

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 * Question 3**
 * __Solution 3__**

math \left[ {\begin{array}{*{20}{c}} {30}&{15}&{22} \\ {27}&{10}&{14} \\ {18}&8&{12} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {10}&8&6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\boxed{552}}&{434}&{316} \end{array}} \right] math

math {\text{The answer is 552}} math


 * __Error Explanation 3__**

A) 750.................Error. Student set up matrices incorrectly and solved for the total income generated by blue tuna B) 552.................Correct! C) 434.................Error. Student selected the income gained by Vendor 2 D) No Solution.....Error. Student failed to set up matrices in a way that allows division.

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Student assumes that because of the difference in dimensions that they cannot be multiplied, and therefore the answer is D. D is wrong because the dimensions of the matrices are 5X2 and 2X1. Since the inner dimensions match, there is a real solution with dimensions 5X1. C is the correct answer.
 * Question 4**
 * __Solution 4__**
 * __Error Explanation__** __4__

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Student solves for the determinant as 1/bc-ad, and therefore the answer must be d __**Error Explanation** 5__ A is the correct answer. the equation for the determinant is 1/ad-bc. then this is multiplied by the matrix math \[\begin{bmatrix} d& -b\\ -c& a \end{bmatrix}\] math The determinant alone can knock out all but one answer but if the denominator is switched around to something like bc-ad, like the student did, then the answer will change Click here to return to OFSA.
 * Question 5**
 * __Solution 5__**

math \frac{-x-2}{x^2+10x+24} math Factor the original fraction and divide it into to parts math \frac{-x-2}{(x+6)(x+4)} =\frac{A}{x+6}+\frac{B}{x+4} math Solve for A using the //Cover Up Method// math A=\frac{-(-6)-2}{-6+4}=-2 math Solve for B using the //Cover Up Method// math B=\frac{-(-4)-2}{-4+6}=1 math Final Solution math \frac{-x-2}{x^2+10x+24}=\frac{-2}{x+6}+\frac{1}{x+4} math
 * Question 6**
 * __Solution 6__**

Thus, Choice C is the correct solution.
 * __Error Explanation 6__**

math a)\[\frac{-x}{x+6}-\frac{2}{x+4}\] math Student tried to separate the fraction without using the //Cover Up Method// math b)\[\frac{-x}{x+4}-\frac{2}{x+6}\] math Student once again tried to separate the fraction without using the //Cover Up Method// math c)\[\frac{-2}{x+6}+\frac{1}{x+4}\] math Correct

d) No Solution - Student forgot to cover up the term he/she was solving for

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The student starts to reduce matrix using REF to solve the matrix. math \[\begin{bmatrix} 1& -2& 1& 1\\ 0& 1& 2& 5\\ 0& 1& 3& 8 \end{bmatrix}\] R3-1->R3 \[\begin{bmatrix} 1& -2& 1& 1\\ 0& 1& 2& 5\\ 0& 0& 2& 8 \end{bmatrix}\] math Here the student uses the new matrix to solve for the variables. z=4 y+2z=5 y=-3 x-2y+z=1 x=-9 D is the correct answer. First, the student did not do the operation R3-1 for all values in the row. since the matrix is based off of a system, the student separated the last column because they represented the answers for the system equations. The 8 should have become 7. Next, the student did not reduce the matrix to proper REF form. the last row was left as it was, when it should've been divided by 2. so the new z value would be 3.5. the solution for the problem is (-6.5,-2,3.5). B is the correct answer. Click here to return to OFSA.
 * Question 7**
 * __Solution 7__**
 * __Error Explanation 7__**


 * Question 8**
 * __Solution 8__**

For the inverse of the matrix to be undefined: math \[(n-3)(n+5)=-12\] math Expand left hand side math \[n^2+2n-15=-12\] math Add 12 to both sides math \[n^2+2n-3=0\] math Factor left hand side math \[(n+3)(n-1)=0\] math Solve for n math \[n=-3,1\] math Thus, n cannot equal -3 or 1 and the correct answer is Choice B
 * __Error Explanation 8__**

a) n must equal -3 and 1 - Student forgot that he/she was solving for values that n cannot be b) n must not equal -3 and 1 - Correct c) The inverse is defined for all values of n - Student did not recognize that the matrix is undefined when the determinant=0 d) The inverse is undefined for all values of n - Student once again did not recognize that the matrix is undefined when the determinant=0

For a matrix's inverse to exist, the determinant, or (ad-bc) cannot equal 0. Click here to return to OFSA.


 * Question 9**
 * __Solution 9__**

Solution B is the correct answer as it has 1's for every entry along the major diagonal and 0's for every other entry.


 * __Error Explanation 9__**

math a) \[\begin{bmatrix}

0 &1 \\

1 &0

\end{bmatrix}\] math Student believed that the main diagonal covered entries from top right to bottom left as opposed to top left to bottom right math b) \[\begin{bmatrix}

1 &0 &0 \\

0 &1 &0 \\

0 &0 &1

\end{bmatrix}\] math Correct math c) \[\begin{bmatrix}

0 &0 \\

0& 0

\end{bmatrix}\] math Student did not recognize that an identity matrix must have 1's as the entries along the major diagonal math d) \[\begin{bmatrix}

1 &0 \\

0& 1\\

0&0

\end{bmatrix}\] math Student did not recognize that an identity matrix must be a square matrix

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 * Question 10**
 * __Solution 10__**
 * __Error Explanation 10__**

The student split up the equation to math \[\frac{A}{x}+\frac{B}{x+3}+\frac{C}{x+2}\] math the student then solved for the zero of each factor in the denominator and substituted it back into the original expression. This gave the student undefined solutions, so the answer is D. The student did not use the cover up method when substituting zeros into the original expression. If they had done this they would've gotten A as the correct answer Click here to return to OFSA.
 * Question 11**
 * __Solution 11__**
 * __Error Explanation 11__**

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 * Question 12**
 * __Solution 12__**
 * __Error Explanation 12__**

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 * Question 13**
 * __Solution 13__**
 * __Error Explanation 13__**

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 * Question 14**
 * __Solution 14__**
 * __Error Explanation 14__**

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 * Question 15**
 * __Solution 15__**
 * __Error Explanation 15__**

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 * Question 16**
 * __Solution 16__**
 * __Error Explanation 16__**

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 * Question 17**
 * __Solution 17__**
 * __Error Explanation 17__**

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 * Question 18**
 * __Solution 18__**
 * __Error Explanation 18__**

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 * Question 19**
 * __Solution 19__**
 * __Error Explanation 19__**

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 * Question 20**
 * __Solution 20__**
 * __Error Explanation 20__**

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 * Question 21**
 * __Solution 21__**
 * __Error Explanation 21__**

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 * Question 22**
 * __Solution 22__**
 * __Error Explanation 22__**

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 * Question 23**
 * __Solution 23__**
 * __Error Explanation 23__**

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 * Question 24**
 * __Solution 24__**
 * __Error Explanation 24__**