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Linear Algebra : Linear Algebra

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Example Questions

Example Question #11 : Non Homogeneous Cases

Given a minimization linear programming problem, the objective function of its dual maximization problem is $4y_{1} + 3y_{2}+ 6y_{3}$ .

True or false: It follows that in the original minimization problem, the objective function is $4x_{1} + 3x_{2}+ 6x_{3}$ .

Possible Answers:

True

False

Correct answer:

False

Explanation:

The statement is false. To show the reason for this, let the original minimization problem be as follows - and note that the objective function has three variables, so there will be three linear constraints other than the trivial ones:

Minimize $n_{1}x_{1}+n_{2} x_{2}+ n_{3}x_{3}$ subject to:

$x_{1}, x_{2}, x_{3}\ge 0$

$a_{11}x_{1}+ a_{12}x_{2}+ a_{13}x_{3} \ge b_{1}$

$a_{11}x_{1}+ a_{22}x_{2}+ a_{23}x_{3} \ge b_{2}$

$a_{31}x_{1}+ a_{32}x_{2}+ a_{33}x_{3} \ge b_{3}$

To find the dual maximization problem:

First, form the matrix of coefficients of the constraints (other than the positivity constraints) and the objective function, with the objective function on the bottom:

$\begin{bmatrix} a_{11} & a_{12} & a_{13} & b_{1} \\ a_{21} & a_{22} & a_{23}& b_{2} \\ a_{31} & a_{32} & a_{33} & b_{3} \\ n_{1} & n_{2} & n_{3}&0\end{bmatrix}$

Next, transpose the matrix:

$\begin{bmatrix} a_{11} & a_{21} & a_{31} & n_{1} \\ a_{12} & a_{22} & a_{32}& n_{2} \\ a_{13} & a_{23} & a_{33} & n_{3} \\ b_{1} &b_{2} & b_{3}&0\end{bmatrix}$

This is the coefficient matrix of the dual maximization problem. The bottom row represents the coefficients of the objective function. This function is given to be $4y_{1} + 3y_{2}+ 6y_{3}$ , so

$(b_{1}, b_{2}, b_{3}) = (4,3,6)$ .

The coefficient matrix of the original problem is

$\begin{bmatrix} a_{11} & a_{12} & a_{13} & 4\\ a_{21} & a_{22} & a_{23}& 3 \\ a_{31} & a_{32} & a_{33} & 6 \\ n_{1} & n_{2} & n_{3}&0\end{bmatrix}$

Since the entries in the bottom row are still unknown, no information about the objective function of the original minimization problem has been revealed.

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Example Question #12 : Non Homogeneous Cases

Solve the linear system:

x+ 2y + z = 9

2x + y - z = 9

4y + 7z= 6

Possible Answers:

$\left ( \frac{1}{2}t , \frac{5}{2}t,t\right ) , t \in \mathbb{R}$

$\left (-\frac{1}{2}t , -\frac{5}{2}t,t\right ) , t \in \mathbb{R}$

(-1,-5, 2 )

(1,5, -2 )

The system has no solution.

Correct answer:

(1,5, -2 )

Explanation:

Write the augmented matrix of the system using the coefficients of the equations:

$\begin{matrix} x+ 2y + z = 9 \\ 2x + y - z = 9 \\ 0x+ 4y + 7z= 6 \end{matrix} \Rightarrow \begin{bmatrix} 1 & 2 & 1 & 9 \\ 2 & 1 & -1 & 9 \\ 0 & 4 & 7 & 6 \end{bmatrix}$

Perform the Gauss-Jordan elimination method on this matrix to get it in reduced row-echelon form. Use the following row operations:

$-2R1 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 2 & 1 & 9 \\ 0 & -3 & -3 &- 9 \\ 0 & 4 & 7 & 6 \end{bmatrix}$

$-\frac{1}{3}R2 \rightarrow R2$

$\begin{bmatrix} 1 & 2 & 1 & 9 \\ 0 & 1 &1&3 \\ 0 & 4 & 7 & 6 \end{bmatrix}$

$-2R2 +R1 \rightarrow R1$

$-4R2 +R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & -1 & 3 \\ 0 & 1 &1&3 \\ 0 & 0 & 3 &- 6 \end{bmatrix}$

$\frac{1}{3} R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & -1 & 3 \\ 0 & 1 &1&3 \\ 0 & 0 & 1 &- 2 \end{bmatrix}$

$R3 + R1 \rightarrow R1$

$-R3 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 &0&5 \\ 0 & 0 & 1 &- 2 \end{bmatrix}$

The matrix is now in reduced row-echelon form. This matrix can be interpreted to mean:

x= 1, y = 5, y = -2 .

The only solution is (1,5, -2 ) .

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Example Question #13 : Non Homogeneous Cases

Give the partial fractions decomposition of $\frac{4}{x^{3}+ 4x^{2}- 3x -12 }$ .

Possible Answers:

$\frac{-4x+ 16 }{13 (x^{2}-3)}+ \frac{4}{13(x+ 4)}$

$\frac{ 4x+ 16 }{13 (x^{2}-3)}+ \frac{4}{13(x+ 4)}$

$\frac{4x- 16 }{13 (x^{2}-3)}+ \frac{4}{13(x+ 4)}$

$\frac{-4x+ 4 }{13 (x^{2}-3)}+ \frac{16}{13(x+ 4)}$

$\frac{4x-4 }{13 (x^{2}-3)}+ \frac{16}{13(x+ 4)}$

Correct answer:

$\frac{-4x+ 16 }{13 (x^{2}-3)}+ \frac{4}{13(x+ 4)}$

Explanation:

The denominator of the fraction, $x^{3}+ 4x^{2}- 3x -12$ , can be factored as

$(x^{2} - 3 )(x+4)$ .

The partial fractions decomposition of a rational expression is the sum of fractions with these factors, with the numerator above each denominator being degree one less. Thus, the partial fractions decomposition of the given expression takes the form

$\frac{4}{x^{3}+ 4x^{2}- 3x -12 }$

$= \frac{4}{ (x^{2}-3)(x+ 4)}$

$= \frac{Ax+B}{x^{2}-3}+ \frac{C}{x+ 4}$

for some $A,B, C \in \mathbb{R}$ .

Express the fractions with a common denominator:

$\frac{4}{ (x^{2}-3) (x+ 4)} = \frac{(Ax+B)(x+ 4)}{(x^{2}-3) (x+ 4)}+ \frac{C(x^{2}-3)}{(x^{2}-3) (x+ 4)}$

Eliminate the denominators and perform some algebra:

$4 = (Ax+B)(x+ 4) + C(x^{2}-3)$

$4 = Ax^{2} +4Ax+ Bx + 4B + Cx^{2} -3C$

$4 = (A+C)x^{2} +(4A + B)x +( 4B -3C)$

Set the coefficients equal to form the linear system

A+C = 0

4A+B = 0

4B-3C = 4

This can be solved using Gauss-Jordan elimination on the augmented matrix

$\begin{bmatrix} 1 & 0 & 1 & 0 \\ 4 & 1 & 0 & 0 \\ 0 & 4 & -3 &4 \end{bmatrix}$

Perform the following row operations on the matrix to get it into reduced row-echelon form:

$-4R1+R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & -4 & 0 \\ 0 & 4 & -3 &4 \end{bmatrix}$

$-4R2+R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & -4 & 0 \\ 0 & 0 & 13 &4 \end{bmatrix}$

$\frac{1}{13}R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 1 & 0 \\ \\ 0 & 1 & -4 & 0 \\ \\ 0 & 0 & 1 & \frac{4}{13} \end{bmatrix}$

$-R3 + R1 \rightarrow R1$

$4R3 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 & - \frac{4}{13} \\ \\ 0 & 1 & 0 & \frac{16}{13} \\ \\ 0 & 0 & 1 & \frac{4}{13} \end{bmatrix}$

$A = - \frac{4}{13} ,B = \frac{16}{13} , C= \frac{4}{13}$ , so the partial fractions decomposition is

$\frac{4}{x^{3}+ 4x^{2}- 3x -12 } = \frac{- \frac{4}{13}x+ \frac{16}{13} }{x^{2}-3}+ \frac{\frac{4}{13}}{x+ 4}$ ,

or, simplified,

$\frac{4}{x^{3}+ 4x^{2}- 3x -12 } = \frac{-4x+ 16 }{13 (x^{2}-3)}+ \frac{4}{13(x+ 4)}$ .

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Example Question #14 : Non Homogeneous Cases

Give the partial fractions decomposition of $\frac{4}{x^{3}+ 3x^{2}- 4x -12 }$

Possible Answers:

$\frac{ 1 }{5(x-2)}- \frac{1}{x+2}+ \frac{ 4}{5(x+3)}$

$\frac{ 4 }{5(x-2)}-\frac{1}{x+2}+ \frac{ 1}{5(x+3)}$

$\frac{ 4 }{5(x-2)}-\frac{1}{x+2}- \frac{ 1}{5(x+3)}$

$\frac{ 1 }{5(x-2)}+\frac{1}{x+2}+ \frac{ 4}{5(x+3)}$

$\frac{ 4 }{5(x-2)}+\frac{1}{x+2}+ \frac{ 1}{5(x+3)}$

Correct answer:

$\frac{ 1 }{5(x-2)}- \frac{1}{x+2}+ \frac{ 4}{5(x+3)}$

Explanation:

The denominator of the fraction, $x^{3}+ 3x^{2}- 4x -12$ , can be factored as

$x^{3}+ 3x^{2}- 4x -12 = (x-2) (x+2)(x-3)$

$\frac{4}{x^{3}+ 3x^{2}- 4x -12 }$

$=\frac{4}{(x-2)(x+2)(x+3) }$

$= \frac{A}{x-2}+ \frac{B}{x+2}+ \frac{C}{x+3}$

for some $A,B, C \in \mathbb{R}$ .

Express the fractions with a common denominator: $\frac{4}{ (x-2) (x+2)(x+3) }$

$= \frac{A(x+2)(x+3)}{ (x-2) (x+2)(x+3)}+ \frac{B(x-2)(x+3)}{ (x-2) (x+2) (x+3)}+ \frac{C(x-2) (x+2)}{ (x-2) (x+2)(x+3)}$

Eliminate the denominators and perform some algebra:

4 = A(x+2)(x+3) + B(x-2)(x+3) + C(x-2) (x+2)

$4 = Ax^{2}+ 5Ax+ 6A + B x^{2} +Bx - 6B + Cx^{2} -4C$

$4 = (A+B+C) x^{2}+( 5A+B) x+ (6A - 6B -4C)$

Set the coefficients equal to form the linear system

A+B+C = 0

5A+B = 0

6A-6B-4C = 4

This can be solved using Gauss-Jordan elimination on the augmented matrix

$\begin{bmatrix} 1 & 1 & 1 & 0 \\ 5 & 1 & 0 & 0 \\ 6 & -6 & -4 & 4 \end{bmatrix}$

Perform the following row operations on the matrix to get it into reduced row-echelon form:

$-5R1+R2 \rightarrow R2$

$-6R1+R3 \rightarrow R3$

$\begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & -4 & -5 & 0 \\ 0 & -12 & -10 & 4 \end{bmatrix}$

$-\frac{1}{4} R2 \rightarrow R2$

$\begin{bmatrix} 1 & 1 & 1 & 0 \\ \\0 &1 & \frac{5}{4} & 0 \\ \\ 0 & -12 & -10 & 4 \end{bmatrix}$

$-R2+R1 \rightarrow R1$

$12R2 + R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & -\frac{1}{4} & 0 \\ \\0 &1 & \frac{5}{4} & 0 \\ \\ 0 & 0 & 5 & 4 \end{bmatrix}$

$\frac{1}{5} R32 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & -\frac{1}{4} & 0 \\ \\0 &1 & \frac{5}{4} & 0 \\ \\ 0 & 0 & 1 & \frac{4}{5} \end{bmatrix}$

$\frac{1}{4}R3 + R1 \rightarrow R1$

$-\frac{5}{4}R3 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 &0& \frac{1}{5}\\ \\0 &1 &0 & -1 \\ \\ 0 & 0 & 1 & \frac{4}{5} \end{bmatrix}$

$A= \frac{1}{5}, B= -1, C= \frac{4}{5}$ , so the partial fractions decomposition is

$\frac{4}{x^{3}+ 3x^{2}- 4x -12 } = \frac{ \frac{1}{5}}{x-2}+ \frac{-1}{x+2}+ \frac{ \frac{4}{5}}{x+3}$ ,

or, simplified,

$\frac{4}{x^{3}+ 3x^{2}- 4x -12 } = \frac{ 1 }{5(x-2)}- \frac{1}{x+2}+ \frac{ 4}{5(x+3)}$ .

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Example Question #1 : Criteria For Uniqueness And Consistency

$\begin{align*}&\text{Decide if there is a unique solution for the equation: }\\&\begin{bmatrix}-6&-7&-19\\-8&19&-14\\0&5&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-21\\-36\\-149\end{bmatrix}\end{align*}$

Possible Answers:

$\text{There is not a unique solution.}$

$\text{There is a unique solution.}$

Correct answer:

$\text{There is not a unique solution.}$

Explanation:

$\begin{align*}&\text{To determine whether or not the system of equations represented}\\&\text{by a matrix equation has a unique solution, one should calculate}\\&\text{the determinant. If the determinant is a nonzero value, then}\\&\text{the system of equations has a unique solution. Otherwise, if}\\&\text{the determinant is zero, the system of equation may have no}\\&\text{solution, or it may have an infinite number of them. To refresh,}\\&\text{to calculate a matrix of the given dimensions we use the formula:}\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&det\begin{bmatrix}-6&-7&-19\\-8&19&-14\\0&5&2\end{bmatrix}=0\\&\text{There is not a unique solution.}\end{align*}$

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Example Question #671 : Linear Algebra

$\begin{align*}&\text{Is there a unique solution for the equation }\\&\begin{bmatrix}-20&12\\7&16\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-171\\149\end{bmatrix}?\end{align*}$

Possible Answers:

$\text{There is a unique solution.}$

$\text{There is not a unique solution.}$

Correct answer:

$\text{There is a unique solution.}$

Explanation:

$\begin{align*}&\text{To determine whether or not the system of equations represented}\\&\text{by a matrix equation has a unique solution, one should calculate}\\&\text{the determinant. If the determinant is a nonzero value, then}\\&\text{the system of equations has a unique solution. Otherwise, if}\\&\text{the determinant is zero, the system of equation may have no}\\&\text{solution, or it may have an infinite number of them. To refresh,}\\&\text{to calculate a matrix of the given dimensions we use the formula:}\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&det\begin{bmatrix}-20&12\\7&16\end{bmatrix}=-404\\&\text{There is a unique solution.}\end{align*}$

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Example Question #672 : Linear Algebra

$\begin{align*}&\text{Determine whether or not the equation }\\&\begin{bmatrix}6&-8\\3&12\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-188\\-101\end{bmatrix}\\&\text{has a unique solution.}\end{align*}$

Possible Answers:

$\text{There is a unique solution.}$

$\text{There is not a unique solution.}$

Correct answer:

$\text{There is a unique solution.}$

Explanation:

$\begin{align*}&\text{To determine whether or not the system of equations represented}\\&\text{by a matrix equation has a unique solution, one should calculate}\\&\text{the determinant. If the determinant is a nonzero value, then}\\&\text{the system of equations has a unique solution. Otherwise, if}\\&\text{the determinant is zero, the system of equation may have no}\\&\text{solution, or it may have an infinite number of them. To refresh,}\\&\text{to calculate a matrix of the given dimensions we use the formula:}\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&det\begin{bmatrix}6&-8\\3&12\end{bmatrix}=96\\&\text{There is a unique solution.}\end{align*}$

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Example Question #114 : Linear Equations

$\begin{align*}&\text{Is there a unique solution for the equation }\\&\begin{bmatrix}0&-8&2\\-5&3&2\\20&12&-14\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-163\\36\\22\end{bmatrix}?\end{align*}$

Possible Answers:

$\text{There is a unique solution.}$

$\text{There is not a unique solution.}$

Correct answer:

$\text{There is not a unique solution.}$

Explanation:

$\begin{align*}&\text{To determine whether or not the system of equations represented}\\&\text{by a matrix equation has a unique solution, one should calculate}\\&\text{the determinant. If the determinant is a nonzero value, then}\\&\text{the system of equations has a unique solution. Otherwise, if}\\&\text{the determinant is zero, the system of equation may have no}\\&\text{solution, or it may have an infinite number of them. To refresh,}\\&\text{to calculate a matrix of the given dimensions we use the formula:}\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&det\begin{bmatrix}0&-8&2\\-5&3&2\\20&12&-14\end{bmatrix}=0\\&\text{There is not a unique solution.}\end{align*}$

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Example Question #115 : Linear Equations

$\begin{align*}&\text{Is there a unique solution for the equation }\\&\begin{bmatrix}13&13&-2\\12&12&12\\-16&-16&-11\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}28\\125\\107\end{bmatrix}?\end{align*}$

Possible Answers:

$\text{There is a unique solution.}$

$\text{There is not a unique solution.}$

Correct answer:

$\text{There is not a unique solution.}$

Explanation:

$\begin{align*}&\text{To determine whether or not the system of equations represented}\\&\text{by a matrix equation has a unique solution, one should calculate}\\&\text{the determinant. If the determinant is a nonzero value, then}\\&\text{the system of equations has a unique solution. Otherwise, if}\\&\text{the determinant is zero, the system of equation may have no}\\&\text{solution, or it may have an infinite number of them. To refresh,}\\&\text{to calculate a matrix of the given dimensions we use the formula:}\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&det\begin{bmatrix}13&13&-2\\12&12&12\\-16&-16&-11\end{bmatrix}=0\\&\text{There is not a unique solution.}\end{align*}$

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Example Question #116 : Linear Equations

$\begin{align*}&\text{Decide if there is a unique solution for the equation: }\\&\begin{bmatrix}1&11&14\\-1&10&-12\\11&-5&16\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}59\\4\\-32\end{bmatrix}\end{align*}$

Possible Answers:

$\text{There is a unique solution.}$

$\text{There is not a unique solution.}$

Correct answer:

$\text{There is a unique solution.}$

Explanation:

$\begin{align*}&\text{To determine whether or not the system of equations represented}\\&\text{by a matrix equation has a unique solution, one should calculate}\\&\text{the determinant. If the determinant is a nonzero value, then}\\&\text{the system of equations has a unique solution. Otherwise, if}\\&\text{the determinant is zero, the system of equation may have no}\\&\text{solution, or it may have an infinite number of them. To refresh,}\\&\text{to calculate a matrix of the given dimensions we use the formula:}\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&det\begin{bmatrix}1&11&14\\-1&10&-12\\11&-5&16\end{bmatrix}=-2646\\&\text{There is a unique solution.}\end{align*}$

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #11 : Non Homogeneous Cases

Example Question #12 : Non Homogeneous Cases

Example Question #13 : Non Homogeneous Cases

Example Question #14 : Non Homogeneous Cases

Example Question #1 : Criteria For Uniqueness And Consistency

Example Question #671 : Linear Algebra

Example Question #672 : Linear Algebra

Example Question #114 : Linear Equations

Example Question #115 : Linear Equations

Example Question #116 : Linear Equations

All Linear Algebra Resources

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