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Linear Algebra : Linear Equations

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Example Questions

Example Question #111 : Linear Equations

Solve the linear system:

x+ 2y + z = 9

2x + y - z = 9

4y + 7z= 6

Possible Answers:

(-1,-5, 2 )

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

(1,5, -2 )

The system has no solution.

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

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 #11 : Non Homogeneous Cases

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

Possible Answers:

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

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

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

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

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

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 #113 : Linear Equations

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

Possible Answers:

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

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

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

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

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

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 #111 : Linear Equations

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

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 #112 : Linear Equations

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

Possible Answers:

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

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

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 #113 : Linear Equations

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

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:

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

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

Correct answer:

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

Explanation:

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

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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:

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

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

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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Example Question #117 : Linear Equations

$\begin{align*}&\text{Determine whether or not the equation }\\&\begin{bmatrix}9&-12\\-3&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-57\\182\end{bmatrix}\\&\text{has a unique solution.}\end{align*}$

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:

$\begin{align*}&\text{To determine whether or not the system of equations represented}\\&\text{by a matrix equation has a unique solution, one should calculate}\\&\text{the determinant. If the determinant is a nonzero value, then}\\&\text{the system of equations has a unique solution. Otherwise, if}\\&\text{the determinant is zero, the system of equation may have no}\\&\text{solution, or it may have an infinite number of them. To refresh,}\\&\text{to calculate a matrix of the given dimensions we use the formula:}\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&det\begin{bmatrix}9&-12\\-3&4\end{bmatrix}=0\\&\text{There is not a unique solution.}\end{align*}$

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Linear Algebra : Linear Equations

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #111 : Linear Equations

Example Question #11 : Non Homogeneous Cases

Example Question #113 : Linear Equations

Example Question #111 : Linear Equations

Example Question #112 : Linear Equations

Example Question #113 : Linear Equations

Example Question #114 : Linear Equations

Example Question #115 : Linear Equations

Example Question #116 : Linear Equations

Example Question #117 : Linear Equations

All Linear Algebra Resources

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