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Linear Algebra : Non-Homogeneous Cases

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Example Questions

Example Question #91 : Linear Equations

$\begin{align*}&\text{Find }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}-3&-15&6\\15&-9&14\\-18&-5&-10\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-14\\5\\-8\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}11.086\\3.1452\\-0.92742\end{bmatrix}$

$\begin{bmatrix}5.086\\-2.8548\\-6.9274\end{bmatrix}$

$\begin{bmatrix}8.086\\0.14516\\-3.9274\end{bmatrix}$

$\begin{bmatrix}-2.914\\-10.8548\\-14.9274\end{bmatrix}$

Correct answer:

$\begin{bmatrix}5.086\\-2.8548\\-6.9274\end{bmatrix}$

Explanation:

$- ¬\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-3&-15&6\\15&-9&14\\-18&-5&-10\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-14\\5\\-8\end{bmatrix}\\&\text{an approach often taught is to manipulate coefficients to eliminate}\\&\text{variables. There is another to consider.}\\&\text{Note that if a matrix has an inverse, }\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{Knowing this, it stands to reason that if given an equation like: }Ax=B\\&\text{We can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this property we can solve our system:}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-0.43011&0.48387&0.41935\\0.27419&-0.37097&-0.35484\\0.6371&-0.68548&-0.67742\end{bmatrix}\begin{bmatrix}-14\\5\\-8\end{bmatrix}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5.086\\-2.8548\\-6.9274\end{bmatrix}\end{align*}$

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Example Question #2 : Non Homogeneous Cases

$\begin{align*}&\text{Find the values of }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}7&15\\0&-18\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-8\\6\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}-8.43\\-8.33\end{bmatrix}$

$\begin{bmatrix}2.57\\2.67\end{bmatrix}$

$\begin{bmatrix}-0.43\\-0.33\end{bmatrix}$

$\begin{bmatrix}-5.43\\-5.33\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-0.43\\-0.33\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}7&15\\0&-18\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-8\\6\end{bmatrix}\\&\text{an approach often taught is to manipulate coefficients to eliminate}\\&\text{variables. This approach works, though there is another to consider.}\\&\text{If a matrix has an inverse, i.e. it has a determinant,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{It then stands to reason that if given an equation like: }Ax=B\\&\text{We can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{Using this, we can solve for our values:}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0.143&0.119\\0.000&-0.056\end{bmatrix}\begin{bmatrix}-8\\6\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-0.43\\-0.33\end{bmatrix}\end{align*}$

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Example Question #652 : Linear Algebra

$\begin{align*}&\text{Find }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}12&-6\\-2&-5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\3\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}5.24\\4.31\end{bmatrix}$

$\begin{bmatrix}7.24\\6.31\end{bmatrix}$

$\begin{bmatrix}0.24\\-0.69\end{bmatrix}$

$\begin{bmatrix}2.24\\1.31\end{bmatrix}$

Correct answer:

$\begin{bmatrix}0.24\\-0.69\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}12&-6\\-2&-5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\3\end{bmatrix}\\&\text{an approach often taught is to manipulate coefficients to eliminate}\\&\text{variables. This approach works, though there is another to consider.}\\&\text{If a matrix has an inverse, i.e. it has a determinant,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{It then stands to reason that if given an equation like: }Ax=B\\&\text{We can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this property we can solve our system:}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0.069&-0.083\\-0.028&-0.167\end{bmatrix}\begin{bmatrix}7\\3\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0.24\\-0.69\end{bmatrix}\end{align*}$

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Example Question #93 : Linear Equations

$\begin{align*}&\text{Find }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}12&18\\-20&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\-10\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}3.51\\3.05\end{bmatrix}$

$\begin{bmatrix}4.51\\4.05\end{bmatrix}$

$\begin{bmatrix}-8.49\\-8.95\end{bmatrix}$

$\begin{bmatrix}0.51\\0.05\end{bmatrix}$

Correct answer:

$\begin{bmatrix}0.51\\0.05\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}12&18\\-20&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\-10\end{bmatrix}\\&\text{an approach often taught is to manipulate coefficients to eliminate}\\&\text{variables. This approach works, though there is another to consider.}\\&\text{If a matrix has an inverse, i.e. it has a determinant,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{It then stands to reason that if given an equation like: }Ax=B\\&\text{We can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this property we can solve our system:}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0.014&-0.042\\0.046&0.028\end{bmatrix}\begin{bmatrix}7\\-10\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0.51\\0.05\end{bmatrix}\end{align*}$

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Example Question #94 : Linear Equations

$\begin{align*}&\text{Find }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}13&-19&-4\\-10&-1&16\\-9&4&-19\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-138\\134\\-122\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}7.87\\13.81\\16.97\end{bmatrix}$

$\begin{bmatrix}1.87\\7.81\\10.97\end{bmatrix}$

$\begin{bmatrix}-1.13\\4.81\\7.97\end{bmatrix}$

$\begin{bmatrix}4.87\\10.81\\13.97\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-1.13\\4.81\\7.97\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}13&-19&-4\\-10&-1&16\\-9&4&-19\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-138\\134\\-122\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{It then stands to reason that if given an equation like: }Ax=B\\&\text{We can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this property we can solve our system:}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-0.008&-0.063&-0.052\\-0.056&-0.048&-0.028\\-0.008&0.020&-0.034\end{bmatrix}\begin{bmatrix}-138\\134\\-122\end{bmatrix}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-1.13\\4.81\\7.97\end{bmatrix}\end{align*}$

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Example Question #653 : Linear Algebra

$\begin{align*}&\text{Solve the system of equations: }\begin{bmatrix}-5&-16\\19&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-167\\87\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}7.93\\10.58\end{bmatrix}$

$\begin{bmatrix}-0.07\\2.58\end{bmatrix}$

$\begin{bmatrix}5.93\\8.58\end{bmatrix}$

$\begin{bmatrix}-1.07\\1.58\end{bmatrix}$

Correct answer:

$\begin{bmatrix}5.93\\8.58\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-5&-16\\19&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-167\\87\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this, we can solve our system of equations:}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-0.009&0.050\\-0.060&-0.016\end{bmatrix}\begin{bmatrix}-167\\87\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5.93\\8.58\end{bmatrix}\end{align*}$

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Example Question #96 : Linear Equations

$\begin{align*}&\text{Find }\begin{bmatrix}x\\y\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}4&18\\-16&16\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}20\\-29\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}3.39\\1.58\end{bmatrix}$

$\begin{bmatrix}2.39\\0.58\end{bmatrix}$

$\begin{bmatrix}-5.61\\-7.42\end{bmatrix}$

$\begin{bmatrix}-2.61\\-4.42\end{bmatrix}$

Correct answer:

$\begin{bmatrix}2.39\\0.58\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}4&18\\-16&16\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}20\\-29\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this property we can solve our system:}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0.045&-0.051\\0.045&0.011\end{bmatrix}\begin{bmatrix}20\\-29\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2.39\\0.58\end{bmatrix}\end{align*}$

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Example Question #8 : Non Homogeneous Cases

$\begin{align*}&\text{Find the values of }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}-4&4&6\\-15&0&-8\\3&18&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-187\\172\\170\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}4.74\\3.59\\-30.40\end{bmatrix}$

$\begin{bmatrix}-3.26\\-4.41\\-38.40\end{bmatrix}$

$\begin{bmatrix}-1.26\\-2.41\\-36.40\end{bmatrix}$

$\begin{bmatrix}6.74\\5.59\\-28.40\end{bmatrix}$

Correct answer:

$\begin{bmatrix}4.74\\3.59\\-30.40\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-4&4&6\\-15&0&-8\\3&18&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-187\\172\\170\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{Using this, we can solve for our values:}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-0.058&-0.049&0.013\\0.028&0.002&0.049\\0.109&-0.034&-0.024\end{bmatrix}\begin{bmatrix}-187\\172\\170\end{bmatrix}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4.74\\3.59\\-30.40\end{bmatrix}\end{align*}$

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Example Question #1 : Non Homogeneous Cases

$\begin{align*}&\text{Solve the system of equations: }\begin{bmatrix}18&-17&14\\17&-20&1\\6&-5&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}105\\30\\53\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}54.31\\44.23\\-8.62\end{bmatrix}$

$\begin{bmatrix}53.31\\43.23\\-9.62\end{bmatrix}$

$\begin{bmatrix}49.31\\39.23\\-13.62\end{bmatrix}$

$\begin{bmatrix}63.31\\53.23\\0.38\end{bmatrix}$

Correct answer:

$\begin{bmatrix}54.31\\44.23\\-8.62\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}18&-17&14\\17&-20&1\\6&-5&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}105\\30\\53\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this, we can solve our system of equations:}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-2.212&0.615&5.058\\-1.846&0.462&4.231\\0.673&-0.231&-1.365\end{bmatrix}\begin{bmatrix}105\\30\\53\end{bmatrix}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}54.31\\44.23\\-8.62\end{bmatrix}\end{align*}$

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Example Question #2 : Non Homogeneous Cases

$\begin{align*}&\text{Find the values of }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}-2&-17&10\\-19&-3&-19\\20&1&17\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-47\\154\\-98\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}5.98\\-10.63\\-18.88\end{bmatrix}$

$\begin{bmatrix}3.98\\-12.63\\-20.88\end{bmatrix}$

$\begin{bmatrix}8.98\\-7.63\\-15.88\end{bmatrix}$

$\begin{bmatrix}0.98\\-15.63\\-23.88\end{bmatrix}$

Correct answer:

$\begin{bmatrix}8.98\\-7.63\\-15.88\end{bmatrix}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-2&-17&10\\-19&-3&-19\\20&1&17\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-47\\154\\-98\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{Using this, we can solve for our values:}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-0.022&0.207&0.245\\-0.040&-0.162&-0.158\\0.028&-0.234&-0.220\end{bmatrix}\begin{bmatrix}-47\\154\\-98\end{bmatrix}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}8.98\\-7.63\\-15.88\end{bmatrix}\end{align*}$

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Linear Algebra : Non-Homogeneous Cases

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #91 : Linear Equations

Example Question #2 : Non Homogeneous Cases

Example Question #652 : Linear Algebra

Example Question #93 : Linear Equations

Example Question #94 : Linear Equations

Example Question #653 : Linear Algebra

Example Question #96 : Linear Equations

Example Question #8 : Non Homogeneous Cases

Example Question #1 : Non Homogeneous Cases

Example Question #2 : Non Homogeneous Cases

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