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Linear Algebra : Linear Algebra

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Example Questions

Example Question #801 : Linear Algebra

$\begin{align*}&\text{Find the value of }C\text{ if }C=A\times B \\&\text{Where }A=\begin{bmatrix}-11&-1&0&2\\-16&8&0&-5\\-16&3&-18&-3\\0&8&-15&0\end{bmatrix}\text{ and }B=\begin{bmatrix}-15\\8\\-11\\-8\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}141\\344\\486\\229\end{bmatrix}$

$\text{The multiplication cannot be performed.}$

$\begin{bmatrix}132\\335\\477\\220\end{bmatrix}$

$\begin{bmatrix}233\\436\\578\\321\end{bmatrix}$

Correct answer:

$\begin{bmatrix}141\\344\\486\\229\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In order to multiply two matrices, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }m\times n\text{ and }n\times p\text{ to create an}\\&m\times p\text{ matrix. Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }4\times4\\&\text{and B has dimensions: }4\times1\\&\text{The resultant matrix C has dimensions: }4\times1\\&C=\begin{bmatrix}-11&-1&0&2\\-16&8&0&-5\\-16&3&-18&-3\\0&8&-15&0\end{bmatrix}\times\begin{bmatrix}-15\\8\\-11\\-8\end{bmatrix}\\&C=\begin{bmatrix}(-11)(-15)+(-1)(8)+(0)(-11)+(2)(-8)\\(-16)(-15)+(8)(8)+(0)(-11)+(-5)(-8)\\(-16)(-15)+(3)(8)+(-18)(-11)+(-3)(-8)\\(0)(-15)+(8)(8)+(-15)(-11)+(0)(-8)\end{bmatrix}\\&C=\begin{bmatrix}141\\344\\486\\229\end{bmatrix}\end{align*}$

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Example Question #802 : Linear Algebra

$\begin{align*}&\text{Find the product }A\times B \\&\text{Where }A=\begin{bmatrix}-15&16&13&-5\\10&1&6&-6\end{bmatrix}\text{ and }B=\begin{bmatrix}19\\6\\20\\0\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}-14\\231\end{bmatrix}$

$\begin{bmatrix}71\\316\end{bmatrix}$

$\text{The multiplication cannot be performed.}$

$\begin{bmatrix}100\\345\end{bmatrix}$

Correct answer:

$\begin{bmatrix}71\\316\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In order to multiply two matrices, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }m\times n\text{ and }n\times p\text{ to create an}\\&m\times p\text{ matrix. Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }2\times4\\&\text{and B has dimensions: }4\times1\\&\text{The resultant matrix has dimensions: }2\times1\\&\begin{bmatrix}-15&16&13&-5\\10&1&6&-6\end{bmatrix}\times\begin{bmatrix}19\\6\\20\\0\end{bmatrix}\\&\begin{bmatrix}(-15)(19)+(16)(6)+(13)(20)+(-5)(0)\\(10)(19)+(1)(6)+(6)(20)+(-6)(0)\end{bmatrix}\\&\begin{bmatrix}71\\316\end{bmatrix}\end{align*}$

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Example Question #803 : Linear Algebra

$\begin{align*}&\text{Find the matrix product }\\&\begin{bmatrix}2&11&13\\-1&-3&16\\-13&17&-7\\10&-12&-13\end{bmatrix}\times\begin{bmatrix}-11\\-3\\2\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}71\\152\\178\\0\end{bmatrix}$

$\begin{bmatrix}-29\\52\\78\\-100\end{bmatrix}$

$\text{The multiplication cannot be performed.}$

$\begin{bmatrix}8\\89\\115\\-63\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-29\\52\\78\\-100\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In order to multiply two matrices, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }m\times n\text{ and }n\times p\text{ to create an}\\&m\times p\text{ matrix. Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our matrices have dimensions: }4\times3\text{ and }3\times1\\&\text{The resultant matrix has dimensions: }4\times1\\&\begin{bmatrix}2&11&13\\-1&-3&16\\-13&17&-7\\10&-12&-13\end{bmatrix}\times\begin{bmatrix}-11\\-3\\2\end{bmatrix}\\&\begin{bmatrix}(2)(-11)+(11)(-3)+(13)(2)\\(-1)(-11)+(-3)(-3)+(16)(2)\\(-13)(-11)+(17)(-3)+(-7)(2)\\(10)(-11)+(-12)(-3)+(-13)(2)\end{bmatrix}\\&\begin{bmatrix}-29\\52\\78\\-100\end{bmatrix}\end{align*}$

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Example Question #804 : Linear Algebra

$\begin{align*}&\text{Find the value of }C\text{ if }C=A\times B \\&\text{Where }A=\begin{bmatrix}10&0&-11\\-16&12&17\\-20&-20&-2\\17&5&6\end{bmatrix}\text{ and }B=\begin{bmatrix}6\\-18\\-17\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}247\\-601\\274\\-90\end{bmatrix}$

$\text{The multiplication cannot be performed.}$

$\begin{bmatrix}253\\-595\\280\\-84\end{bmatrix}$

$\begin{bmatrix}343\\-505\\370\\6\end{bmatrix}$

Correct answer:

$\begin{bmatrix}247\\-601\\274\\-90\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In order to multiply two matrices, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }m\times n\text{ and }n\times p\text{ to create an}\\&m\times p\text{ matrix. Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }4\times3\\&\text{and B has dimensions: }3\times1\\&\text{The resultant matrix C has dimensions: }4\times1\\&C=\begin{bmatrix}10&0&-11\\-16&12&17\\-20&-20&-2\\17&5&6\end{bmatrix}\times\begin{bmatrix}6\\-18\\-17\end{bmatrix}\\&C=\begin{bmatrix}(10)(6)+(0)(-18)+(-11)(-17)\\(-16)(6)+(12)(-18)+(17)(-17)\\(-20)(6)+(-20)(-18)+(-2)(-17)\\(17)(6)+(5)(-18)+(6)(-17)\end{bmatrix}\\&C=\begin{bmatrix}247\\-601\\274\\-90\end{bmatrix}\end{align*}$

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Example Question #805 : Linear Algebra

$\begin{align*}&\text{Find the value of }C\text{ if }C=A\times B \\&\text{Where }A=\begin{bmatrix}-7&15\\3&-8\end{bmatrix}\text{ and }B=\begin{bmatrix}16\\4\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}-52\\16\end{bmatrix}$

$\text{The multiplication cannot be performed.}$

$\begin{bmatrix}-19\\49\end{bmatrix}$

$\begin{bmatrix}-115\\-47\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-52\\16\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In order to multiply two matrices, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }m\times n\text{ and }n\times p\text{ to create an}\\&m\times p\text{ matrix. Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }2\times2\\&\text{and B has dimensions: }2\times1\\&\text{The resultant matrix C has dimensions: }2\times1\\&C=\begin{bmatrix}-7&15\\3&-8\end{bmatrix}\times\begin{bmatrix}16\\4\end{bmatrix}\\&C=\begin{bmatrix}(-7)(16)+(15)(4)\\(3)(16)+(-8)(4)\end{bmatrix}\\&C=\begin{bmatrix}-52\\16\end{bmatrix}\end{align*}$

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Example Question #806 : Linear Algebra

$\begin{align*}&\text{Find the matrix product }\\&\begin{bmatrix}-16&16\\3&19\\3&-1\\7&-13\end{bmatrix}\times\begin{bmatrix}-1\\-20\end{bmatrix}\end{align*}$

Possible Answers:

$\text{The multiplication cannot be performed.}$

$\begin{bmatrix}-402\\-481\\-81\\155\end{bmatrix}$

$\begin{bmatrix}-304\\-383\\17\\253\end{bmatrix}$

$\begin{bmatrix}-267\\-346\\54\\290\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-304\\-383\\17\\253\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In order to multiply two matrices, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }m\times n\text{ and }n\times p\text{ to create an}\\&m\times p\text{ matrix. Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our matrices have dimensions: }4\times2\text{ and }2\times1\\&\text{The resultant matrix has dimensions: }4\times1\\&\begin{bmatrix}-16&16\\3&19\\3&-1\\7&-13\end{bmatrix}\times\begin{bmatrix}-1\\-20\end{bmatrix}\\&\begin{bmatrix}(-16)(-1)+(16)(-20)\\(3)(-1)+(19)(-20)\\(3)(-1)+(-1)(-20)\\(7)(-1)+(-13)(-20)\end{bmatrix}\\&\begin{bmatrix}-304\\-383\\17\\253\end{bmatrix}\end{align*}$

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Example Question #807 : Linear Algebra

$\begin{align*}&\text{Find the matrix product }\\&\begin{bmatrix}9&8\\13&2\end{bmatrix}\times\begin{bmatrix}20\\-7\end{bmatrix}\end{align*}$

Possible Answers:

$\text{The multiplication cannot be performed.}$

$\begin{bmatrix}173\\295\end{bmatrix}$

$\begin{bmatrix}124\\246\end{bmatrix}$

$\begin{bmatrix}220\\342\end{bmatrix}$

Correct answer:

$\begin{bmatrix}124\\246\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In order to multiply two matrices, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }m\times n\text{ and }n\times p\text{ to create an}\\&m\times p\text{ matrix. Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our matrices have dimensions: }2\times2\text{ and }2\times1\\&\text{The resultant matrix has dimensions: }2\times1\\&\begin{bmatrix}9&8\\13&2\end{bmatrix}\times\begin{bmatrix}20\\-7\end{bmatrix}\\&\begin{bmatrix}(9)(20)+(8)(-7)\\(13)(20)+(2)(-7)\end{bmatrix}\\&\begin{bmatrix}124\\246\end{bmatrix}\end{align*}$

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Example Question #808 : Linear Algebra

Let be a matrix and $\bf{x}$ be a vector defined by:

$A = \begin{bmatrix} 2 & 6 & 0 \\ 7 & -1 & 9 \end{bmatrix}$

$\bf{x} = \begin{bmatrix} 2\\ 4 \\-3 \end{bmatrix}$

Find the product $A\bf{x}$ .

Possible Answers:

$A\bf{x} = \begin{bmatrix} 32 & -17 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 24 \\ -7 \end{bmatrix}$

The product does not exist because the dimensions do not match.

$A\bf{x} = \begin{bmatrix} 28 & -17 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 28 \\ -17 \end{bmatrix}$

Correct answer:

$A\bf{x} = \begin{bmatrix} 28 \\ -17 \end{bmatrix}$

Explanation:

First we check the dimensions. The matrix has 3 columns and the vector $\bf{x}$ has three rows. The dimensions match and the product exists.

$A\bf{x} = \begin{bmatrix} 2 & 6 & 0 \\ 7 & -1 & 9 \end{bmatrix} \cdot\begin{bmatrix} 2\\ 4 \\-3 \end{bmatrix}$

Now we take the dot product of rows in the matrix and the vector $\bf{x}$ .

$\begin{bmatrix} 2 & 6 & 0 \\ 7 & -1 & 9 \end{bmatrix} \cdot\begin{bmatrix} 2\\ 4 \\-3 \end{bmatrix} = \begin{bmatrix} 2\cdot 2 + 6\cdot 4 + 0\cdot -3 \\ 7\cdot 2 + -1\cdot 4 + 9\cdot -3 \end{bmatrix} = \begin{bmatrix} 28 \\ -17 \end{bmatrix}$

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Example Question #11 : Matrix Vector Product

Let be a matrix and $\bf{x}$ be a vector defined as

$A = \begin{bmatrix} 9 & 0 & 3 \\ 7 & 2 & 9 \\ 5 & -9 & -4 \end{bmatrix}$

$\bf{x} = \begin{bmatrix} -9\\ 3 \\ -2 \end{bmatrix}$

Find the product $A\bf{x}$ .

Possible Answers:

$A\bf{x}=\begin{bmatrix} 2 & -3 & 9 \end{bmatrix}$

$A\bf{x}=\begin{bmatrix}-87 & -75 & -64 \end{bmatrix}$

$A\bf{x}=\begin{bmatrix}-75\\ -87 \\ 64 \end{bmatrix}$

$A\bf{x}=\begin{bmatrix}-87\\ -75 \\ -64 \end{bmatrix}$

The product does not exist because the dimensions do not match.

Correct answer:

$A\bf{x}=\begin{bmatrix}-87\\ -75 \\ -64 \end{bmatrix}$

Explanation:

First we check that the dimensions match. Matrix has 3 columns and vector $\bf{x}$ has three rows. The dimensions match and the product exists.

$A\bf{x}= \begin{bmatrix} 9 & 0 & 3 \\ 7 & 2 & 9 \\ 5 & -9 & -4 \end{bmatrix} \cdot \begin{bmatrix} -9\\ 3 \\ -2 \end{bmatrix} =\begin{bmatrix} (9)(-9) + (0)(3) + (3)(-2)\\ (7)(-9)+(2)(3)+(9)(-2)\\ (5)(-9)+(-9)(3)+(-4)(-2) \end{bmatrix} =\begin{bmatrix} -87\\ -75\\ -64 \end{bmatrix}$

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Example Question #809 : Linear Algebra

Let be a matrix and $\bf{x}$ be a vector defined by

$A = \begin{bmatrix} 0 & 8 & 3 & 6\\ 9 & 0 & 8 & -4 \end{bmatrix}$

$\bf{x} = \begin{bmatrix} 5 \\ 8 \\ -2 \\ 7\end{bmatrix}$

Find the product $A\bf{x}$ .

Possible Answers:

$A\bf{x} = \begin{bmatrix} 100 \\ 1 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 100 & 99 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 10 \\ 10 \end{bmatrix}$

The product does not exist because the dimensions do not match.

$A\bf{x} = \begin{bmatrix} 100 & 1 \end{bmatrix}$

Correct answer:

$A\bf{x} = \begin{bmatrix} 100 \\ 1 \end{bmatrix}$

Explanation:

Matrix has 4 columns and vector $\bf{x}$ has 4 rows. The dimensions match and the product exists.

$A\bf{x} = \begin{bmatrix} 0 & 8 & 3 & 6\\ 9 & 0 & 8 & -4 \end{bmatrix} \cdot \begin{bmatrix} 5 \\ 8 \\ -2 \\ 7\end{bmatrix} = \begin{bmatrix} (0)(5)+(8)(8)+(3)(-2)+(6)(7) \\ (9)(5)+(0)(8)+(8)(-2)+(-4)(7)\end{bmatrix}$

$A\bf{x}= \begin{bmatrix} 100 \\ 1 \end{bmatrix}$

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #801 : Linear Algebra

Example Question #802 : Linear Algebra

Example Question #803 : Linear Algebra

Example Question #804 : Linear Algebra

Example Question #805 : Linear Algebra

Example Question #806 : Linear Algebra

Example Question #807 : Linear Algebra

Example Question #808 : Linear Algebra

Example Question #11 : Matrix Vector Product

Example Question #809 : Linear Algebra

All Linear Algebra Resources

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