Matrix-Vector Product

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Linear Algebra › Matrix-Vector Product

Questions 1 - 10

Rewrite the system of equations:

into a matrix vector product:

where is a 3x3 matrix and are vectors in $\mathbb{R}^3$ .

$\begin{pmatrix} 1& 2 &3 \ -1&5 &7 \ 2&3 &6\end{pmatrix}\begin{pmatrix} x\ y\z \end{pmatrix}=\begin{pmatrix} 3\ 10\2 \end{pmatrix}$

$\begin{pmatrix} -1&5 &7 \ 1& 2 &3 \ 2&3 &6\end{pmatrix}\begin{pmatrix} x\ y\z \end{pmatrix}=\begin{pmatrix} 3\ 10\2 \end{pmatrix}$

$\begin{pmatrix} 1& -1 &2 \ 2&5 &3 \ 3&7 &6\end{pmatrix}\begin{pmatrix} x\ y\z \end{pmatrix}=\begin{pmatrix} 3\ 10\2 \end{pmatrix}$

$\begin{pmatrix} 1& 2 &3 \ -1&5 &7 \ 2&3 &6\end{pmatrix}\begin{pmatrix} x\ y\z \end{pmatrix}=\begin{pmatrix} 10\ 3\2 \end{pmatrix}$

Explanation

To write

into matrix vector form, we recall that matrix multiplication with a vector is done such that the first element in the resulting vector is the dot product of the first row of with the vector , the second element is the dot product of the second row with , and so on. The first row is thus , the second row is , and the third row is . So the left side of the equality is

$\begin{pmatrix} 1& 2 &3 \ -1&5 &7 \ 2&3 &6\end{pmatrix}\begin{pmatrix} x\ y\z \end{pmatrix}$

The right side is the vector , so the final answer is

$\begin{pmatrix} 1& 2 &3 \ -1&5 &7 \ 2&3 &6\end{pmatrix}\begin{pmatrix} x\ y\z \end{pmatrix}=\begin{pmatrix} 3\ 10\2 \end{pmatrix}$

which is equivalent to

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

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

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

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

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

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 eq 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*}$

Multiply $\begin{bmatrix} 5 &3 &-1 \ 0&6 &3 \1 &-2 &1 \end{bmatrix} \cdot \begin{bmatrix} 1\5 \-2 \end{bmatrix}$

$\begin{bmatrix} 22\24 \-11 \end{bmatrix}$

$\begin{bmatrix} 18\ 36\ -8\end{bmatrix}$

$\begin{bmatrix} 3\ 37\12 \end{bmatrix}$

$\begin{bmatrix} 7\29 \16 \end{bmatrix}$

Explanation

To multiply, add:

$1 \cdot \begin{bmatrix} 5\ 0\ 1\end{bmatrix} + 5 \cdot \begin{bmatrix} 3\ 6\ -2\end{bmatrix} + -2 \cdot \begin{bmatrix} -1\3 \1 \end{bmatrix}$

$\begin{bmatrix} 5\ 0\ 1\end{bmatrix} + \begin{bmatrix} 15\ 30\ -10\end{bmatrix} + \begin{bmatrix} 2\ -6\ -2\end{bmatrix} = \begin{bmatrix} 22\ 24\ -11\end{bmatrix}$

Multiply: $\begin{bmatrix} 2 & 0& -5\ -1& 3& 1\end{bmatrix} \cdot \begin{bmatrix} 2\ 2\ 1\end{bmatrix}$

$\begin{bmatrix} -1\5 \end{bmatrix}$

$\begin{bmatrix} 9\8 \end{bmatrix}$

$\begin{bmatrix} 9\5 \end{bmatrix}$

$\begin{bmatrix} -1\8 \end{bmatrix}$

$\begin{bmatrix} -1\ -3\end{bmatrix}$

Explanation

To multiply, add:

$2 \begin{bmatrix} 2\ -1\end{bmatrix} + 2 \begin{bmatrix} 0\3 \end{bmatrix} + 1 \begin{bmatrix} -5\ 1\end{bmatrix}$

$= \begin{bmatrix} 4\-2 \end{bmatrix} + \begin{bmatrix} 0\ 6\end{bmatrix} + \begin{bmatrix} -5\ 1\end{bmatrix} = \begin{bmatrix} -1\ 5\end{bmatrix}$

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

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

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

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

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

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 eq 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*}$

Let $A = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \ - \sqrt{2} & 2\sqrt{3} \end{bmatrix}$ and $B = \begin{bmatrix} \sqrt{2} \ \sqrt{3} \end{bmatrix}$ .

Find .

$\begin{bmatrix}7 \ 4 \end{bmatrix}$

$\begin{bmatrix} 4 & 3 \ -2 & 6 \end{bmatrix}$

$\begin{bmatrix} 4 & \sqrt{6} \ - \sqrt{6} &6\end{bmatrix}$

$\begin{bmatrix} 4 - \sqrt{6} \ 6+ \sqrt{6} \end{bmatrix}$

is not defined.

Explanation

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has two rows, is defined.

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$AB = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \ - \sqrt{2} & 2\sqrt{3} \end{bmatrix} \begin{bmatrix} \sqrt{2} \ \sqrt{3} \end{bmatrix}$

$= \begin{bmatrix} 2\sqrt{2} (\sqrt{2})+ \sqrt{3} (\sqrt{3} )\ - \sqrt{2} (\sqrt{2})+ 2\sqrt{3}(\sqrt{3} ) \end{bmatrix}$

$= \begin{bmatrix}7 \ 4 \end{bmatrix}$

$\begin{align*}&\text{Find the matrix product }\&\begin{bmatrix}9&-11\14&-3\end{bmatrix}\times\begin{bmatrix}7\11\end{bmatrix}\end{align*}$

$\begin{bmatrix}-58\65\end{bmatrix}$

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

$\begin{bmatrix}-33\90\end{bmatrix}$

$\begin{bmatrix}-124\-1\end{bmatrix}$

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 eq 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&-11\14&-3\end{bmatrix}\times\begin{bmatrix}7\11\end{bmatrix}\&\begin{bmatrix}(9)(7)+(-11)(11)\(14)(7)+(-3)(11)\end{bmatrix}\&\begin{bmatrix}-58\65\end{bmatrix}\end{align*}$

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

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

$\bf{x} = \begin{bmatrix} -5 \ 4 \ 8 \ 3 \ 8\end{bmatrix}$

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

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

$A\bf{x} = \begin{bmatrix} 47 \ 61 \ -52\end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 47 & 61 & -52\end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 35 \ 0 \ -72\end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 42 \ 9 \ -7\end{bmatrix}$

Explanation

The matrix has 3 columns and the vector $\bf {x}$ has 5 rows. The dimensions do not match and the product does not exist.

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

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

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

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

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

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 eq 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*}$

Compute AB.

$A=\begin{bmatrix} 1&3 \2 &5 \end{bmatrix}$

$B = \begin{bmatrix} 4& 1& 3\ 6& -2&0 \end{bmatrix}$

$AB = \begin{bmatrix} 22& -5&3 \ 38&-8 &6 \end{bmatrix}$

None of the other answers.

$AB = \begin{bmatrix} 4& 3&3 \ 12&-10 &0 \end{bmatrix}$

$AB = \begin{bmatrix} -14&7 & 3\ -22& 12&6 \end{bmatrix}$

$AB = \begin{bmatrix} 5&4 &3 \ 8&3 &0 \end{bmatrix}$

Explanation

Because the number of columns in matrix A and the number of rows in matrix B are equal, we know that product AB does in fact exist. Matrix AB should have the same number of rows as A and the same number of columns as B. In this case, AB is a 2x3 matrix:

$AB=\begin{bmatrix} (1\cdot 4+3\cdot 6) & (1\cdot 1+3\cdot -2) &(1\cdot 3+3\cdot 0) \ (2\cdot 4+5\cdot 6) & (2\cdot 1+5\cdot -2) &(2\cdot 3+5\cdot 0) \end{bmatrix}$

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