Matrices - Linear Algebra

Card 1 of 696

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

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

Tap to reveal answer

Answer

To multiply, add:

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

To multiply, add:

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

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

← Didn't Know|Knew It →

Question

Compute AB.

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

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

Tap to reveal answer

Answer

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:

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

← Didn't Know|Knew It →

Question

Compute AB

$A=\begin{bmatrix} 5\end{bmatrix}$

$B = \begin{bmatrix} -5&8 &12 &7 \end{bmatrix}$

Tap to reveal answer

Answer

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 1x4 matrix:

$AB = \begin{bmatrix} 5\cdot -5& 5\cdot 8&5\cdot 12 &5\cdot 7 \end{bmatrix}$

← Didn't Know|Knew It →

Question

Calculate $A\bf{x}$ , given

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

Tap to reveal answer

Answer

By definition,

$A\bf{x}=\begin{bmatrix} 3&4 \ 0&-2 \ 1& 3 \end{bmatrix} \begin{bmatrix} 6\ 2 \end{bmatrix}= \begin{bmatrix} 3(6)+4(2)\ 0(6)-2(2)\ 1(6)+3(2) \end{bmatrix}= \begin{bmatrix} 26\ -4\ 12 \end{bmatrix}$ .

← Didn't Know|Knew It →

Question

Calculate , given

$A = \begin{bmatrix} 1&4 \end{bmatrix}$

$B=\begin{bmatrix} 3\ 4 \end{bmatrix}$

Tap to reveal answer

Answer

By definition,

$AB = \begin{bmatrix} 1& 4 \end{bmatrix} \begin{bmatrix} 3\ 4 \end{bmatrix} = [3(1)+4(4)]= [19]=19$ . A matrix with only one entry is simply a scalar.

← Didn't Know|Knew It →

Question

Calculate , given

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

Tap to reveal answer

Answer

By definition,

$cA = 3\begin{bmatrix} 3&1 \ 5& -2 \end{bmatrix}= \begin{bmatrix} 3(3)&3(1) \ 3(5)&3(-2) \end{bmatrix}= \begin{bmatrix} 9&3 \ 15&-6 \end{bmatrix}$ .

← Didn't Know|Knew It →

Question

$\begin{align}&$\text{Find the product }$A\times B \&$\text{Where }$A=\begin{bmatrix}3&10\3&-10\end{bmatrix}$\text{ and }$B=\begin{bmatrix}2&-8\11&-19\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\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\times2\&\text{The resultant matrix has dimensions: }2\times2\&\begin{bmatrix}3&10\3&-10\end{bmatrix}\times\begin{bmatrix}2&-8\11&-19\end{bmatrix}\&\begin{bmatrix}(3)(2)+(10)(11)&(3)(-8)+(10)(-19)\(3)(2)+(-10)(11)&(3)(-8)+(-10)(-19)\end{bmatrix}\&\begin{bmatrix}116&-214\-104&166\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&$\text{Find the value of }$C$\text{ if }$C=A\times B \&$\text{Where }$A=\begin{bmatrix}-1&-12&-13\-12&14&-11\19&-7&1\end{bmatrix}$\text{ and }$B=\begin{bmatrix}-17\-19\-12\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\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: }3\times3\&\text{and B has dimensions: }3\times1\&\text{The resultant matrix C has dimensions: }3\times1\&C=\begin{bmatrix}-1&-12&-13\-12&14&-11\19&-7&1\end{bmatrix}\times\begin{bmatrix}-17\-19\-12\end{bmatrix}\&C=\begin{bmatrix}(-1)(-17)+(-12)(-19)+(-13)(-12)\(-12)(-17)+(14)(-19)+(-11)(-12)\(19)(-17)+(-7)(-19)+(1)(-12)\end{bmatrix}\&C=\begin{bmatrix}401\70\-202\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&$\text{Find the matrix product }$\&\begin{bmatrix}-19&4\-6&-12\-18&4\end{bmatrix}\times\begin{bmatrix}-10\-19\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\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: }3\times2\&\text{and }2\times1\&\text{The resultant matrix has dimensions: }3\times1\&\begin{bmatrix}-19&4\-6&-12\-18&4\end{bmatrix}\times\begin{bmatrix}-10\-19\end{bmatrix}\&\begin{bmatrix}(-19)(-10)+(4)(-19)\(-6)(-10)+(-12)(-19)\(-18)(-10)+(4)(-19)\end{bmatrix}\&\begin{bmatrix}114\288\104\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&$\text{Find the product }$A\times B \&$\text{Where }$A=\begin{bmatrix}17&7&-7&2\10&0&12&-2\-8&-4&20&-3\end{bmatrix}$\text{ and }$B=\begin{bmatrix}5\20\-19\-2\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\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: }3\times4\&\text{and B has dimensions: }4\times1\&\text{The resultant matrix has dimensions: }3\times1\&\begin{bmatrix}17&7&-7&2\10&0&12&-2\-8&-4&20&-3\end{bmatrix}\times\begin{bmatrix}5\20\-19\-2\end{bmatrix}\&\begin{bmatrix}(17)(5)+(7)(20)+(-7)(-19)+(2)(-2)\(10)(5)+(0)(20)+(12)(-19)+(-2)(-2)\(-8)(5)+(-4)(20)+(20)(-19)+(-3)(-2)\end{bmatrix}\&\begin{bmatrix}354\-174\-494\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\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&-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}$

← Didn't Know|Knew It →

Question

$\begin{align}&$\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 }$B=\begin{bmatrix}-15\8\-11\-8\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\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}$

← Didn't Know|Knew It →

Question

$\begin{align}&$\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 }$B=\begin{bmatrix}6\-18\-17\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\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}$

← Didn't Know|Knew It →

Question

$\begin{align}&$\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 }$B=\begin{bmatrix}16\4\end{bmatrix}\end{align}$

Tap to reveal answer

Answer

$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\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}$

← Didn't Know|Knew It →

Question

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 $A\bf{x}$ .

Tap to reveal answer

Answer

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 $\bf{x}$ .

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

← Didn't Know|Knew It →

0

Knew It