1, 3, or 5 Minute Drill Multiplication Worksheets Number Range (0 - 12) A timed drill is a multiplication worksheet with all of the single digit multiplication problems on one page. A student should be able to work all of the problems on the multiplication worksheets correctly in the allowed time. These multiplication worksheets are appropriate. Motrix 1.5.15 add to watchlist send us an update. 5 screenshots: runs on: Windows 10 32/64 bit Windows 8 32/64 bit Windows 7 32/64 bit file size: 94.6 MB filename: Motrix-1.5.15.exe main.
If (mathbf{A}) has dimensions (mtimes n) and (mathbf{B}) has dimensions (ntimes p), then the product (mathbf{AB}) is defined, and has dimensions (mtimes p).
The entry ((ab)_{ij}) is obtained by multiplying row (i) of (mathbf{A}) by column (j) of (mathbf{B}), which is done by multiplying corresponding entries together and then adding the results:
Example (PageIndex{1})
Calculate the product
[begin{pmatrix} 1 &-2 &4 5 &0 &3 0 & 1/2 &9 end{pmatrix}begin{pmatrix} 1 &0 5 &3 -1 &0 end{pmatrix} nonumber]
Solution
We need to multiply a (3times 3) matrix by a (3times 2) matrix, so we expect a (3times 2) matrix as a result.
[begin{pmatrix} 1 &-2 &4 5 &0 &3 0 & 1/2 &9 end{pmatrix}begin{pmatrix} 1 &0 5 &3 -1 &0 end{pmatrix}=begin{pmatrix} a&b c&d e &f end{pmatrix} nonumber]
To calculate (a), which is entry (1,1), we use row 1 of the matrix on the left and column 1 of the matrix on the right:
[begin{pmatrix} {color{red}1} &{color{red}-2} &{color{red}4} 5 &0 &3 0 & 1/2 &9 end{pmatrix}begin{pmatrix} {color{red}1} &0 {color{red}5} &3 {color{red}-1} &0 end{pmatrix}=begin{pmatrix} {color{red}a}&b c&d e &f end{pmatrix}rightarrow a=1times 1+(-2)times 5+4times (-1)=-13 nonumber]
To calculate (b), which is entry (1,2), we use row 1 of the matrix on the left and column 2 of the matrix on the right:
[begin{pmatrix} {color{red}1} &{color{red}-2} &{color{red}4} 5 &0 &3 0 & 1/2 &9 end{pmatrix}begin{pmatrix} 1&{color{red}0} 5&{color{red}3} -1&{color{red}0} end{pmatrix}=begin{pmatrix} a&{color{red}b} c&d e &f end{pmatrix}rightarrow b=1times 0+(-2)times 3+4times 0=-6 nonumber]
To calculate (c), which is entry (2,1), we use row 2 of the matrix on the left and column 1 of the matrix on the right:
[begin{pmatrix} 1&-2&4 {color{red}5} &{color{red}0} &{color{red}3} 0 & 1/2 &9 end{pmatrix}begin{pmatrix} {color{red}1} &0 {color{red}5} &3 {color{red}-1} &0 end{pmatrix}=begin{pmatrix} a&b {color{red}c}&d e &f end{pmatrix}rightarrow c=5times 1+0times 5+3times (-1)=2 nonumber]
To calculate (d), which is entry (2,2), we use row 2 of the matrix on the left and column 2 of the matrix on the right:
[begin{pmatrix} 1&-2&4 {color{red}5} &{color{red}0} &{color{red}3} 0 & 1/2 &9 end{pmatrix}begin{pmatrix} 1&{color{red}0} 5&{color{red}3} -1&{color{red}0} end{pmatrix}=begin{pmatrix} a&b c&{color{red}d} e &f end{pmatrix}rightarrow d=5times 0+0times 3+3times 0=0 nonumber] Maccleanse 8 0 1.
To calculate (e), which is entry (3,1), we use row 3 of the matrix on the left and column 1 of the matrix on the right:
[begin{pmatrix} 1&-2&4 5&0&3 {color{red}0} &{color{red}1/2} &{color{red}9} end{pmatrix}begin{pmatrix} {color{red}1} &0 {color{red}5} &3 {color{red}-1} &0 end{pmatrix}=begin{pmatrix} a&b c&d {color{red}e} &f end{pmatrix}rightarrow e=0times 1+1/2times 5+9times (-1)=-13/2 nonumber]
To calculate (f), which is entry (3,2), we use row 3 of the matrix on the left and column 2 of the matrix on the right:
[begin{pmatrix} 1&-2&4 5&0&3 {color{red}0} &{color{red}1/2} &{color{red}9} end{pmatrix}begin{pmatrix} 1&{color{red}0} 5&{color{red}3} -1&{color{red}0} end{pmatrix}=begin{pmatrix} a&b c&d e&{color{red}f} end{pmatrix}rightarrow f=0times 0+1/2times 3+9times 0=3/2 nonumber]
The result is:
[displaystyle{color{Maroon}begin{pmatrix} 1 &-2 &4 5 &0 &3 0 & 1/2 &9 end{pmatrix}begin{pmatrix} 1 &0 5 &3 -1 &0 end{pmatrix}=begin{pmatrix} -13&-6 2&0 -13/2 &3/2 end{pmatrix}} nonumber]
Example (PageIndex{2})
Calculate
[begin{pmatrix} 1 &-2 &4 5 &0 &3 end{pmatrix}begin{pmatrix} 1 5 -1 end{pmatrix}nonumber]
Solution
We are asked to multiply a (2times 3) matrix by a (3times 1) matrix (a column vector). The result will be a (2times 1) matrix (a vector).
[begin{pmatrix} 1 &-2 &4 5 &0 &3 end{pmatrix}begin{pmatrix} 1 5 -1 end{pmatrix}=begin{pmatrix} a b end{pmatrix}nonumber]
[a=1times1+(-2)times 5+ 4times (-1)=-13nonumber]
[b=5times1+0times 5+ 3times (-1)=2nonumber]
The solution is:
[displaystyle{color{Maroon}begin{pmatrix} 1 &-2 &4 5 &0 &3 end{pmatrix}begin{pmatrix} 1 5 -1 end{pmatrix}=begin{pmatrix} -13 2 end{pmatrix}}nonumber]
Need help? The link below contains solved examples: Multiplying matrices of different shapes (three examples): http://tinyurl.com/kn8ysqq
External links:
Matrix multiplication is not, in general, commutative. For example, we can perform
[begin{pmatrix} 1 &-2 &4 5 &0 &3 end{pmatrix}begin{pmatrix} 1 5 -1 end{pmatrix}=begin{pmatrix} -13 2 end{pmatrix} nonumber]
but cannot perform
[begin{pmatrix} 1 5 -1 end{pmatrix}begin{pmatrix} 1 &-2 &4 5 &0 &3 end{pmatrix} nonumber]
Even with square matrices, that can be multiplied both ways, multiplication is not commutative. In this case, it is useful to define the commutator, defined as:
[[mathbf{A},mathbf{B}]=mathbf{A}mathbf{B}-mathbf{B}mathbf{A} nonumber]
Example (PageIndex{3})
Given (mathbf{A}=begin{pmatrix} 3&1 2&0 end{pmatrix}) and (mathbf{B}=begin{pmatrix} 1&0 -1&2 end{pmatrix})
Calculate the commutator ([mathbf{A},mathbf{B}])
Solution
[[mathbf{A},mathbf{B}]=mathbf{A}mathbf{B}-mathbf{B}mathbf{A}nonumber]
[mathbf{A}mathbf{B}=begin{pmatrix} 3&1 2&0 end{pmatrix}begin{pmatrix} 1&0 -1&2 end{pmatrix}=begin{pmatrix} 3times 1+1times (-1)&3times 0 +1times 2 2times 1+0times (-1)&2times 0+ 0times 2 end{pmatrix}=begin{pmatrix} 2&2 2&0 end{pmatrix}nonumber]
[mathbf{B}mathbf{A}=begin{pmatrix} 1&0 -1&2 end{pmatrix}begin{pmatrix} 3&1 2&0 end{pmatrix}=begin{pmatrix} 1times 3+0times 2&1times 1 +0times 0 -1times 3+2times 2&-1times 1+2times 0 end{pmatrix}=begin{pmatrix} 3&1 1&-1 end{pmatrix}nonumber]
[[mathbf{A},mathbf{B}]=mathbf{A}mathbf{B}-mathbf{B}mathbf{A}=begin{pmatrix} 2&2 2&0 end{pmatrix}-begin{pmatrix} 3&1 1&-1 end{pmatrix}=begin{pmatrix} -1&1 1&1 end{pmatrix}nonumber]
[displaystyle{color{Maroon}[mathbf{A},mathbf{B}]=begin{pmatrix} -1&1 1&1 end{pmatrix}}nonumber]
The multiplication of a vector (vec{v_1}) by a scalar (n) produces another vector of the same dimensions that lies in the same direction as (vec{v_1});
[nbegin{pmatrix} x y end{pmatrix}=begin{pmatrix} nx ny end{pmatrix} nonumber]
Focus 1 9 14 esv. The scalar can stretch or compress the length of the vector, but cannot rotate it (figure [fig:vector_by_scalar]).
The multiplication of a vector (vec{v_1}) by a square matrix produces another vector of the same dimensions of (vec{v_1}). For example, we can multiply a (2times 2) matrix and a 2-dimensional vector:
[begin{pmatrix} a&b c&d end{pmatrix}begin{pmatrix} x y end{pmatrix}=begin{pmatrix} ax+by cx+dy end{pmatrix} nonumber]
For example, consider the matrix
[mathbf{A}=begin{pmatrix} -2 &0 0 &1 end{pmatrix} nonumber]
The product
[begin{pmatrix} -2&0 0&1 end{pmatrix}begin{pmatrix} x y end{pmatrix} nonumber]
Sketchup pro 2017 17 2 2554 en ru. is
[begin{pmatrix} -2x y end{pmatrix} nonumber]
We see that (2times 2) matrices act as operators that transform one 2-dimensional vector into another 2-dimensional vector. This particular matrix keeps the value of (y) constant and multiplies the value of (x) by -2 (Figure (PageIndex{3})).
Notice that matrices are useful ways of representing operators that change the orientation and size of a vector. An important class of operators that are of particular interest to chemists are the so-called symmetry operators.
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