Question 255238
See this <a href=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/matrix-multiplication.solver>solver</a> for more help with multiplying matrices.





Since the first matrix is a <font color=red>3</font> by 3 matrix and the second matrix is a 3 by <font color=green>2</font> matrix, this means that the resulting matrix will be a <font color=red>3</font> by <font color=green>2</font> matrix. The final matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix.


So the final resulting matrix will look like:



{{{(matrix(3,3,x,x,x,x,x,x))}}}



note: the "x"s are just placeholders for now




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Multiply the corresponding entries from the <font size="4" color="red">1st</font> row of the first matrix by the <font size="4" color="green">1st</font> column of the second matrix. After multiplying, add the values:



<font size="4" color="red">1st</font> row, <font size="4" color="green">1st</font> column: {{{(matrix(3,3,highlight(-2),highlight(10),highlight(2),-10,9,-1,-3,-6,10))(matrix(3,2,highlight(1),6,highlight(-9),-2,highlight(10),-2))}}}

{{{(-2)*(1)+(10)*(-9)+(2)*(10)=-2+-90+20=-72}}}



 So the element in the <font size="4" color="red">1st</font> row, <font size="4" color="green">1st</font> column of the resulting matrix is {{{-72}}}. Now let's update the matrix:

 

{{{(matrix(3,3,-72,x,x,x,x,x))}}}

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Multiply the corresponding entries from the <font size="4" color="red">1st</font> row of the first matrix by the <font size="4" color="green">2nd</font> column of the second matrix. After multiplying, add the values:



<font size="4" color="red">1st</font> row, <font size="4" color="green">2nd</font> column: {{{(matrix(3,3,highlight(-2),highlight(10),highlight(2),-10,9,-1,-3,-6,10))(matrix(3,2,1,highlight(6),-9,highlight(-2),10,highlight(-2)))}}}

{{{(-2)*(6)+(10)*(-2)+(2)*(-2)=-12+-20+-4=-36}}}



 So the element in the <font size="4" color="red">1st</font> row, <font size="4" color="green">2nd</font> column of the resulting matrix is {{{-36}}}. Now let's update the matrix:

 

{{{(matrix(3,3,-72,-36,x,x,x,x))}}}





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Multiply the corresponding entries from the <font size="4" color="red">2nd</font> row of the first matrix by the <font size="4" color="green">1st</font> column of the second matrix. After multiplying, add the values:



<font size="4" color="red">2nd</font> row, <font size="4" color="green">1st</font> column: {{{(matrix(3,3,-2,10,2,highlight(-10),highlight(9),highlight(-1),-3,-6,10))(matrix(3,2,highlight(1),6,highlight(-9),-2,highlight(10),-2))}}}

{{{(-10)*(1)+(9)*(-9)+(-1)*(10)=-10+-81+-10=-101}}}



 So the element in the <font size="4" color="red">2nd</font> row, <font size="4" color="green">1st</font> column of the resulting matrix is {{{-101}}}. Now let's update the matrix:

 

{{{(matrix(3,3,-72,-36,-101,x,x,x))}}}

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Multiply the corresponding entries from the <font size="4" color="red">2nd</font> row of the first matrix by the <font size="4" color="green">2nd</font> column of the second matrix. After multiplying, add the values:



<font size="4" color="red">2nd</font> row, <font size="4" color="green">2nd</font> column: {{{(matrix(3,3,-2,10,2,highlight(-10),highlight(9),highlight(-1),-3,-6,10))(matrix(3,2,1,highlight(6),-9,highlight(-2),10,highlight(-2)))}}}

{{{(-10)*(6)+(9)*(-2)+(-1)*(-2)=-60+-18+2=-76}}}



 So the element in the <font size="4" color="red">2nd</font> row, <font size="4" color="green">2nd</font> column of the resulting matrix is {{{-76}}}. Now let's update the matrix:

 

{{{(matrix(3,3,-72,-36,-101,-76,x,x))}}}





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Multiply the corresponding entries from the <font size="4" color="red">3rd</font> row of the first matrix by the <font size="4" color="green">1st</font> column of the second matrix. After multiplying, add the values:



<font size="4" color="red">3rd</font> row, <font size="4" color="green">1st</font> column: {{{(matrix(3,3,-2,10,2,-10,9,-1,highlight(-3),highlight(-6),highlight(10)))(matrix(3,2,highlight(1),6,highlight(-9),-2,highlight(10),-2))}}}

{{{(-3)*(1)+(-6)*(-9)+(10)*(10)=-3+54+100=151}}}



 So the element in the <font size="4" color="red">3rd</font> row, <font size="4" color="green">1st</font> column of the resulting matrix is {{{151}}}. Now let's update the matrix:

 

{{{(matrix(3,3,-72,-36,-101,-76,151,x))}}}

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Multiply the corresponding entries from the <font size="4" color="red">3rd</font> row of the first matrix by the <font size="4" color="green">2nd</font> column of the second matrix. After multiplying, add the values:



<font size="4" color="red">3rd</font> row, <font size="4" color="green">2nd</font> column: {{{(matrix(3,3,-2,10,2,-10,9,-1,highlight(-3),highlight(-6),highlight(10)))(matrix(3,2,1,highlight(6),-9,highlight(-2),10,highlight(-2)))}}}

{{{(-3)*(6)+(-6)*(-2)+(10)*(-2)=-18+12+-20=-26}}}



 So the element in the <font size="4" color="red">3rd</font> row, <font size="4" color="green">2nd</font> column of the resulting matrix is {{{-26}}}. Now let's update the matrix:

 

{{{(matrix(3,3,-72,-36,-101,-76,151,-26))}}}









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Answer:



So the solution is  {{{(matrix(3,2,-72,-36,-101,-76,151,-26))}}}


In other words,


{{{(matrix(3,3,-2,10,2,-10,9,-1,-3,-6,10))*(matrix(3,3,1,6,-9,-2,10,-2))=(matrix(3,2,-72,-36,-101,-76,151,-26))}}}



Once again, see this <a href=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/matrix-multiplication.solver>solver</a> for more help with multiplying matrices.