How do you multiply #((4, 0), (-1, 3), (2, -5))# with #((1),( -3))#?

2 Answers
Dec 14, 2016

#[(4),(-10),(17)]#

Explanation:

#color(white)(aaa)[A]color(white)(aaaaaaa)[B]color(white)(aaaaa)[C]#

#[(4,0),(-1,3),(2,-5)] * [(1),(-3)] = [(a), (b),(c)]#

#color(white)(aa)color(red)3" x "color(blue)2color(white)(aaaa)color(blue)2" x "color(red)1color(white)(aa)color(red)3" x "color(red)1#

The size of the first matrix #A# is #"2 x 2"# because it has 2 rows and 2 columns. The size of the second matrix #B# is #"2 x 1"# because it has 2 rows and 1 column.

When the sizes are written next to each other, the "inner" numbers must match. The "outer" numbers determine the size of the resultant matrix.

The resultant matrix #C# has been labeled #[(a),(b),(c)]# and has size #3" x "1#.

Position #a# is in the first row and first column. It's value is obtained by multplying the first row of matrix #A# by the first column of matrix #B#.

#a=4*1 +0*3=4#

Position #b# is in the second row and first column. It is the result of multiplying the second row of #A# by the first column of #B#.

#b=-1*1+3*-3=-10#

Position #c# is obtained by multiplying the third row of #A# by the first column of #B#.

#c=2*1+ -5*-3=17#

The answer is then #[(4),(-10),(17)]#

Dec 14, 2016

Shell wrote a good answer while I was writing mine.