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anonymous · Answer

attachment

anonymous · Answer

Find -5A+4B

ganeshie8 · Answer

To multiply a matrix by a number. 
simply multiply each element by that number

anonymous · Answer

That's what I thought, but the answer I got wasn't one of my choices. Can you help me figure out the answer?

ganeshie8 · Answer

$$

A =   \left[ \begin{array}{cc}
6&1 \
-4&-6 \
7&-7

 \end{array} \right]

$$

anonymous · Answer

Would I do: -5(6) -5(1)
                 -5(-4) -5(-6)
                 -5(7) -5(-7)

anonymous · Answer

Since A=-5?

ganeshie8 · Answer

Yes,


to get  $\large  -5A$, multiply each element by -5 :

$$

-5A =   \left[ \begin{array}{cc}
6(-5)&1(-5) \
-4(-5)&-6(-5) \
7(-5)&-7(-5)

 \end{array} \right]

$$

anonymous · Answer

ok i got 1 -4
            -9 -11
            2 -12
is that right?

ganeshie8 · Answer

$$

-5A  + 4B  =   \left[ \begin{array}{cc}
6(-5)&1(-5) \
-4(-5)&-6(-5) \
7(-5)&-7(-5)

 \end{array} \right]

+
 \left[ \begin{array}{cc}
-5(4)&-1(4) \
-3(4)&-8(4) \
6(4)&8(4)

 \end{array} \right]


$$

ganeshie8 · Answer

simplify first before adding...

anonymous · Answer

and for the second one i got -20 -4
                                           -12 -32
                                          24 32

anonymous · Answer

so now i add the two together?

anonymous · Answer

1+(-20)
-4+(-4)
-9+(-12)
-11+(-32)
2+24
-12+32

ganeshie8 · Answer

$$

-5A  + 4B  =   \left[ \begin{array}{cc}
6(-5)&1(-5) \
-4(-5)&-6(-5) \
7(-5)&-7(-5)

 \end{array} \right]

+
 \left[ \begin{array}{cc}
-5(4)&-1(4) \
-3(4)&-8(4) \
6(4)&8(4)

 \end{array} \right]


$$ 

$$\large 

\qquad  =   \left[ \begin{array}{cc}
-30&-5 \
20&30 \
-35&35

 \end{array} \right]

+
 \left[ \begin{array}{cc}
-20&-4 \
-23&-32 \
24&32

 \end{array} \right]


$$

anonymous · Answer

ok thanks so much!!

ganeshie8 · Answer

$$

-5A  + 4B  =   \left[ \begin{array}{cc}
6(-5)&1(-5) \
-4(-5)&-6(-5) \
7(-5)&-7(-5)

 \end{array} \right]

+
 \left[ \begin{array}{cc}
-5(4)&-1(4) \
-3(4)&-8(4) \
6(4)&8(4)

 \end{array} \right]


$$ 

$$\large 

\qquad  =   \left[ \begin{array}{cc}
-30&-5 \
20&30 \
-35&35

 \end{array} \right]

+
 \left[ \begin{array}{cc}
-20&-4 \
-23&-32 \
24&32

 \end{array} \right]


$$ 


$$\large 

\qquad  =   \left[ \begin{array}{cc}
-30 + (-20)&-5+(-4) \
20+(-3)&30+(-32) \
-35+24&35+32

 \end{array} \right]

$$

$$\large 

\qquad  =   \left[ \begin{array}{cc}
-50&-9\
17&-2 \
-11&67

 \end{array} \right]

$$

anonymous · Answer

could you help me with a couple more questions?

ganeshie8 · Answer

shoot...

anonymous · Answer

attachment

ganeshie8 · Answer

$$

2X  + 2 \left[ \begin{array}{cc}
2&-8 \
-4& 2\

 \end{array} \right]
=
 \left[ \begin{array}{cc}
4&-6\
2&-8\

 \end{array} \right]


$$

ganeshie8 · Answer

divide the equation thru by 2 first

ganeshie8 · Answer

$$

2X  + 2 \left[ \begin{array}{cc}
2&-8 \
-4& 2\

 \end{array} \right]
=
 \left[ \begin{array}{cc}
4&-6\
2&-8\

 \end{array} \right]


$$ 


$$

X  +  \left[ \begin{array}{cc}
2&-8 \
-4& 2\

 \end{array} \right]
=
 \left[ \begin{array}{cc}
2&-3\
1&-4\

 \end{array} \right]


$$

ganeshie8 · Answer

next, isolate the X by subtract that matrix

anonymous · Answer

how do you subtract them?

ganeshie8 · Answer

$$

2X  + 2 \left[ \begin{array}{cc}
2&-8 \
-4& 2\

 \end{array} \right]
=
 \left[ \begin{array}{cc}
4&-6\
2&-8\

 \end{array} \right]


$$ 


$$

X  +  \left[ \begin{array}{cc}
2&-8 \
-4& 2\

 \end{array} \right]
=
 \left[ \begin{array}{cc}
2&-3\
1&-4\

 \end{array} \right]


$$ 


$$

X  =  

 \left[ \begin{array}{cc}
2&-3\
1&-4\

 \end{array} \right]

-

 \left[ \begin{array}{cc}
2&-8 \
-4& 2\

 \end{array} \right]

$$

ganeshie8 · Answer

subtract the corresponding elements... just like u did the addition...

ganeshie8 · Answer

$$

2X  + 2 \left[ \begin{array}{cc}
2&-8 \
-4& 2\

 \end{array} \right]
=
 \left[ \begin{array}{cc}
4&-6\
2&-8\

 \end{array} \right]


$$ 


$$

X  +  \left[ \begin{array}{cc}
2&-8 \
-4& 2\

 \end{array} \right]
=
 \left[ \begin{array}{cc}
2&-3\
1&-4\

 \end{array} \right]


$$ 


$$

X  =  

 \left[ \begin{array}{cc}
2&-3\
1&-4\

 \end{array} \right]

-

 \left[ \begin{array}{cc}
2&-8 \
-4& 2\

 \end{array} \right]

$$ 

$$

X  =  

 \left[ \begin{array}{cc}
2-2&-3-(-8)\
1-(-4)&-4-2\

 \end{array} \right]

$$

ganeshie8 · Answer

simplify

anonymous · Answer

0 5 
-3 -2?