how do you solve the simultaneous equations 2x + y = 9 and x - 2y = -8 using the matrix method?
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i first convert the bases into a 2x2 matrix and then i multiplied the inverse matrix by the 2x1 e/f matrix of 9 (on top) and -8 (on the bottom)
which gave me x = 5.4 and y = 5
however the answer sheets states that x= 2 and y=5
May i please have some help on where I'm going wrong? Am i not multiplying the signs correctly? thank you for your help.
matrices systems-of-equations matrix-equations
asked Apr 24 '16 at 17:02
m.am.a
21
$endgroup$
add a comment |
$begingroup$
i first convert the bases into a 2x2 matrix and then i multiplied the inverse matrix by the 2x1 e/f matrix of 9 (on top) and -8 (on the bottom)
which gave me x = 5.4 and y = 5
however the answer sheets states that x= 2 and y=5
May i please have some help on where I'm going wrong? Am i not multiplying the signs correctly? thank you for your help.
matrices systems-of-equations matrix-equations
asked Apr 24 '16 at 17:02
m.am.a
21
$endgroup$
$begingroup$
your approach is right , just look at the value of inverse , whether it is right , or whether you wrote the matrix wrong
$endgroup$
– avz2611
Apr 24 '16 at 17:04
add a comment |
$begingroup$
i first convert the bases into a 2x2 matrix and then i multiplied the inverse matrix by the 2x1 e/f matrix of 9 (on top) and -8 (on the bottom)
which gave me x = 5.4 and y = 5
however the answer sheets states that x= 2 and y=5
May i please have some help on where I'm going wrong? Am i not multiplying the signs correctly? thank you for your help.
matrices systems-of-equations matrix-equations
asked Apr 24 '16 at 17:02
m.am.a
21
$endgroup$
i first convert the bases into a 2x2 matrix and then i multiplied the inverse matrix by the 2x1 e/f matrix of 9 (on top) and -8 (on the bottom)
which gave me x = 5.4 and y = 5
however the answer sheets states that x= 2 and y=5
May i please have some help on where I'm going wrong? Am i not multiplying the signs correctly? thank you for your help.
matrices systems-of-equations matrix-equations
matrices systems-of-equations matrix-equations
asked Apr 24 '16 at 17:02
m.am.a
21
asked Apr 24 '16 at 17:02
m.am.a
21
asked Apr 24 '16 at 17:02
m.am.a
21
asked Apr 24 '16 at 17:02
m.am.a
21
asked Apr 24 '16 at 17:02
m.am.a
21
21
$begingroup$
your approach is right , just look at the value of inverse , whether it is right , or whether you wrote the matrix wrong
$endgroup$
– avz2611
Apr 24 '16 at 17:04
add a comment |
$begingroup$
your approach is right , just look at the value of inverse , whether it is right , or whether you wrote the matrix wrong
$endgroup$
– avz2611
Apr 24 '16 at 17:04
$begingroup$
your approach is right , just look at the value of inverse , whether it is right , or whether you wrote the matrix wrong
$endgroup$
– avz2611
Apr 24 '16 at 17:04
$begingroup$
your approach is right , just look at the value of inverse , whether it is right , or whether you wrote the matrix wrong
$endgroup$
– avz2611
Apr 24 '16 at 17:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
In matrix notation we have:
$$
Ax=b
$$
with
$$
A=begin{bmatrix}
2&1\1&-2
end{bmatrix}
qquad x=begin{bmatrix}
x\y
end{bmatrix}qquad
b=begin{bmatrix}
9\-8
end{bmatrix}
$$
so $$x=A^{-1}b$$
with
$$
A^{-1}=frac{1}{5}begin{bmatrix}
2&1\1&-2
end{bmatrix}
$$
answered Apr 24 '16 at 17:11
Emilio NovatiEmilio Novati
51.6k43473
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
In matrix notation we have:
$$
Ax=b
$$
with
$$
A=begin{bmatrix}
2&1\1&-2
end{bmatrix}
qquad x=begin{bmatrix}
x\y
end{bmatrix}qquad
b=begin{bmatrix}
9\-8
end{bmatrix}
$$
so $$x=A^{-1}b$$
with
$$
A^{-1}=frac{1}{5}begin{bmatrix}
2&1\1&-2
end{bmatrix}
$$
answered Apr 24 '16 at 17:11
Emilio NovatiEmilio Novati
51.6k43473
$endgroup$
add a comment |
$begingroup$
Hint:
In matrix notation we have:
$$
Ax=b
$$
with
$$
A=begin{bmatrix}
2&1\1&-2
end{bmatrix}
qquad x=begin{bmatrix}
x\y
end{bmatrix}qquad
b=begin{bmatrix}
9\-8
end{bmatrix}
$$
so $$x=A^{-1}b$$
with
$$
A^{-1}=frac{1}{5}begin{bmatrix}
2&1\1&-2
end{bmatrix}
$$
answered Apr 24 '16 at 17:11
Emilio NovatiEmilio Novati
51.6k43473
$endgroup$
add a comment |
$begingroup$
Hint:
In matrix notation we have:
$$
Ax=b
$$
with
$$
A=begin{bmatrix}
2&1\1&-2
end{bmatrix}
qquad x=begin{bmatrix}
x\y
end{bmatrix}qquad
b=begin{bmatrix}
9\-8
end{bmatrix}
$$
so $$x=A^{-1}b$$
with
$$
A^{-1}=frac{1}{5}begin{bmatrix}
2&1\1&-2
end{bmatrix}
$$
answered Apr 24 '16 at 17:11
Emilio NovatiEmilio Novati
51.6k43473
$endgroup$
Hint:
In matrix notation we have:
$$
Ax=b
$$
with
$$
A=begin{bmatrix}
2&1\1&-2
end{bmatrix}
qquad x=begin{bmatrix}
x\y
end{bmatrix}qquad
b=begin{bmatrix}
9\-8
end{bmatrix}
$$
so $$x=A^{-1}b$$
with
$$
A^{-1}=frac{1}{5}begin{bmatrix}
2&1\1&-2
end{bmatrix}
$$
answered Apr 24 '16 at 17:11
Emilio NovatiEmilio Novati
51.6k43473
answered Apr 24 '16 at 17:11
Emilio NovatiEmilio Novati
51.6k43473
answered Apr 24 '16 at 17:11
Emilio NovatiEmilio Novati
51.6k43473
answered Apr 24 '16 at 17:11
Emilio NovatiEmilio Novati
51.6k43473
51.6k43473
add a comment |
add a comment |
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$begingroup$
your approach is right , just look at the value of inverse , whether it is right , or whether you wrote the matrix wrong
$endgroup$
– avz2611
Apr 24 '16 at 17:04