Algebra 2 Quiz
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SYSTEMS OF EQUATIONS
Question 1
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Convert the system to
A
X
=
B
AX=B
A
X
=
B
(use variable order
x
,
y
,
z
x, y, z
x
,
y
,
z
):
{
5
x
−
2
y
+
z
=
0
−
x
+
3
y
+
4
z
=
10
2
x
+
y
−
3
z
=
−
6
\begin{cases} 5x-2y+z=0\\ -x+3y+4z=10\\ 2x+y-3z=-6 \end{cases}
⎩
⎨
⎧
5
x
−
2
y
+
z
=
0
−
x
+
3
y
+
4
z
=
10
2
x
+
y
−
3
z
=
−
6
A
[
5
−
2
1
−
1
3
4
2
1
−
3
]
[
x
y
z
]
=
[
0
10
−
6
]
\begin{bmatrix}5&-2&1\\-1&3&4\\2&1&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\10\\-6\end{bmatrix}
5
−
1
2
−
2
3
1
1
4
−
3
x
y
z
=
0
10
−
6
B
[
5
−
2
1
−
1
3
4
2
1
−
3
]
[
x
y
z
]
=
[
0
−
6
10
]
\begin{bmatrix}5&-2&1\\-1&3&4\\2&1&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\-6\\10\end{bmatrix}
5
−
1
2
−
2
3
1
1
4
−
3
x
y
z
=
0
−
6
10
C
[
5
−
2
1
1
3
4
2
1
−
3
]
[
x
y
z
]
=
[
0
10
−
6
]
\begin{bmatrix}5&-2&1\\1&3&4\\2&1&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\10\\-6\end{bmatrix}
5
1
2
−
2
3
1
1
4
−
3
x
y
z
=
0
10
−
6
D
[
5
−
2
1
−
1
3
4
2
1
−
3
]
[
x
y
z
]
=
[
0
10
−
6
]
\begin{bmatrix}5&-2&1\\-1&3&4\\2&1&-3\end{bmatrix}\begin{bmatrix}x&y&z\end{bmatrix}=\begin{bmatrix}0\\10\\-6\end{bmatrix}
5
−
1
2
−
2
3
1
1
4
−
3
[
x
y
z
]
=
0
10
−
6
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