Answer on Question #82899 – Math – Linear Algebra
Question
solve the following equation using matrix algebra
2 x + y − z = 11 2x + y - z = 11 2 x + y − z = 11 x − 2 y + 2 z = 2 x - 2y + 2z = 2 x − 2 y + 2 z = 2 3 x − y + 3 z = 5 3x - y + 3z = 5 3 x − y + 3 z = 5
Solution
( 2 1 − 1 1 − 2 2 3 − 1 3 ) ( x y z ) = ( 11 2 5 ) \left( \begin{array}{rrr}
2 & 1 & -1 \\
1 & -2 & 2 \\
3 & -1 & 3
\end{array} \right)
\left( \begin{array}{c}
x \\
y \\
z
\end{array} \right)
=
\left( \begin{array}{c}
11 \\
2 \\
5
\end{array} \right) ⎝ ⎛ 2 1 3 1 − 2 − 1 − 1 2 3 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 11 2 5 ⎠ ⎞ ( x y z ) = ( 2 1 − 1 1 − 2 2 3 − 1 3 ) − 1 ( 11 2 5 ) \left( \begin{array}{c}
x \\
y \\
z
\end{array} \right)
=
\left( \begin{array}{rrr}
2 & 1 & -1 \\
1 & -2 & 2 \\
3 & -1 & 3
\end{array} \right)^{-1}
\left( \begin{array}{c}
11 \\
2 \\
5
\end{array} \right) ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 2 1 3 1 − 2 − 1 − 1 2 3 ⎠ ⎞ − 1 ⎝ ⎛ 11 2 5 ⎠ ⎞ ∣ 2 1 − 1 1 − 2 2 3 − 1 3 ∣ = 2 ⋅ ( − 2 ) ⋅ 3 + 1 ⋅ 2 ⋅ 3 + ( − 1 ) ⋅ 1 ⋅ ( − 1 ) − ( − 1 ) ⋅ ( − 2 ) ⋅ 3 − 2 ⋅ 2 ⋅ ( − 1 ) − 1 ⋅ 1 ⋅ 3 = − 12 + 6 + 1 − 6 + 4 − 3 = − 10 \begin{array}{l}
\left| \begin{array}{rrr}
2 & 1 & -1 \\
1 & -2 & 2 \\
3 & -1 & 3
\end{array} \right| = 2 \cdot (-2) \cdot 3 + 1 \cdot 2 \cdot 3 + (-1) \cdot 1 \cdot (-1) - (-1) \cdot (-2) \cdot 3 - 2 \cdot 2 \cdot (-1) - 1 \cdot 1 \cdot 3 \\
= -12 + 6 + 1 - 6 + 4 - 3 = -10
\end{array} ∣ ∣ 2 1 3 1 − 2 − 1 − 1 2 3 ∣ ∣ = 2 ⋅ ( − 2 ) ⋅ 3 + 1 ⋅ 2 ⋅ 3 + ( − 1 ) ⋅ 1 ⋅ ( − 1 ) − ( − 1 ) ⋅ ( − 2 ) ⋅ 3 − 2 ⋅ 2 ⋅ ( − 1 ) − 1 ⋅ 1 ⋅ 3 = − 12 + 6 + 1 − 6 + 4 − 3 = − 10 \begin{array}{l}
\left( \begin{array}{rrr}
2 & 1 & -1 \\
1 & -2 & 2 \\
3 & -1 & 3
\end{array} \right)^{-1}
=
\frac{1}{\left| \begin{array}{rrr}
2 & 1 & -1 \\
1 & -2 & 2 \\
3 & -1 & 3
\end{array} \right|}
\left( \begin{array}{rrr}
\left| \begin{array}{rrr}
-2 & 2 & -1 \cdot \left| \begin{array}{rrr}
1 & -1 & 1 \\
-1 & 3 & 3 \\
-1 \cdot \left| \begin{array}{rr}
1 & 2 & -1 \\
3 & 3 & 3 \\
\end{array} \right| \\
\left| \begin{array}{rrr}
1 & -2 & -1 \cdot \left| \begin{array}{rrr}
2 & 1 & 2 \\
3 & -1 & 3 \\
\end{array} \right| \\
\end{array} \right|}
\left| \begin{array}{rrr}
1 & -1 & -1 \\
1 & -2 & 1 \\
3 & -1 & 3
\end{array} \right| \\
= \frac{1}{-10}
\left( \begin{array}{rrr}
-4 & -2 & 0 \\
3 & 9 & -5 \\
5 & 5 & -5
\end{array} \right)
=
\left( \begin{array}{rrr}
2/5 & 1/5 & 0 \\
-3/10 & -9/10 & 1/2 \\
-1/2 & -1/2 & 1/2
\end{array} \right) \\
\end{array} ( x y z ) = ( 2 1 − 1 1 − 2 2 3 − 1 3 ) − 1 ( 11 2 5 ) = ( 2 / 5 1 / 5 0 − 3 / 10 − 9 / 10 1 / 2 − 1 / 2 − 1 / 2 1 / 2 ) ( 11 2 5 ) = ( 11 ⋅ 2 / 5 + 2 ⋅ 1 / 5 + 5 ⋅ 0 11 ⋅ ( − 3 / 10 ) + 2 ⋅ ( − 9 / 10 ) + 5 ⋅ 1 / 2 11 ⋅ ( − 1 / 2 ) + 2 ⋅ ( − 1 / 2 ) + 5 ⋅ 1 / 2 ) = ( 24 / 5 − 13 / 5 − 4 ) \begin{array}{l}
\left( \begin{array}{c}
x \\
y \\
z \\
\end{array} \right)
=
\left( \begin{array}{rrr}
2 & 1 & -1 \\
1 & -2 & 2 \\
3 & -1 & 3
\end{array} \right)^{-1}
\left( \begin{array}{c}
11 \\
2 \\
5
\end{array} \right)
=
\left( \begin{array}{rrr}
2/5 & 1/5 & 0 \\
-3/10 & -9/10 & 1/2 \\
-1/2 & -1/2 & 1/2
\end{array} \right)
\left( \begin{array}{c}
11 \\
2 \\
5
\end{array} \right) \\
= \left( \begin{array}{c}
11 \cdot 2/5 + 2 \cdot 1/5 + 5 \cdot 0 \\
11 \cdot (-3/10) + 2 \cdot (-9/10) + 5 \cdot 1/2 \\
11 \cdot (-1/2) + 2 \cdot (-1/2) + 5 \cdot 1/2
\end{array} \right)
=
\left( \begin{array}{c}
24/5 \\
-13/5 \\
-4
\end{array} \right)
\end{array} ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 2 1 3 1 − 2 − 1 − 1 2 3 ⎠ ⎞ − 1 ⎝ ⎛ 11 2 5 ⎠ ⎞ = ⎝ ⎛ 2/5 − 3/10 − 1/2 1/5 − 9/10 − 1/2 0 1/2 1/2 ⎠ ⎞ ⎝ ⎛ 11 2 5 ⎠ ⎞ = ⎝ ⎛ 11 ⋅ 2/5 + 2 ⋅ 1/5 + 5 ⋅ 0 11 ⋅ ( − 3/10 ) + 2 ⋅ ( − 9/10 ) + 5 ⋅ 1/2 11 ⋅ ( − 1/2 ) + 2 ⋅ ( − 1/2 ) + 5 ⋅ 1/2 ⎠ ⎞ = ⎝ ⎛ 24/5 − 13/5 − 4 ⎠ ⎞
Answer: x = 4.8 , y = − 2.6 , z = − 4 x = 4.8, y = -2.6, z = -4 x = 4.8 , y = − 2.6 , z = − 4
Answer provided by https://www.AssignmentExpert.com
#340153 on Dec 2023