Answer on Question #84623 – Math – Linear Algebra
Question
Check whether the following system of equations has a solution:
4 x + 2 y + 8 z + 6 z = 3 , 2 x + 2 y + 2 z + 2 w = 1 , x + 3 z + 2 w = 3 ? 4x + 2y + 8z + 6z = 3, 2x + 2y + 2z + 2w = 1, x + 3z + 2w = 3? 4 x + 2 y + 8 z + 6 z = 3 , 2 x + 2 y + 2 z + 2 w = 1 , x + 3 z + 2 w = 3 ? Solution
Find ( x , y , z , w ) (x, y, z, w) ( x , y , z , w )
{ x + 3 z + 2 w = 3 , 2 x + 2 y + 2 z + 2 w = 1 , 4 x + 2 y + 8 z + 6 z = 3. \left\{
\begin{array}{l}
x + 3z + 2w = 3, \\
2x + 2y + 2z + 2w = 1, \\
4x + 2y + 8z + 6z = 3.
\end{array}
\right. ⎩ ⎨ ⎧ x + 3 z + 2 w = 3 , 2 x + 2 y + 2 z + 2 w = 1 , 4 x + 2 y + 8 z + 6 z = 3.
Construct the following matrix and reducing it to a triangular form
[ 1 0 3 2 3 2 2 2 2 1 4 2 14 0 3 ] 2 row + 1 row × ( − 2 ) , 3 row + 1 row × ( − 4 ) → [ 1 0 3 2 3 0 2 − 4 − 2 − 5 0 2 2 − 8 − 9 ] 3 row + 2 row × ( − 1 ) → [ 1 0 3 2 3 0 2 − 4 − 2 − 5 0 0 6 − 6 − 4 ] . \begin{array}{l}
\left[
\begin{array}{cccc}
1 & 0 & 3 & 2 & 3 \\
2 & 2 & 2 & 2 & 1 \\
4 & 2 & 14 & 0 & 3
\end{array}
\right] 2 \text{ row} + 1 \text{ row} \times (-2), 3 \text{ row} + 1 \text{ row} \times (-4) \\
\rightarrow \left[
\begin{array}{cccc}
1 & 0 & 3 & 2 & 3 \\
0 & 2 & -4 & -2 & -5 \\
0 & 2 & 2 & -8 & -9
\end{array}
\right] 3 \text{ row} + 2 \text{ row} \times (-1) \rightarrow \left[
\begin{array}{cccc}
1 & 0 & 3 & 2 & 3 \\
0 & 2 & -4 & -2 & -5 \\
0 & 0 & 6 & -6 & -4
\end{array}
\right].
\end{array} ⎣ ⎡ 1 2 4 0 2 2 3 2 14 2 2 0 3 1 3 ⎦ ⎤ 2 row + 1 row × ( − 2 ) , 3 row + 1 row × ( − 4 ) → ⎣ ⎡ 1 0 0 0 2 2 3 − 4 2 2 − 2 − 8 3 − 5 − 9 ⎦ ⎤ 3 row + 2 row × ( − 1 ) → ⎣ ⎡ 1 0 0 0 2 0 3 − 4 6 2 − 2 − 6 3 − 5 − 4 ⎦ ⎤ .
The system that corresponds to the last matrix has the form
x + 3 z + 2 w = 3 2 y − 4 z − 2 w = − 5 6 z − 6 w = − 4. } ⇒ x = 3 − 3 z − 2 w y = − 5 + 4 z + 2 w 2 z = 3 w − 2 3 w = w } ⇒ x = 5 − 5 w y = 18 w − 23 6 z = 3 w − 2 3 w = w } \left.\begin{array}{l}
x + 3z + 2w = 3 \\
2y - 4z - 2w = -5 \\
6z - 6w = -4.
\end{array}
\right\}
\Rightarrow
\left.\begin{array}{l}
x = 3 - 3z - 2w \\
y = \dfrac{-5 + 4z + 2w}{2} \\
z = \dfrac{3w - 2}{3} \\
w = w
\end{array}
\right\}
\Rightarrow
\left.\begin{array}{l}
x = 5 - 5w \\
y = \dfrac{18w - 23}{6} \\
z = \dfrac{3w - 2}{3} \\
w = w
\end{array}
\right\} x + 3 z + 2 w = 3 2 y − 4 z − 2 w = − 5 6 z − 6 w = − 4. ⎭ ⎬ ⎫ ⇒ x = 3 − 3 z − 2 w y = 2 − 5 + 4 z + 2 w z = 3 3 w − 2 w = w ⎭ ⎬ ⎫ ⇒ x = 5 − 5 w y = 6 18 w − 23 z = 3 3 w − 2 w = w ⎭ ⎬ ⎫
Answer: yes, x = 5 − 5 w x = 5 - 5w x = 5 − 5 w y = 18 w − 23 6 y = \dfrac{18w - 23}{6} y = 6 18 w − 23 z = 3 w − 2 3 z = \dfrac{3w - 2}{3} z = 3 3 w − 2 w = w w = w w = w w ∈ R w \in \mathbb{R} w ∈ R
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#339914 on Dec 2023