[tex]{\huge{\boxed{\sf{\green{Answer}}}}}[/tex] Given: 3x+2y+z=10, 4x+y+3z=15, x+y+z=6 To find: Find the values of x, y, and z. Solution: [tex]\begin{gathered}A = \left[\begin{array}{ccc}3&2&1\\4&1&3\\1&1&1\end{array}\right]\end{gathered} A= ⎣⎢⎡ 341 211 131 ⎦⎥⎤ \begin{gathered}X = \left[\begin{array}{ccc}x\\y\\z\end{array}\right]\end{gathered} X= ⎣⎢⎡ xyz ⎦⎥⎤[/tex] [tex]\begin{gathered}B = \left[\begin{array}{ccc}10\\15\\6\end{array}\right]\end{gathered} B= ⎣⎢⎡ 10156 ⎦⎥⎤ [/tex] AX = B A⁻¹AX = A⁻¹B IX = A⁻¹B X = A⁻¹B To find inverse of A use AA⁻¹ = I So, [tex]\begin{gathered}A^{-1} = \left[\begin{array}{ccc}2/5&1/5&-1\\1/5&-2/5&1\\-3/5&1/5&1\end{array}\right]\end{gathered} A −1 = ⎣⎢⎡ 2/51/5−3/5 1/5−2/51/5 −111 ⎦⎥⎤ X = A⁻¹B[/tex] [tex]\begin{gathered}\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}2/5&1/5&-1\\1/5&-2/5&1\\-3/5&1/5&1\end{array}\right]\left[\begin{array}{ccc}10\\15\\6\end{array}\right]\end{gathered} ⎣⎢⎡ xyz ⎦⎥⎤ = ⎣⎢⎡ 2/51/5−3/5 1/5−2/51/5 −111 ⎦⎥⎤ ⎣⎢⎡ 10156 ⎦⎥⎤[/tex] [tex]\begin{gathered}\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}2/5*10+1/5*15+(-1)*6\\1/5*10+(-2/5)*15+1*6\\(-3/5)*10+1/5*15+1*6\end{array}\right]\end{gathered} ⎣⎢⎡ xyz ⎦⎥⎤ = ⎣⎢⎡ 2/5∗10+1/5∗15+(−1)∗61/5∗10+(−2/5)∗15+1∗6(−3/5)∗10+1/5∗15+1∗6 ⎦⎥⎤[/tex] [tex]\begin{gathered}\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}1\\2\\3\end{array}\right]\end{gathered} ⎣⎢⎡ xyz ⎦⎥⎤ = ⎣⎢⎡ 123 ⎦⎥⎤[/tex] So, x = 1 y = 2 z = 3 Therefore, the value of x = 1, y = 2, and z = 3 Reply
[tex]{\huge{\boxed{\sf{\green{Answer}}}}}[/tex]
Given:
3x+2y+z=10,
4x+y+3z=15,
x+y+z=6
To find:
Find the values of x, y, and z.
Solution:
[tex]\begin{gathered}A = \left[\begin{array}{ccc}3&2&1\\4&1&3\\1&1&1\end{array}\right]\end{gathered} A= ⎣⎢⎡ 341 211 131 ⎦⎥⎤ \begin{gathered}X = \left[\begin{array}{ccc}x\\y\\z\end{array}\right]\end{gathered} X= ⎣⎢⎡ xyz ⎦⎥⎤[/tex]
[tex]\begin{gathered}B = \left[\begin{array}{ccc}10\\15\\6\end{array}\right]\end{gathered} B= ⎣⎢⎡ 10156 ⎦⎥⎤ [/tex]
AX = B
A⁻¹AX = A⁻¹B
IX = A⁻¹B
X = A⁻¹B
To find inverse of A use AA⁻¹ = I
So,
[tex]\begin{gathered}A^{-1} = \left[\begin{array}{ccc}2/5&1/5&-1\\1/5&-2/5&1\\-3/5&1/5&1\end{array}\right]\end{gathered} A −1 = ⎣⎢⎡ 2/51/5−3/5 1/5−2/51/5 −111 ⎦⎥⎤ X = A⁻¹B[/tex]
[tex]\begin{gathered}\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}2/5&1/5&-1\\1/5&-2/5&1\\-3/5&1/5&1\end{array}\right]\left[\begin{array}{ccc}10\\15\\6\end{array}\right]\end{gathered} ⎣⎢⎡ xyz ⎦⎥⎤ = ⎣⎢⎡ 2/51/5−3/5 1/5−2/51/5 −111 ⎦⎥⎤ ⎣⎢⎡ 10156 ⎦⎥⎤[/tex]
[tex]\begin{gathered}\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}2/5*10+1/5*15+(-1)*6\\1/5*10+(-2/5)*15+1*6\\(-3/5)*10+1/5*15+1*6\end{array}\right]\end{gathered} ⎣⎢⎡ xyz ⎦⎥⎤ = ⎣⎢⎡ 2/5∗10+1/5∗15+(−1)∗61/5∗10+(−2/5)∗15+1∗6(−3/5)∗10+1/5∗15+1∗6 ⎦⎥⎤[/tex]
[tex]\begin{gathered}\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}1\\2\\3\end{array}\right]\end{gathered} ⎣⎢⎡ xyz ⎦⎥⎤ = ⎣⎢⎡ 123 ⎦⎥⎤[/tex]
So,
x = 1
y = 2
z = 3
Therefore, the value of x = 1, y = 2, and z = 3
Step-by-step explanation:
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