Question

Solve by matrix method 3x + 2y +z = 10, 4x + y + 3z = 15, x + y + z = 6.
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Answer 1

${\huge{\boxed{\sf{\green{Answer}}}}}$

Given:

3x+2y+z=10,

4x+y+3z=15,

x+y+z=6

To find:

Find the values of x, y, and z.

Solution:

$\begin{gathered}A = \left[\begin{array}{ccc}3&2&1\\4&1&3\\1&1&1\end{array}\right]\end{gathered} A= ⎣⎢⎡​ 341​ 211​ 131​ ⎦⎥⎤​ ​ \begin{gathered}X = \left[\begin{array}{ccc}x\\y\\z\end{array}\right]\end{gathered} X= ⎣⎢⎡​ xyz​ ⎦⎥⎤$

$\begin{gathered}B = \left[\begin{array}{ccc}10\\15\\6\end{array}\right]\end{gathered} B= ⎣⎢⎡​ 10156​ ⎦⎥⎤​ ​$

AX = B

A⁻¹AX = A⁻¹B

IX = A⁻¹B

X = A⁻¹B

To find inverse of A use AA⁻¹ = I

So,

$\begin{gathered}A^{-1} = \left[\begin{array}{ccc}2/5&1/5&-1\\1/5&-2/5&1\\-3/5&1/5&1\end{array}\right]\end{gathered} A −1 = ⎣⎢⎡​ 2/51/5−3/5​ 1/5−2/51/5​ −111​ ⎦⎥⎤​ ​ X = A⁻¹B$

$\begin{gathered}\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}2/5&1/5&-1\\1/5&-2/5&1\\-3/5&1/5&1\end{array}\right]\left[\begin{array}{ccc}10\\15\\6\end{array}\right]\end{gathered} ⎣⎢⎡​ xyz​ ⎦⎥⎤​ = ⎣⎢⎡​ 2/51/5−3/5​ 1/5−2/51/5​ −111​ ⎦⎥⎤​ ⎣⎢⎡​ 10156​ ⎦⎥⎤$

$\begin{gathered}\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}2/5*10+1/5*15+(-1)*6\\1/5*10+(-2/5)*15+1*6\\(-3/5)*10+1/5*15+1*6\end{array}\right]\end{gathered} ⎣⎢⎡​ xyz​ ⎦⎥⎤​ = ⎣⎢⎡​ 2/5∗10+1/5∗15+(−1)∗61/5∗10+(−2/5)∗15+1∗6(−3/5)∗10+1/5∗15+1∗6​ ⎦⎥⎤$

$\begin{gathered}\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}1\\2\\3\end{array}\right]\end{gathered} ⎣⎢⎡​ xyz​ ⎦⎥⎤​ = ⎣⎢⎡​ 123​ ⎦⎥⎤​$

So,

x = 1

y = 2

z = 3

Therefore, the value of x = 1, y = 2, and z = 3

Answer 2

Step-by-step explanation:

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