Question

Using matrices, solve the following system of equations :
3x +2y+z=10,
4x+y+3z =15,
x+y+z=6.
ed by the system o
cu​

Answer 1

Given:

3x+2y+z=10,

4x+y+3z=15,

x+y+z=6

To find:

Find the values of x, y, and z.

Solution:

$A = \left[\begin{array}{ccc}3&2&1\\4&1&3\\1&1&1\end{array}\right]$

$X = \left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

$B = \left[\begin{array}{ccc}10\\15\\6\end{array}\right]$

AX = B

A⁻¹AX = A⁻¹B

IX = A⁻¹B

X = A⁻¹B

To find inverse of A use AA⁻¹ = I

So,

$A^{-1} = \left[\begin{array}{ccc}2/5&1/5&-1\\1/5&-2/5&1\\-3/5&1/5&1\end{array}\right]$

X = A⁻¹B

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}2/5&1/5&-1\\1/5&-2/5&1\\-3/5&1/5&1\end{array}\right]\left[\begin{array}{ccc}10\\15\\6\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}2/5*10+1/5*15+(-1)*6\\1/5*10+(-2/5)*15+1*6\\(-3/5)*10+1/5*15+1*6\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}1\\2\\3\end{array}\right]$

So,

x = 1

y = 2

z = 3

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