Question

41) Solve the system of linear equations by matrix method
2x + 3y + 3z = 5
x-2y+z=-4
3x - y - 2z=3​

Answer 1

Given :

2x + 3y + 3z = 5

x-2y+z=-4

3x - y - 2z=3​

To Find : Solve for x , y & z

Solution:

2x + 3y + 3z = 5

x-2y+z=-4

3x - y - 2z=3​

$\left[\begin{array}{ccc}2&3&3\\1&-2&1\\3&-1&-2\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}5\\4\\3\end{array}\right]$

AX = Y

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

X = A⁻¹Y

| A | =  2(4 + 1)  - 3(-2 - 3)  + 3(-1 + 6)

= 10 + 15 + 15

= 40

A⁻¹ = AdjA / | A |

AdjA =

5   3  9

5 -13  1

5  11 -7

X = A⁻¹Y

  $X= \dfrac{1}{40} \left[\begin{array}{ccc}5&3&9\\5&-13&1\\5&11&-7\end{array}\right] \left[\begin{array}{ccc}5\\4\\3\end{array}\right]$

$X= \dfrac{1}{40} \left[\begin{array}{ccc}64\\-24\\48\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]= \dfrac{1}{40} \left[\begin{array}{ccc}64\\-24\\48\end{array}\right]$

x = 64/40   =  8/5

y = -24/40  =  -3/5

z = 48/40  = 6/5

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