Question

solve the following system of linear equations by adjoint method 5x + y = 8 2x + 3y = 11​

Answer 1

Answer:

Explanation:given equation

2x+3y=11. (2)

5x+y=8

Now 5x+y=8

y=8-5x. (3)

from second equation

2x+3y=11

2x+3(8-5x)=11

2x+24-15x=11

24-11=13x

13=13x

x=1

from equation (3)

y=8-5x

=8-5(1)

=8-5

=3

Hence x=1. and y= 3

Answer 2

Answer:

x = 1 and y = 3

Explanation:

Given equation, 5x + y = 8 and 2x + 3y = 11.

To find the given solution of the linear equations by adjoint method.

So,

AX = B

where,

 $A=\left[\begin{array}{cc}5&1\\2&3\\\end{array}\right]$

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

$B=\left[\begin{array}{ccc}8\\11\\\end{array}\right]$

here, X = A⁻¹B

    and A⁻¹ = $\frac{adjA}{\left[\begin{array}{}A\\\\\end{array}\right] }$

adj A = The adjoint of a square matrix A is transpose of the matrix which obtained by cofactors of each element of a determinant corresponding to the given matrix. It is denoted by adj(A).

By solving we get adj(A) = $\left[\begin{array}{ccc}3&-1\\-2&5\\\end{array}\right]$

    and   I A I = 13.

                             A⁻¹ =   $\frac{1}{13}\left[\begin{array}{ccc}3&-1\\-2&5\\\end{array}\right]$

 X =  A⁻¹B

 X =    $\frac{1}{13}\left[\begin{array}{ccc}3&-1\\-2&5\\\end{array}\right]$ $\left[\begin{array}{ccc}8\\11\\\end{array}\right]$

 

 X =    $\frac{1}{13} \left[\begin{array}{ccc}24-11\\-16+55\\\end{array}\right]$

X = $\frac{1}{13}\left[\begin{array}{ccc}13\\39\\\end{array}\right]$

X = $\left[\begin{array}{ccc}1\\3\\\end{array}\right]$

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

Therefore, from the above equation it is clear that

x = 1 and y = 3.

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