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solve by using matrix inversion method
2x+y^3,6x+5y=1


Sagot :

Matrix inversion method can be used when the matrix A is square and non singular.

Square matrix is one which has equal number of rows and columns.

Non singular matrix is one whose determinant value is not equal to zero.

x= 14/4 and y= -16/4

2x+y=3 ---------eq. A

6x+5y=1--------eq.B

The above equations can be written in the form of matrices as follows

[tex]\left[\begin{array}{cc}2&1\\6&5\\\end{array}\right][/tex]  [tex]\left[\begin{array}{c}x\\y\\\end{array}\right][/tex] = [tex]\left[\begin{array}{c}3\\1\\\end{array}\right][/tex]

Let  A=[tex]\left[\begin{array}{cc}2&1\\6&5\\\end{array}\right][/tex]     X=  [tex]\left[\begin{array}{c}x\\y\\\end{array}\right][/tex]  and B= [tex]\left[\begin{array}{c}3\\1\\\end{array}\right][/tex]

Then

AX= B

X= A⁻¹ B

where A⁻¹ = [tex]\frac{Adj A}{mod of A}[/tex]  where A mod ≠ 0

Adj A= [tex]\left[\begin{array}{cc} 5&-1\\-6&2\\\end{array}\right][/tex]

A mod = 10-6=4

A⁻¹ = 1/4  [tex]\left[\begin{array}{cc} 5&-1\\-6&2\\\end{array}\right][/tex]

X= A⁻¹ B= 1/4  [tex]\left[\begin{array}{cc} 5&-1\\-6&2\\\end{array}\right][/tex] [tex]\left[\begin{array}{c}3\\1\\\end{array}\right][/tex]

X= 1/4  [tex]\left[\begin{array}{c} 5x3+-1x1\\-6x3+2x1\\\end{array}\right][/tex]

X= 1/4  [tex]\left[\begin{array}{c} 15-1\\-18+2\\\end{array}\right][/tex]

X= 1/4  [tex]\left[\begin{array}{c} 14\\-16\\\end{array}\right][/tex]

From the above x= 14/4 and y= -16/4

Matrix inversion method can also be understood

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