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Sagot :
Answer:
[tex]\left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}3\\2\\\end{array}\right][/tex]
Solution of the given system of equations
x = 3 and y= 2
Step-by-step explanation:
Step(i):-
Given system of equations are
4x - y = 10 ...(i)
8x +5y = 34...(ii)
The matrix form
A X = B
[tex]\left[\begin{array}{ccc}4&-1\\8&5\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}10\\34\\\end{array}\right][/tex]
Step(ii):-
By using matrix inversion method
[tex]A^{-1} = \frac{1}{|A|} adj A[/tex]
|A| = ad-b c = 4(5) - 8(-1) = 20+8 = 28
Given matrix
[tex]A = \left[\begin{array}{ccc}4&-1\\8&5\\\end{array}\right][/tex]
[tex]AdjA = \left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right][/tex]
[tex]A djA = \left[\begin{array}{ccc}5&1\\-8&4\\\end{array}\right][/tex]
Inverse of the matrix
[tex]A^{-1} = \frac{1}{|A|} adj A[/tex]
[tex]A^{-1} = \frac{1}{|A|} A djA =\frac{1}{28} \left[\begin{array}{ccc}5&1\\-8&4\\\end{array}\right][/tex]
Step(iii):-
The solution of the given system of equations by using matrix inversion method
X = A⁻¹ B
[tex]X = A^{-1}B =\frac{1}{28} \left[\begin{array}{ccc}5&1\\-8&4\\\end{array}\right]\left[\begin{array}{ccc}10\\34\\\end{array}\right][/tex]
[tex]X = A^{-1}B =\frac{1}{28} \left[\begin{array}{ccc}5X10+1X34\\-8X10+4X34\\\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}x\\y\\\end{array}\right] =\frac{1}{28} \left[\begin{array}{ccc}84\\56\\\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}3\\2\\\end{array}\right][/tex]
Final answer:-
Solution of the given system x = 3 and y= 2
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