Question

[tex] Obtain a,b,c,d if \left[\begin{array}{ccc}a-ab&c+d\\2a-b&3a-c\end{array}\right]=\left[\begin{array}{ccc}2&0\\7&10\end{array}\right] [\tex]

Answer 1

Answer:

$a=\frac{1}{2-\sqrt{3}}and\frac{1}{2+\sqrt{3}}$

$b=-3+2\sqrt{3}and-3-2\sqrt{3}$

$c=\frac{3}{2-\sqrt{3}}-10and\frac{3}{2+\sqrt{3}}-10$

$d=\10-\frac{3}{2-\sqrt{3}}and10-\frac{3}{2+\sqrt{3}}$

Step-by-step explanation:

$\left[\begin{array}{ccc}a-ab&c+d\\2a-b&3a-c\\\end{array}\right]=\left[\begin{array}{ccc}2&0\\7&10\\\end{array}\right]$

$\textrm{equating both the sides we get}$

a-ab=2;c+d=0;2a-b=7;3a-c=10

a-ab=2,a(1-b)=2 implies $a=\frac{2}{1-b}$

c+d=0 imples c=-d

$\textrm{putting the values of a in 2a-b=7 we get the quadratic equation}$

$b^{2}+6b-3=0$

$\textrm{using sridhanacharya formula we get}$

$b=-3+2\sqrt{3}and-3-2\sqrt{3}$

$\textrm{putting the value of b in 2a-b=7 to get the value of a}$

therefore $a=\frac{1}{2-\sqrt{3}}and\frac{1}{2+\sqrt{3}}$

$\textrm{putting the value of a in 3a-c=10 we get}$

$c=\frac{3}{2-sqrt{3}}-10and\frac{3}{2+\sqrt{3}}-10$

$\textrm{putting thr value of c n c+d=0 we get}$

$d=10-\frac{3}{2-\sqrt{3}}and10-\frac{3}{2+\sqrt{3}}$