Question

Can someone please help me to solve this sum? ​

Answer 1

$\huge \underline \bold \pink{Solution : }$

Given that :

A = [ 1 2 ]

[ 1 3 ]

As we know that : For 2×2 matrix the value of I

=> I = [ 1 0 ]

[ 0 1 ]

Also, Given that :

AB = I

Let, B = [ a b ]

[ c d ]

AB = [ 1 2 ] [ a b ]

[ 1 3 ] [ c d ]

=> AB = [ a+2c b+2d ]

[ a+3c b+3d ]

According to question :

[ a+2c b+2d ] = [ 1 0 ]

[ a+3c b+3d ] [ 0 1 ]

On comparing :

a+2c = 1 ... (1)

b+2d = 0 ... (2)

a+3c = 0 ... (3)

b+3d = 1 ... (4)

From equation (3) :

a = -3c

Now, from equation (1) :

-3c + 2c = 1

=> -c = 1 => c = -1

=> a = -3c = -3×(-1) = 3

Again, From equation (2) :

b = -2d

Now, from equation (4) :

-2d + 3d = 1

=> d = 1

=> b = -2d => -2×1 = -2

Hence, B = [ 3 -2 ]

[ -1 1 ]

___________________________

$\huge \fbox \green{Hope it helps ☺}$

Answer 2

$\huge \underline \bold \pink{Solution : } [tex] \mid \orange{On \: comparing :}$

a+2c = 1 ... (1)

b+2d = 0 ... (2)

a+3c = 0 ... (3)

b+3d = 1 ... (4)

$\mid \blue{From \: equation \: (3) :}$

a = -3c

$\mid \pink{Now, from \: equation (1) :}$

-3c + 2c = 1

=> -c = 1 => c = -1

=> a = -3c = -3×(-1) = 3

$\mid \purple{Again, From \: equation (2) :}$

b = -2d

Now, from equation (4) :

-2d + 3d = 1

=> d = 1

=> b = -2d => -2×1 = -2

Hence, B = [ 3 -2 ]

[ -1 1 ]

___________________________

$\huge \fbox \green{Hope it helps ☺}$