Question

$If space straight A equals space open square brackets table row 3 cell negative 4 end cell row cell negative 1 end cell cell space space space space 2 end cell end table close square brackets space find space straight a space matrix space straight B space such space that space AB space equals space straight I.$

Solution

Here space straight A equals space open square brackets table row 3 cell negative 4 end cell row cell negative 1 end cell cell space space space space 2 end cell end table close square brackets Let space space space straight B equals space open square brackets table row straight x cell space space straight y end cell row straight y cell space space space straight u end cell end table close square brackets NOW AB equals straight I space space space rightwards double arrow space space space space open square brackets table row 3 cell negative 4 end cell row cell negative 1 end cell cell space space space space 2 end cell end table close square brackets space open square brackets table row straight x cell space space straight y end cell row straight y cell space space space straight u end cell end table close square brackets equals space open square brackets table row 1 cell space space 0 end cell row 0 cell space space 1 end cell end table close square brackets

rightwards double arrow space space space space space open square brackets table row cell 3 x minus 4 z end cell cell space space 3 y minus 4 u end cell row cell negative x plus 2 z end cell cell space minus y plus 2 u end cell end table close square brackets equals open square brackets table row cell 1 space end cell cell space space 0 end cell row 0 cell space 1 end cell end table close square brackets

By definition of equality of matrices, we get,
3 x – 4z = 1    ...(1)
x + 2 z = 0    ...(2)
3 y – 4 u= 0        ...(3)
– y + 2u = l    ...(4)
Multiplying (1) by 1 and (2) by 2, we get,
3 x – 4 z = 1    ...(5)
– 2x + 4z = 0    ...(6)
Adding (5) and (6), we get,
x = 1Putting x = 1 in (5), we get,
$3 minus 4 straight z equals 1 space or space 4 straight z equals 2 space rightwards double arrow space straight z equals 1 half$
Multiplying (3) by 1 and (4) by 2, we get,
3y – 4 u = 0    ...(7)
–2y + 4 u = 2    ..(8)
Adding (7) and (8), we get,
y = 2
Putting y = 2 in (3), we get,
$6 minus 4 straight u equals 0 space space space space space space rightwards double arrow space 4 straight u equals 6 space space space space rightwards double arrow space straight u equals 3 over 2$
$therefore space space space space space space space space space space space space space space straight B equals open square brackets table row 1 2 row cell 1 half end cell cell 3 over 2 end cell end table close square brackets$

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