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Matrices

Question
CBSEENMA12032973

Given 3 open square brackets table row straight x straight y row straight z straight w end table close square brackets space equals space open square brackets table row straight x 6 row cell negative 1 end cell cell 2 space straight w end cell end table close square brackets plus open square brackets table row 4 cell straight x plus straight y end cell row cell straight z plus straight w end cell 3 end table close square bracketsfind the values of .x, y, z and w.

Solution
The given matrix equation is
3 space open square brackets table row straight x straight y row straight z straight w end table close square brackets equals open square brackets table row straight x 6 row cell negative 1 end cell cell 2 straight w end cell end table close square brackets plus open square brackets table row 4 cell straight x plus straight y end cell row cell straight z plus straight w end cell 3 end table close square brackets
or space open square brackets table row cell 3 straight x end cell cell 3 straight y end cell row cell 3 straight z end cell cell 3 straight w end cell end table close square brackets space equals space open square brackets table row cell straight x plus 4 end cell cell 6 plus straight x plus straight y end cell row cell negative 1 plus straight z plus straight w end cell cell 2 space straight w plus 3 end cell end table close square brackets

By definition of equality of matrices, we get,
3 x = .x + 4    ...(1)
3y = 6+x+ y    -(2)
3: = – I +z + w    ’ ...(3)
3 w’ = 2 w + 3    ....(4)
From (1), 3 .x – .x = 4 or 2 .x = 4 ⇒ x = 2
From (4), 3 w – 2 w = 3 or w = 3
Putting x = 2 in (2), we get,
3 y = 6 + 2 + y or 2 y = 8 ⇒ y = 4
Putting w = 3 in (3), we get,
3z= – l + z + 3 or 2 : = 2 ⇒ z  = l
∴ we have
x = 2, y = 4, z = 1, w = 3.