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Matrices

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111.

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.

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112. Find a 2 X 2 matrix B such that

open square brackets table row 2 5 row cell negative 3 end cell 7 end table close square brackets space space straight B equals open square brackets table row 17 cell negative 1 end cell row 47 cell negative 13 end cell end table close square brackets

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113. Without using'the concept of inverse of a matrix, find the matrix $open square brackets table row straight x straight y row straight z straight u end table close square brackets$ such that $open square brackets table row 5 cell negative 7 end cell row cell negative 2 end cell 3 end table close square brackets space open square brackets table row x y row z u end table close square brackets equals space open square brackets table row cell negative 16 end cell cell negative 6 end cell row 7 2 end table close square brackets$

we space have space open square brackets table row 5 cell negative 7 end cell row cell negative 2 end cell 3 end table close square brackets space open square brackets table row straight x straight y row straight z straight u end table close square brackets equals open square brackets table row cell negative 16 end cell cell negative 6 end cell row 7 2 end table close square brackets or space open square brackets table row cell 5 straight x minus 7 straight z end cell cell 5 straight y minus 7 straight u end cell row cell negative 2 straight x plus 3 straight z end cell cell negative 2 straight y plus 3 straight u end cell end table close square brackets equals space open square brackets table row cell negative 16 end cell cell negative 6 end cell row 7 2 end table close square brackets

by definition of equality of matrices,
5 x – 7 z = – 16        ...(1)
–2x + 3z = 7    ...(2)
5y–7u = -6    .... ..(3)
–2y’ + 3u = 2    ...(4)
Equations (1) and (2) can be written as
5 x – 7 z + 16 = 0
–2x – 3z + 7 = 0 $therefore fraction numerator straight x over denominator negative 49 plus 48 end fraction equals fraction numerator straight u over denominator 12 minus 10 end fraction equals fraction numerator 1 over denominator negative 15 plus 14 end fraction space space space space rightwards double arrow space space space space space space straight y over 4 equals straight u over 2 equals fraction numerator 1 over denominator negative 1 end fraction therefore space space straight x equals 1 comma straight z equals 3 Equations space left parenthesis 5 right parenthesis space and space left parenthesis 4 right parenthesis space can space be space written space as space space 5 y – space 7 u space plus space 6 space equals space 0 comma space 2 y space – space 3 space u space plus space 2 space equals space 0 therefore fraction numerator straight x over denominator negative 49 plus 48 end fraction equals fraction numerator straight u over denominator 12 minus 10 end fraction equals fraction numerator 1 over denominator negative 15 plus 14 end fraction space space space space rightwards double arrow space space space space space space straight y over 4 equals straight u over 2 equals fraction numerator 1 over denominator negative 1 end fraction therefore space space space space space space space straight y equals negative 4 comma space straight u equals negative 2 therefore space open square brackets table row straight x straight y row straight z straight u end table close square brackets equals open square brackets table row 1 cell negative 4 end cell row 3 cell negative 2 end cell end table close square brackets$