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Compute (AB)¹ where:
$straight A space equals space open square brackets table row 1 cell space space space space 1 end cell cell space space space 2 end cell row 0 cell space space space 2 end cell cell space minus 3 end cell row 3 cell space minus 2 end cell cell space space space space 4 end cell end table close square brackets space space space space space space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space space space space 2 end cell cell space space space space space 0 end cell row 0 cell space space space 3 end cell cell space space minus 1 end cell row 1 cell space space 0 end cell cell space space space space 2 end cell end table close square brackets$

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222.

If $straight A space equals space open square brackets table row 2 cell space space space 2 end cell cell space space space 1 end cell row cell negative 2 end cell cell space space space 1 end cell cell space space 2 end cell row 1 cell space minus 2 end cell cell space space 2 end cell end table close square brackets space space space and space straight B space equals space open square brackets table row 1 cell space space 3 end cell cell space space 2 end cell row 1 cell space space space 1 end cell cell space 1 end cell row 2 cell negative 3 end cell cell space 1 end cell end table close square brackets comma$ verify that (AB)^–1 = B"^–1 A^–1.

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223.

If $straight A to the power of negative 1 end exponent space equals space open square brackets table row 3 cell space space minus 1 end cell cell space space space space space 1 end cell row cell negative 15 end cell cell space space space space space 6 end cell cell space minus 5 end cell row 5 cell space minus 2 end cell cell space space space space 2 end cell end table close square brackets space space space and space straight B space equals space open square brackets table row cell space space space 1 end cell cell space space space space 2 end cell cell space space space minus 2 end cell row cell negative 1 end cell cell space space space space space 3 end cell cell space space space space space 0 end cell row cell space space 0 end cell cell space space minus 2 end cell cell space space space space space 1 end cell end table close square brackets comma$ find (AB)^–1.

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Multiple Choice Questions

224. Let A be a non-singular square matrix of order 3 × 3. Then | adj A| j is equal to

| A |
| A |²
| A |³
| A |³

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225.

If A is an invertible matrix of order 2, then det (A^–1) is equal to

det (A)
$fraction numerator 1 over denominator det space left parenthesis straight A right parenthesis end fraction$
1
1

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226. If x, y, z are non-zero real numbers, then the inverse of matrix

straight A space equals open square brackets table row straight x cell space space space space 0 end cell cell space space space 0 end cell row 0 cell space space space straight y end cell cell space space space 0 end cell row 0 cell space space space 0 end cell cell space space space straight z end cell end table close square brackets space is

$open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space 0 end cell row 0 cell space space straight y to the power of negative 1 space end exponent end cell cell space space 0 end cell row 0 0 cell space straight z to the power of negative 1 end exponent end cell end table close square brackets$
$xyz open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space space 0 end cell row 0 cell space space straight y to the power of negative 1 end exponent space end cell cell space space 0 end cell row 0 0 cell space straight z to the power of negative 1 end exponent end cell end table close square brackets$
$1 over xyz open square brackets table row straight x cell space space 0 end cell cell space space 0 end cell row 0 cell space straight y end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space straight z end cell end table close square brackets$
$1 over xyz open square brackets table row straight x cell space space 0 end cell cell space space 0 end cell row 0 cell space straight y end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space straight z end cell end table close square brackets$

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Short Answer Type

227.

Examine the consistency of the system of equations:
x + 2y = 2
2x + 3y = 3

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228.
Examine the consistency of the system of equations:
2x – y = 5
x + y = 4

The given equations are
2x – y = 5
x + y = 4
These equations can be written as
$open square brackets table row 2 cell space space minus 1 end cell row 1 cell space space space space 1 end cell end table close square brackets space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row 5 row 4 end table close square brackets$
or $AX space equals space straight B space where space straight A space equals space open square brackets table row 2 cell space space minus 1 end cell row 1 cell space space space space 1 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y end table close square brackets comma space space space straight B space equals space open square brackets table row 5 row 4 end table close square brackets space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space minus 1 end cell row 1 cell space space space space space 1 end cell end table close vertical bar space equals space 2 plus 1 space equals space 3 space not equal to space 0$
∴ A^–1 exists
∴ system of equations has a unique solution, system of equations is consistent.