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Let A = $open square brackets table row 4 cell space space space minus 6 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space minus 1 end cell cell space space 1 end cell row cell negative 4 end cell cell space space space 11 end cell cell negative 1 end cell end table close square brackets.$
Show that A is inevitable. Find adj. A and A Also verify that AA^–1 = A^–1 A = I.

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212.

Verify AA^–1 = A^–1 A = I where
$straight A space equals open square brackets table row 0 cell space space space 0 end cell cell space space minus 1 end cell row 3 cell space space space space 4 end cell cell space space space space space 5 end cell row cell negative 2 end cell cell negative 4 end cell cell space space minus 7 end cell end table close square brackets.$

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Short Answer Type

213.

If $straight A space equals space open square brackets table row 3 cell space space minus 3 end cell cell space space space 4 end cell row 2 cell space space minus 3 end cell cell space space 4 end cell row 0 cell space minus 1 end cell cell space space 1 end cell end table close square brackets.$ then A³ = A^–1What is adj. A ?

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Long Answer Type

214.

If $straight A space equals space open square brackets table row 1 cell space space minus 1 end cell cell space space 1 end cell row 2 cell space minus 1 end cell cell space space 0 end cell row 1 cell space space space space 0 end cell cell space space 0 end cell end table close square brackets comma$ find A^–1 and show that A^–1 = A².

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215.

If $straight A space equals space open square brackets table row 3 cell space space minus 3 end cell cell space space space 4 end cell row 2 cell space space minus 3 end cell cell space space space 4 end cell row 0 cell space space minus 1 end cell cell space space space 1 end cell end table close square brackets comma$ show that A⁴ = I. Hence find A^–1

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216.

If A = $open square brackets table row 2 cell space space 0 end cell cell space space minus 1 end cell row 5 cell space space 1 end cell cell space space space space 0 end cell row 0 cell space space 1 end cell cell space space space space 3 end cell end table close square brackets comma$ prove that A^–1 = A^–2 – 6 A + 111.

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217.
If $straight A space equals space open square brackets table row 1 cell space space 2 end cell cell space space 2 end cell row 2 cell space space 1 end cell cell space space 2 end cell row 2 cell space space 2 end cell cell space space 1 end cell end table close square brackets comma space$ Find A^–1 and hence prove that A² – 4A – 5 I = O.

straight A space equals space open square brackets table row 1 cell space space space 2 end cell cell space space space 2 end cell row 2 cell space space space 1 end cell cell space space space 2 end cell row 2 cell space space space 2 end cell cell space space 1 end cell end table close square brackets therefore space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space 2 end cell cell space space space 2 end cell row 2 cell space space space 1 end cell cell space space space 2 end cell row 2 cell space space space 2 end cell cell space space 1 end cell end table close vertical bar space space equals space 1 open vertical bar table row 1 cell space space space 2 end cell row 2 cell space space space 1 end cell end table close vertical bar minus 2 open vertical bar table row 2 cell space space space space 2 end cell row 2 cell space space space space 1 end cell end table close vertical bar plus 2 open vertical bar table row 2 cell space space space space 1 end cell row 2 cell space space space space space 2 end cell end table close vertical bar space space space space space space space space space space space space space space space space space equals 1 left parenthesis 1 minus 4 right parenthesis minus 2 left parenthesis 2 minus 4 right parenthesis plus 2 left parenthesis 4 minus 2 right parenthesis space equals space minus 3 plus 4 plus 4 space equals 5 therefore space space space space space straight A to the power of negative 1 end exponent space exists

Co-factors of the elements of the first row of | A | are

open vertical bar table row 1 cell space space 2 end cell row 2 cell space space 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space space 2 end cell row 2 cell space space space 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space 1 end cell row 2 cell space space space 2 end cell end table close vertical bar

i.e. – 3, 2, 2 respectively
Co-factors of the elements of the second row of | A | are

negative open vertical bar table row 2 cell space space space 2 end cell row 2 cell space space space 1 end cell end table close vertical bar comma space space space space open vertical bar table row 1 cell space space space 2 end cell row 2 cell space space 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space 2 end cell row 2 cell space space space 2 end cell end table close vertical bar

i.e. 2, – 3, 2 respectively.
Co-factors of the elements of the third row of | A | are

open vertical bar table row 2 cell space space 2 end cell row 1 cell space space 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space 2 end cell row 2 cell space space space 2 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space 2 end cell row 2 cell space space 1 end cell end table close vertical bar straight i. straight e. comma space space 2 comma space space 2 comma space – space 3 space respectively

therefore space space space adj. space straight A space equals space open square brackets table row cell negative 3 end cell cell space space space space space space 2 end cell cell space space space space space 2 end cell row 2 cell space space minus 3 end cell cell space space space space space 2 end cell row 2 cell space space space space space 2 end cell cell space space minus 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 3 end cell cell space space space space space space 2 end cell cell space space space space space 2 end cell row 2 cell space space minus 3 end cell cell space space space space space 2 end cell row 2 cell space space space space space 2 end cell cell space space minus 3 end cell end table close square brackets Now space straight A to the power of negative 1 end exponent space equals space fraction numerator adj. space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 fifth open square brackets table row cell negative 3 end cell cell space space 2 end cell cell space space space space 2 end cell row cell space space 2 end cell cell negative 3 end cell cell space space space space 2 end cell row cell space space 2 end cell cell space space 2 end cell cell negative 3 end cell end table close square brackets. space therefore space space space space space space space 5 space straight A to the power of negative 1 end exponent space equals space open square brackets table row cell 1 minus 4 end cell cell space space space space 2 plus 0 end cell cell space space space space space 2 plus 0 end cell row cell 2 plus 0 end cell cell space space space 1 minus 4 end cell cell space space space 2 plus 0 end cell row cell 2 plus 0 end cell cell space space space 2 plus 0 end cell cell space space space 1 minus 4 end cell end table close square brackets rightwards double arrow space space space space space space 5 space straight A to the power of negative 1 end exponent space equals space space open square brackets table row 1 cell space space space 2 end cell cell space space 2 end cell row 2 cell space space 1 end cell cell space space 2 end cell row 2 cell space space 2 end cell cell space space 1 end cell end table close square brackets plus space open square brackets table row cell negative 4 end cell cell space space space 0 end cell cell space space space space space space 0 end cell row 0 cell space minus 4 end cell cell space space space space space 0 end cell row 0 cell space space 0 end cell cell space minus 4 end cell end table close square brackets rightwards double arrow space space space space space 5 space straight A to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space 2 end cell cell space space 2 end cell row 2 cell space 1 end cell cell space space 2 end cell row 2 cell space space 2 end cell cell space 1 end cell end table close square brackets space minus space 4 open square brackets table row 1 cell space 0 end cell cell space 0 end cell row 0 cell space 1 end cell cell space 0 end cell row 0 cell space space 0 end cell cell space 1 end cell end table close square brackets rightwards double arrow space space space space 5 space straight A to the power of negative 1 end exponent space equals space straight A space minus space 4 space straight I rightwards double arrow space space space space 5 space straight A space straight A to the power of negative 1 end exponent space equals space AA space minus space 4 straight A space straight I rightwards double arrow space space space space space space space 5 space straight I space equals space straight A squared minus 4 space straight A rightwards double arrow space space space space space space straight A squared minus 4 straight A space minus space 5 space straight I space equals space 0.

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218.

For the matrix $straight A space equals space open square brackets table row 1 cell space space 1 end cell cell space space space space space 1 end cell row 1 cell space space space 2 end cell cell space minus 3 end cell row 2 cell negative 1 end cell cell space space space space 3 end cell end table close square brackets comma$ show that A³ + 6A² + 5A + 11 I = 0. Hence, find A^–1.

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219.

If $straight A space equals space open square brackets table row cell space space 2 end cell cell space space space minus 1 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell cell space minus 1 end cell row cell space space 1 end cell cell space space minus 1 end cell cell space space space space 2 end cell end table close square brackets comma$
verify that A¹ – 6A² + 9A – 4 I = O and hence find A^–1.

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220.

Compute (AB)^-1 where $straight A space equals space open square brackets table row 5 cell space space space 0 end cell cell space space 4 end cell row 2 cell space space 3 space end cell cell space 2 end cell row 1 cell space 2 end cell cell space 1 end cell end table close square brackets comma space space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space 2 end cell cell space space 3 end cell row 1 cell space space 4 end cell cell space space 3 end cell row 1 cell space space 3 end cell cell space space 4 end cell end table close square brackets$

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217. If Find A–1 and hence prove that A2 – 4A – 5 I = O.