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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.

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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 $straight A space equals space open square brackets table row 5 cell space space space 0 end cell cell space space 4 end cell row 2 cell space space 3 space end cell cell space 2 end cell row 1 cell space 2 end cell cell space 1 end cell end table close square brackets comma space space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space 2 end cell cell space space 3 end cell row 1 cell space space 4 end cell cell space space 3 end cell row 1 cell space space 3 end cell cell space space 4 end cell end table close square brackets$

Here,
$straight A space equals open square brackets table row 5 cell space space 0 end cell cell space space 4 end cell row 2 cell space space 3 end cell cell space space 2 end cell row 1 cell space space 2 end cell cell space 1 space end cell end table close square brackets 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 2 end cell cell space space 3 end cell row 1 cell space 4 end cell cell space 3 end cell row 1 cell space 3 end cell cell space 4 end cell end table close square brackets therefore space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 5 cell space space 0 end cell cell space space 4 end cell row 2 cell space 3 end cell cell space 2 end cell row 1 cell space 2 end cell cell space 1 end cell end table close vertical bar space equals space 5 open vertical bar table row 3 cell space space space 2 end cell row 2 cell space space 1 end cell end table close vertical bar comma space space minus 0 open vertical bar table row 2 cell space space space space 2 end cell row 1 cell space space space 1 end cell end table close vertical bar plus space open vertical bar table row 2 cell space space space 3 end cell row 1 cell 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 equals space 5 left parenthesis 3 minus 4 right parenthesis minus 0 left parenthesis 2 minus 2 right parenthesis plus 4 left parenthesis 4 minus 3 right parenthesis space equals space 5 left parenthesis negative 1 right parenthesis minus 0 left parenthesis 0 right parenthesis space plus space 4 left parenthesis 1 right parenthesis space space space space space space space space space space space space space space space equals negative 5 minus 0 plus 4 space equals space minus 1 not equal to 0 therefore space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row 3 cell space space space 2 end cell row 2 cell space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space 2 end cell row 1 cell space space space space 1 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space space 3 end cell row 1 cell space space space 2 end cell end table close vertical bar$
i.e. 3 – 4. – (2 – 2). 4 – 3 i.e. – 1,0, I respectively.
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row 0 cell space space 4 end cell row 2 cell space space 1 end cell end table close vertical bar comma space space open vertical bar table row 5 cell space space 4 end cell row 1 cell space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 5 cell space space space 0 end cell row 1 cell space space space 2 end cell end table close vertical bar$
i.e. – (0 – 8), 5 – 4 , – (10 – 0) i.e. 8, I, – 10 respectively.
Co-factors of the elements of second row of | A | are
$open vertical bar table row 0 cell space space space 4 end cell row 3 cell space space space 2 end cell end table close vertical bar comma space space minus open vertical bar table row 5 cell space space space 4 end cell row 2 cell space space 2 end cell end table close vertical bar comma space space space open vertical bar table row 5 cell space space space 0 end cell row 2 cell space space space 3 end cell end table close vertical bar$
i.e. 0 – 12, – (10 – 8). 15–0 i.e. – 12, 2, 15 respectively.
$therefore space space space adj. space straight A space equals space open square brackets table row cell negative 1 end cell cell space space 0 end cell cell space space 1 end cell row 8 cell space 1 end cell cell negative 10 end cell row cell negative 12 end cell cell negative 2 end cell 15 end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 1 end cell cell space space 8 end cell cell space space minus 12 end cell row cell space 0 end cell cell space 1 end cell cell space space minus 2 end cell row 1 cell negative 10 end cell cell space 15 end cell end table close square brackets therefore space space space space space 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 open square brackets table row cell negative 1 end cell cell space space 8 end cell cell space space minus 12 end cell row 0 1 cell negative 2 end cell row 1 cell negative 10 end cell cell space 15 end cell end table close square brackets space equals space open square brackets table row 1 cell space space minus 8 end cell cell space space space 12 end cell row 0 cell negative 1 end cell cell space space space 2 end cell row cell negative 1 end cell 10 cell negative 15 end cell end table close square brackets space space space space space space space space space space left parenthesis AB right parenthesis to the power of negative 1 end exponent space equals space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent space equals space minus open square brackets table row 1 cell space 2 end cell cell space 3 end cell row 1 cell space 4 end cell cell space 3 end cell row 1 cell space 3 end cell cell space 4 end cell end table close square brackets open square brackets table row 1 cell space space minus 8 end cell cell space space space 12 end cell row 0 cell space minus 1 end cell cell space space space 2 end cell row cell negative 1 end cell cell space 10 end cell cell negative 15 end cell end table close square brackets space space space space space space space space space space space space space space equals space open square brackets table row cell 1 plus 0 minus 3 end cell cell space space space space minus 8 minus 2 plus 30 end cell cell space space space space 12 plus 4 minus 45 end cell row cell 1 plus 0 minus 3 end cell cell space minus 8 minus 4 plus 30 end cell cell space 12 plus 8 minus 45 end cell row cell 1 plus 0 minus 4 end cell cell space minus 8 minus 3 plus 40 end cell cell 1 space 2 plus 6 minus 60 end cell end table close square brackets space space space space space space space space space space space space space equals space open square brackets table row cell negative 2 end cell cell space space space 20 end cell cell space space minus 29 end cell row cell negative 2 end cell cell space 18 end cell cell negative 25 end cell row cell negative 3 end cell cell space space 29 end cell cell negative 42 end cell end table close square brackets space space space space space space space space space space space$

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220. Compute (AB)-1 where