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If $straight A space equals open square brackets table row 2 cell space space space space space space space 3 end cell row 5 cell space space space minus 2 end cell end table close square brackets$ , show that $straight A to the power of negative 1 end exponent space equals space 1 over 19 straight A$ .

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172.

Let A be the matrix $open square brackets table row 3 cell space space 8 end cell row 2 cell space space 1 end cell end table close square brackets.$ Find A ^–1 and verify that $straight A to the power of negative 1 end exponent space equals space 1 over 13 straight A space minus space 4 over 13 straight I$
where I is 2 × 2 unit matrix.

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173.

If A = $open square brackets table row 2 cell space space space 5 end cell row 1 cell space space space 6 end cell end table close square brackets comma$ find $straight A to the power of negative 1 end exponent$ and verify that $straight A to the power of negative 1 end exponent space equals space minus 1 over 7 straight A plus 8 over 7 straight I$

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174.
Given $straight A space equals space open square brackets table row 2 cell space space space space space minus 3 end cell row cell negative 4 end cell cell space space space space space space space 7 end cell end table close square brackets comma$ compute A^–1 and show that 2A^–1 = 9 I – A.

Here, $straight A space equals space open square brackets table row cell space space space space 2 end cell cell space space space space minus 3 end cell row cell negative 4 end cell cell space space space space space space space space 7 end cell end table close square brackets$

$therefore space space space space space open vertical bar straight A close vertical bar space equals open vertical bar table row cell space space space 2 end cell cell space space space space minus 3 end cell row cell negative 4 end cell cell space space space space space space 7 end cell end table close vertical bar space equals space 14 minus 12 space equals space 2 space not equal to space 0 space space space space space rightwards double arrow space space straight A to the power of negative 1 end exponent space exists$
Co-factors of the elements of first row of | A | are 7, 4 respectively
Co-factors of the elements of second row of | A | are 3, 2 respectively
$therefore space space space space adj. space space straight A space equals space open square brackets table row 7 cell space space space space 4 end cell row 3 cell space space space space 2 end cell end table close square brackets to the power of apostrophe space space equals space open square brackets table row 7 cell space space space 3 space end cell row 4 cell space space space 2 end cell end table close square brackets therefore 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 1 half open square brackets table row 7 cell space space space space 3 end cell row 4 cell space space space space 2 end cell 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 rightwards double arrow space space space 2 straight A to the power of negative 1 end exponent open square brackets table row 7 cell space space space space 3 end cell row 4 cell space space space space space 2 end cell end table close square brackets Also space 9 straight I space minus space straight A space equals space 9 open square brackets table row 1 cell space space space 0 end cell row 0 cell space space space space 1 end cell end table close square brackets space minus space open square brackets table row cell space space space 2 end cell cell space space space space space minus 3 end cell row cell negative 4 end cell cell space space space space space space space 7 end cell 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 equals open square brackets table row 9 cell space space space 0 end cell row 0 cell space space space 9 end cell end table close square brackets space minus space open square brackets table row 2 cell space space space space space space space space minus 3 end cell row cell negative 4 end cell cell space space space space space space space space space space space 7 end cell end table close square brackets space equals space open square brackets table row 7 cell space space space space 3 end cell row 4 cell space space space space 2 end cell end table close square brackets therefore space space space space 2 straight A to the power of negative 1 end exponent space equals space 9 straight I space minus straight A.$

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175.

Find the inverse of $straight A space equals space open square brackets table row 3 cell space space space space space space space 5 end cell row 7 cell space space minus 11 end cell end table close square brackets$ and verify that AA^–1 = I.

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176.

For what values of x, is the following matrix singular?
$open square brackets table row cell 3 minus 2 straight x end cell cell space space space space straight x plus 1 end cell row 2 cell space space space space 4 end cell end table close square brackets$

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177. For what value of ‘a’ is the matrix

open square brackets table row 4 cell space minus 3 end cell cell space space space space space minus 1 end cell row 2 cell space space space space space straight a end cell cell space space space space space space space space 6 end cell row 3 cell space space minus 5 end cell cell space space space minus 4 end cell end table close square brackets

singular?

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Long Answer Type

178.

If $straight A space equals space open square brackets table row 3 cell space space space space space 1 end cell row 7 cell space space space space space 5 end cell end table close square brackets comma$ find x and y find that A² + xI = y A. Hence find A^–1

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179.

If $straight A space equals space open square brackets table row 6 cell space space space space 5 end cell row 7 cell space space space space 6 end cell end table close square brackets comma$ show that A² – 12A + I = O. Hence find A ^–1.

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Short Answer Type

180.

If $straight A space equals space open square brackets table row 2 cell space space space 7 end cell row 1 cell space space space 4 end cell end table close square brackets comma space space$ show that $straight A squared minus 6 straight A space plus space straight I space equals space straight O comma space hence space find space straight A to the power of negative 1 end exponent.$

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174. Given compute A–1 and show that 2A–1 = 9 I – A.

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