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If $straight A space equals open square brackets table row 2 cell space space space space space space space 3 end cell row 1 cell space space minus 4 end cell end table close square brackets comma space straight B space equals space open square brackets table row cell space space space 1 end cell cell space space space space minus 2 end cell row cell negative 1 end cell cell space space space space space space 3 end cell end table close square brackets comma$
verify that (AB)^–1 = B^–1 A^–1

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192.

Let $straight A space equals space open square brackets table row 3 cell space space space 7 end cell row 2 cell space space 5 end cell end table close square brackets space and space straight B space equals space open square brackets table row 6 cell space space 8 end cell row 7 cell space space 9 end cell end table close square brackets comma$ verify that (AB)^–1 = B^–1 A^–1.

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193. Verify that (AB)^–1 = B^–1 A^–1 for the matrices A and B where

straight A space equals space open square brackets table row 3 cell space space 2 end cell row 7 cell space space 5 end cell end table close square brackets comma space space straight B space equals space open square brackets table row 4 cell space space 6 end cell row 3 cell space space 2 end cell end table close square brackets

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194. Verify that (AB)^–1 = B^–1 A^–1 for the matrices A and B where

straight A space equals open square brackets table row 3 cell space space 2 end cell row 7 cell space space 5 end cell end table close square brackets comma space space straight B space equals space open square brackets table row 6 cell space space space space 7 end cell row 8 cell space space space space 9 end cell end table close square brackets

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195. Verify that (AB)^–1 = B^–1 A^–1 for the matrices A and B where $straight A space equals space open square brackets table row 2 cell space space space 1 end cell row 5 cell space space space 3 end cell end table close square brackets comma space space space straight B space equals space open square brackets table row 4 cell space space space 5 end cell row 3 cell space space 4 end cell end table close square brackets$ .

straight A space equals space open square brackets table row 2 cell space space space 1 end cell row 5 cell space space space 3 end cell end table close square brackets comma space space space straight B space equals space open square brackets table row 4 cell space space space 5 end cell row 3 cell space space 4 end cell end table close square brackets open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space 1 end cell row 5 cell space space space 3 end cell end table close vertical bar space equals space 6 minus 5 space equals space 1

Co-factors of elements of first row | A | are 3 and – 5 respectively.
Co-factors of elements of second row of | A | are – 1 and 2 respectively.

adj. space straight A space equals space open square brackets table row 3 cell space space space minus 5 end cell row cell negative 1 end cell cell space space space space space space space 2 end cell end table close square brackets to the power of comma space space equals space open square brackets table row cell space space space 3 end cell cell space space space minus 1 end cell row cell negative 5 end cell cell space space space space space 2 end cell end table close square brackets 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 space space 3 end cell cell space space space space minus 1 end cell row cell negative 5 end cell cell space space space space space space 2 end cell end table close square brackets open vertical bar straight B close vertical bar space equals space open vertical bar table row 4 cell space space space 5 end cell row 3 cell space space space 4 end cell end table close vertical bar space equals space 16 minus 15 space equals space 1

Co-factors of elements of first row of | B | are 4 and – 3 respectively.
Co-factors of element of second row of | B | are – 5 and 4 respectively.

adj. space straight B space equals space open square brackets table row cell space space 4 end cell cell space space space minus 3 end cell row cell negative 5 end cell cell space space space space space 4 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell space space 4 end cell cell space space space space minus 5 end cell row cell negative 3 end cell cell space space 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 fraction numerator adj. space straight B over denominator open vertical bar straight B close vertical bar end fraction space equals space open square brackets table row cell space space 4 end cell cell space space space minus 5 end cell row cell negative 3 end cell cell space space space space space space 4 end cell end table close square brackets therefore space space space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent space equals space open square brackets table row cell space space 4 end cell cell space space space space minus 5 end cell row cell negative 3 end cell cell space space space space space space space 4 end cell end table close square brackets space space open square brackets table row 3 cell space space minus 1 end cell row cell negative 5 end cell cell space space space space 2 end cell end table close square brackets space equals space open square brackets table row cell 12 plus 25 end cell cell space space space minus 4 minus 10 end cell row cell negative 9 minus 20 end cell cell space space space space space space space 3 plus 8 end cell end table close square brackets therefore space space space space space space straight B to the power of negative 1 end exponent straight A to the power of negative 1 end exponent space space equals space open square brackets table row cell space space 37 end cell cell space space space space minus 14 end cell row cell negative 29 end cell cell space space space space space space space space 11 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 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space space space space space space space space space space space space space space space AB space equals space open square brackets table row 2 cell space space space space 1 end cell row 5 cell space space space space 3 end cell end table close square brackets space open square brackets table row 4 cell space space space space space 5 end cell row 3 cell space space space space 4 end cell end table close square brackets space equals space open square brackets table row 11 cell space space space 14 end cell row 29 cell space space space 37 end cell end table close square brackets space space space space space space open vertical bar AB close vertical bar space equals space open square brackets table row 11 cell space space space space space 14 end cell row 29 cell space space space space space 37 end cell end table close square brackets space equals space 407 minus 406 space equals space 1

Co-factors of the first row of | AB | are 37, – 29 respectively.
Co-factors of the second row of | AB | are – 14, 11 respectively.

$adj. space left parenthesis AB right parenthesis space equals space open square brackets table row cell space space space space 37 end cell cell negative 29 end cell row cell negative 14 end cell cell space space space 11 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 37 cell space space minus 14 end cell row cell negative 29 end cell cell space space space space space space space 11 end cell end table close square brackets therefore space space left parenthesis AB right parenthesis to the power of negative 1 end exponent space equals space fraction numerator adj. space left parenthesis AB right parenthesis over denominator open vertical bar AB close vertical bar end fraction space space space space space space space space left parenthesis AB right parenthesis to the power of negative 1 end exponent space equals space open square brackets table row cell space space 37 end cell cell space space space minus 14 end cell row cell negative 29 end cell cell space space space space space space 11 end cell end table close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$
From (1) and (2), we have,

(AB)^–1 = B^–1 A^–1.

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Short Answer Type

196.

Find (AB)^–1 if $straight A space equals space open square brackets table row 3 cell space 4 end cell row 1 cell space 1 end cell end table close square brackets comma space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 4 cell space space 3 end cell row 2 cell space space 1 end cell end table close square brackets$

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197.

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

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Long Answer Type

198.

Find the inverse of the matrix
$straight A space equals space open square brackets table row cell negative 1 end cell cell space space 1 end cell cell space space 2 end cell row cell space space space 3 end cell cell negative 1 end cell cell space space 1 space end cell row cell space minus 1 end cell cell space space 3 end cell cell space 4 end cell end table close square brackets.$

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199. Find the inverse of the matrix:

open square brackets table row 1 cell space space space minus 1 end cell cell space space space space space space 2 end cell row 0 cell space space space space space space 2 end cell cell space space space minus 3 end cell row 3 cell space space space minus 2 end cell cell space space space space space space 4 end cell end table close square brackets.

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200.

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

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195. Verify that (AB)–1 = B–1 A–1 for the matrices A and B where .

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