Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

211.
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.

straight A space equals space open square brackets table row 4 cell space space space minus 6 end cell cell space space space space space space 1 end cell row cell negative 1 end cell cell space space space minus 1 end cell cell space space space space space space space 1 end cell row cell negative 4 end cell cell space space space space 11 end cell cell space space minus 1 end cell end table close square brackets therefore space space space open vertical bar straight A close vertical bar space equals space open vertical bar 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 space space minus 1 end cell cell space space space space space space 1 end cell row cell negative 4 end cell cell space space minus 11 end cell cell space space minus 1 end cell end table close vertical bar space equals space 4 space open vertical bar table row cell negative 1 end cell cell space space space space space space space 1 end cell row 11 cell space space minus 1 end cell end table close vertical bar minus left parenthesis negative 6 right parenthesis space open vertical bar table row cell negative 1 end cell cell space space space space space space space 1 end cell row cell negative 4 end cell cell space space minus 1 end cell end table close vertical bar plus 1 open vertical bar table row cell negative 1 end cell cell space space minus 1 end cell row cell negative 4 end cell cell space space space space 11 end cell end table close vertical bar space space space space space space space space space space space space space space equals 4 left parenthesis 1 minus 11 right parenthesis plus 6 left parenthesis 1 plus 4 right parenthesis plus 1 left parenthesis negative 11 minus 4 right parenthesis space equals space minus 40 plus 30 minus 15 equals space minus 25 not equal to 0. therefore space space space space space straight A space is space non minus singular space and space invertible. space

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

open vertical bar table row cell negative 1 end cell cell space space space space space space 1 end cell row 11 cell space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row cell negative 1 end cell cell space space space space space space space space 1 end cell row cell negative 4 end cell cell space space space space minus 1 end cell end table close vertical bar comma space space open vertical bar table row cell negative 1 end cell cell space space space space minus 1 end cell row cell negative 4 end cell cell space space space space 11 end cell end table close vertical bar

i.e., – 10, – 5, – 15 respectively
Co-factors of the elements of the second row of | A | are
$negative open vertical bar table row cell negative 6 end cell cell space space space space space 1 end cell row 11 cell space space space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 4 cell space space space space space 1 end cell row cell negative 4 end cell cell space space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row cell space 4 end cell cell space space space minus 6 end cell row cell negative 4 end cell cell space space space space space 11 end cell end table close vertical bar$
i.e., 5, 0, – 20 respectively.
Co-factors of the elements of the third row of | A | are
$open vertical bar table row cell negative 6 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 4 cell space space space space space 1 end cell row cell negative 1 end cell cell space space space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row cell space 4 end cell cell space space minus 6 end cell row cell negative 1 end cell cell space space space minus 1 end cell end table close vertical bar$
i.e., – 5, – 5, – 10 respectively.
$therefore space space adj. space straight A space equals space open square brackets table row cell negative 10 end cell cell space space minus 5 end cell cell space space minus 15 end cell row cell space space 5 end cell cell space space space space 0 end cell cell space minus 20 end cell row cell negative 5 end cell cell negative 5 end cell cell space space minus 10 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 10 end cell cell space space 5 end cell cell space space minus 5 end cell row cell negative 5 end cell cell space space space 0 end cell cell space minus 5 end cell row cell negative 15 end cell cell space minus 20 end cell cell space space minus 10 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 minus 1 over 25 open square brackets table row cell negative 10 end cell cell space space space 5 end cell cell space minus 5 end cell row cell negative 5 end cell cell space space 0 end cell cell space minus 5 end cell row cell negative 15 end cell cell space minus 20 end cell cell space minus 10 end cell end table close square brackets space equals space 1 fifth open square brackets table row 2 cell space space minus 1 end cell cell space space 1 end cell row 1 cell space space space 0 end cell cell space space 1 end cell row 3 cell space space 4 end cell cell space 2 end cell end table close square brackets Also space straight A thin space straight A to the power of negative 1 end exponent space equals space 1 fifth 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 space space minus 1 end cell cell space space space space 1 end cell row cell negative 4 end cell cell space space space space 11 end cell cell space space minus 1 end cell end table close square brackets space open square brackets table row 2 cell space space minus 1 end cell cell space space space 1 end cell row 1 cell space space space space 0 end cell cell space 1 end cell row 3 cell space space 4 end cell cell 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 equals space 1 fifth open square brackets table row cell 8 minus 6 plus 3 end cell cell space space space minus 4 plus 0 plus 4 end cell cell space space space space 4 minus 6 plus 2 end cell row cell negative 2 minus 1 plus 3 end cell cell space space 1 plus 0 plus 4 end cell cell space minus 1 minus 1 plus 2 end cell row cell negative 8 plus 11 minus 3 end cell cell space space space 4 plus 0 minus 4 end cell cell space space minus 4 plus 11 plus 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 equals space 1 fifth open square brackets table row 5 cell space space space 0 end cell cell space space space 0 end cell row 0 cell space space space 5 end cell cell space space 0 end cell row 0 cell space space 0 end cell cell space space 5 end cell end table close square brackets space equals space open square brackets table row 1 cell space space 0 end cell cell space space 0 end cell row 0 cell space space 1 end cell cell space space 0 end cell row 0 cell space space 0 end cell cell space 1 end cell end table close square brackets space equals space straight I therefore space space space space AA to the power of negative 1 end exponent space equals space straight I. 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 space space space space space space space space$

Also,
$straight A to the power of negative 1 end exponent straight A space equals 1 fifth open square brackets table row 2 cell space space minus 1 end cell cell space space space 1 end cell row 1 cell space space space space 0 end cell cell space space space 1 end cell row 3 cell space space 4 end cell cell space space 2 end cell end table close square brackets space space open square brackets table row 4 cell space space minus 6 end cell cell space space 1 end cell row cell negative 1 end cell cell negative 1 end cell cell space space space 1 end cell row cell negative 4 end cell cell space 11 end cell cell negative 1 end cell end table close square brackets space space space space space space space space space space space equals 1 fifth open square brackets table row cell 8 plus 1 minus 4 end cell cell space space minus 12 plus 1 plus 11 end cell cell space space space space 2 minus 1 minus 1 end cell row cell 4 plus 0 minus 4 end cell cell space minus 6 plus 0 plus 11 end cell cell space space 1 plus 0 minus 1 end cell row cell 12 minus 4 minus 8 end cell cell space minus 18 minus 4 plus 22 end cell cell space space 3 plus 4 minus 2 end cell end table close square brackets space space space space space space space space space space space equals space 1 fifth open square brackets table row 5 cell space space 0 end cell cell space space 0 end cell row 0 cell space space 5 end cell cell space space 0 end cell row 0 cell space space 0 end cell cell space space 5 end cell end table close square brackets space equals space open square brackets table row 1 cell space space 0 end cell cell space space 0 end cell row 0 cell space space 1 end cell cell space space 0 end cell row 0 cell space space 0 end cell cell space space 1 end cell end table close square brackets space equals space straight I straight A to the power of negative 1 end exponent straight A space equals space straight I 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 space space space space space space space space space space space... left parenthesis 2 right parenthesis space space space space space space space space space space space space$

From (1) and (2), we get,
AA^–1 = I = A^–1A.

72 Views

212.

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

79 Views

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 ?

91 Views

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.

72 Views

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

76 Views

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.

74 Views

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.

76 Views

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.

89 Views

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.

74 Views

220.

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

73 Views

Previous Year Papers

Test Series

Test Yourself

Long Answer Type

211. Let A = Show that A is inevitable. Find adj. A and A Also verify that AA–1 = A–1 A = I.

Short Answer Type

Long Answer Type