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.

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 $straight A to the power of negative 1 end exponent space equals space 1 over 19 straight A$ .

91 Views

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.

73 Views

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 $straight A to the power of negative 1 end exponent space equals space minus 1 over 7 straight A plus 8 over 7 straight I$

72 Views

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.

70 Views

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.

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 space minus 11 end cell end table close square brackets therefore space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space space space space space space space space 5 end cell row 7 cell space space space minus 11 end cell end table close vertical bar space equals space minus 33 minus 35 space equals space minus 68 space not equal to 0 rightwards double arrow space space space space straight A to the power of negative 1 end exponent space exists.

Co-factors of the elements of first row | A | are – 11, – 7 respectively.
Co-factors of the elements of second row of | A | are – 5, 3 respectively.

therefore space space space space space adj. space straight A space equals space open square brackets table row cell negative 11 end cell cell space space space space minus 7 end cell row cell negative 5 end cell cell space space space space space space 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 11 end cell cell space space space space minus 5 end cell row cell negative 7 end cell cell space space space space space space 3 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 minus 1 over 68 open square brackets table row cell negative 11 end cell cell space space space space minus 5 end cell row cell negative 7 end cell cell space space space space space space space 3 end cell end table close square brackets Now comma space space space space AA to the power of negative 1 end exponent space equals space minus 1 over 68 open square brackets table row 3 cell space space space space space space space space 5 end cell row 7 cell space space space minus 11 end cell end table close square brackets space open square brackets table row cell negative 11 end cell cell space space space minus 5 end cell row cell negative 7 end cell cell space space space space space 3 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 equals space minus 1 over 68 open square brackets table row cell negative 33 minus 35 end cell cell space space space space space minus 15 plus 15 end cell row cell negative 77 plus 77 end cell cell space space space space space space space minus 35 minus 33 end cell end table close square brackets space equals space minus 1 over 68 open square brackets table row cell negative 68 end cell cell space space space space 0 end cell row 0 cell space space minus 68 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 equals open square brackets table row 1 cell space space space 0 end cell row 0 cell space space space 1 end cell end table close square brackets space space equals space straight I therefore space space space space space AA to the power of negative 1 end exponent space space equals space space straight I space is space verified.

73 Views

176.

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

85 Views

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?

71 Views

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

73 Views

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.

71 Views

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 $straight A squared minus 6 straight A space plus space straight I space equals space straight O comma space hence space find space straight A to the power of negative 1 end exponent.$