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.

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

$Here space 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 therefore space space space straight A squared 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 5 end cell end table close square brackets space open square brackets table row 3 cell space space space space 1 end cell row 7 cell space space space space 5 end cell end table close square brackets space equals space open square brackets table row cell 9 plus 7 end cell cell space space space space space 3 plus 5 end cell row cell 21 plus 35 end cell cell space space space space space 7 plus 25 end cell end table close square brackets space equals space open square brackets table row 16 cell space space 8 end cell row 56 cell space space 32 end cell end table close square brackets Now space straight A squared plus straight x space straight I space equals space straight y space straight A rightwards double arrow space space space space space open square brackets table row 16 cell space space space space 8 end cell row 56 cell space space space 32 end cell end table close square brackets space plus space straight x space 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 equals space straight y open square brackets table row 3 cell space space space space 1 end cell row 7 cell space space space space 5 end cell end table close square brackets rightwards double arrow space space space space space open square brackets table row 16 cell space space space space space 8 end cell row 56 cell space space space space 32 end cell end table close square brackets space plus space open square brackets table row straight x cell space space space 0 end cell row 0 cell space space space straight x end cell end table close square brackets space equals space open square brackets table row cell 3 straight y end cell cell space space space straight y end cell row cell 7 straight y end cell cell space 5 straight y end cell end table close square brackets rightwards double arrow space space space space space open square brackets table row cell 16 plus straight x end cell cell space space space space space 8 end cell row 56 cell space space space 32 plus straight x end cell end table close square brackets space equals space open square brackets table row cell 3 straight y end cell cell space space space space straight y end cell row cell 7 straight y end cell cell space space space 5 straight y end cell end table close square brackets$

From the definition of equality of matrices,
$space space space space space space space straight y space equals 8 space space space space space space space space space space space space space space space space space space 16 plus straight x space equals space 3 straight y therefore space space space space straight y space equals space 8 comma space space space space space space space space space space space space space space space space 16 plus straight x space equals 24 therefore space space space space straight y space equals space 8 comma 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 straight x space equals space 8 therefore space space space space straight A squared plus 8 thin space straight I space equals space 8 space straight A Post space multiplying space both space sides space by space straight A to the power of negative 1 end exponent comma space space space space space space space space space space space space space space straight A squared space straight A to the power of negative 1 end exponent plus space 8 space straight I thin space straight A to the power of negative 1 end exponent space equals space 8 space straight A thin space straight A to the power of negative 1 end exponent therefore space space space space space space space space space space space space space space space straight A plus space 8 space straight A to the power of negative 1 end exponent space space equals space 8 space straight I therefore space space space space space space space space space space space space space space space 8 space straight A to the power of negative 1 end exponent space space equals space 8 thin space straight I space minus space straight A 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 equals space 8 open square brackets table row 1 cell space space 0 end cell row cell space 0 end cell cell space space 1 end cell end table close square brackets minus space open square brackets table row 3 cell space space space 1 end cell row 7 cell space space space space 5 end cell end table close square brackets space equals space open square brackets table row 8 cell space space space 0 end cell row 0 cell space space space 8 end cell end table close square brackets plus open square brackets table row cell negative 3 end cell cell space space minus 1 end cell row cell negative 7 end cell cell space space space minus 5 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 equals space open square brackets table row cell 8 minus 3 end cell cell space space space 0 minus 1 end cell row cell 0 minus 7 end cell cell space space space 8 minus 5 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 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$

8 thin space straight A to the power of negative 1 end exponent space equals space open square brackets table row 5 cell space space space space space minus 1 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 1 over 8 open square brackets table row 5 cell space space space minus 1 end cell row cell negative 7 end cell cell space space space space space 3 end cell end table close square brackets.

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