If $straight A space equals space open square brackets table row cell space space 3 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space space space 2 end cell end table close square brackets comma$ show that A² – 5A + 7 I = O. Hence find A ^–1.

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Solution

straight A space equals space open square brackets table row 3 cell space space space space space 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell end table close square brackets therefore space space space space straight A squared space equals space space open square brackets table row 3 cell space space space space 1 end cell row cell negative 1 end cell cell space space space space 2 end cell end table close square brackets space space open square brackets table row 3 cell space space space 1 end cell row cell negative 1 end cell cell space space space 2 end cell end table close square brackets space equals space open square brackets table row cell 9 minus 1 end cell cell space space space space space 3 plus 2 end cell row cell negative 3 minus 2 end cell cell space space minus 1 plus 4 end cell end table close square brackets space equals space open square brackets table row 8 cell space space 5 end cell row cell negative 5 end cell cell space space 3 end cell end table close square brackets therefore space space space space straight A squared minus 5 straight A plus 7 straight I space equals open square brackets table row 8 cell space space space 5 end cell row cell negative 5 end cell cell space space 3 end cell end table close square brackets minus 5 open square brackets table row 3 cell space space space 1 end cell row cell negative 1 end cell cell space space space 2 end cell end table close square brackets plus 7 space open square brackets table row 1 cell space space space space 0 end cell row 0 cell space space space 1 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 equals open square brackets table row cell space space 8 end cell cell space space space 5 end cell row cell negative 5 end cell cell space space space space 3 end cell end table close square brackets plus space open square brackets table row cell negative 15 end cell cell space space minus 5 end cell row 5 cell space space minus 10 end cell end table close square brackets space plus space open square brackets table row 7 cell space space space 0 end cell row 0 cell space space space 7 end cell end table close square brackets

equals space open square brackets table row cell 8 minus 15 plus 7 end cell cell space space space space space space space 5 minus 5 plus 0 end cell row cell negative 5 plus 5 plus 0 end cell cell space space space space space space space 3 minus 10 plus 7 end cell end table close square brackets space equals space open square brackets table row 0 cell space space 0 end cell row 0 cell space space 0 end cell end table close square brackets space equals space straight O therefore space space space space straight A squared minus 5 straight A space plus space 7 straight I space equals space straight O rightwards double arrow space space space space space space 7 straight I space equals space minus straight A squared plus 5 straight A space space space space space space rightwards double arrow space space space space space 7 straight A to the power of negative 1 end exponent space equals space minus straight A plus 5 thin space straight I rightwards double arrow space space space space space space space 7 straight A to the power of negative 1 end exponent space equals space minus open square brackets table row cell space space space 3 end cell cell space space space 1 end cell row cell negative 1 end cell cell space space space 2 end cell end table close square brackets space space plus space 5 space open square brackets table row 1 cell space space space 0 end cell row 0 cell space space 1 end cell end table close square brackets space rightwards double arrow space space space space space space space 7 straight A to the power of negative 1 end exponent space equals space open square brackets table row cell negative 3 end cell cell space space space minus 1 end cell row 1 cell space space space minus 2 end cell end table close square brackets space plus space open square brackets table row 5 cell space space 0 end cell row 0 cell space space 5 end cell end table close square brackets rightwards double arrow space space space space space space space space 7 straight A to the power of negative 1 end exponent space equals space open square brackets table row cell negative 3 plus 5 end cell cell space space space space space space minus 1 plus 0 end cell row cell 1 plus 0 end cell cell space space space space minus 2 plus 5 end cell end table close square brackets space space space space space space space rightwards double arrow space space space 7 straight A to the power of negative 1 end exponent space equals space open square brackets table row 2 cell space space space minus 1 end cell row 1 cell space space space space space space space space 3 end cell end table close square brackets therefore space space space space straight A to the power of negative 1 end exponent space equals space 1 over 7 open square brackets table row 2 cell space space space space space minus 1 end cell row 1 cell space space space space space space space space 3 end cell end table close square brackets

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Determinants

If $straight A space equals space open square brackets table row cell space space 3 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space space space 2 end cell end table close square brackets comma$ show that A² – 5A + 7 I = O. Hence find A ^–1.

Some More Questions From Determinants Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

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Determinants

If show that A2 – 5A + 7 I = O. Hence find A –1.

Some More Questions From Determinants Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

If $straight A space equals space open square brackets table row cell space space 3 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space space space 2 end cell end table close square brackets comma$ show that A² – 5A + 7 I = O. Hence find A ^–1.