Show that the matrix satisfies the equation A 2 – 4A + I = O and hence find A –1 . - Wired Faculty

Question

Show that the matrix $straight A space equals space open square brackets table row 2 cell space space space space space 3 end cell row 1 cell space space space space 2 end cell end table close square brackets$ satisfies the equation A² – 4A + I = O and hence find A^–1.

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Solution

straight A space equals space open square brackets table row 2 cell space space space space space space 3 end cell row 1 cell space space space space space space 2 end cell end table close square brackets

therefore space space space space space straight A squared space equals space open square brackets table row 2 cell space space space space space 3 end cell row 1 cell space space space space space 2 end cell end table close square brackets space open square brackets table row 2 cell space space space 3 end cell row 1 cell space space space 2 end cell end table close square brackets space equals space open square brackets table row cell 4 plus 3 end cell cell space space space 6 plus 6 end cell row cell 2 plus 2 end cell cell space space space 3 plus 4 end cell end table close square brackets space equals open square brackets table row 7 cell space space space space 12 end cell row 4 cell space space space space 7 end cell end table close square brackets

therefore space space space space straight A squared minus 4 straight A space plus space straight I space equals space open square brackets table row 7 cell space space space 12 end cell row 4 cell space space space 7 end cell end table close square brackets minus space 4 space open square brackets table row 2 cell space space space space 3 end cell row 1 cell space space space 2 end cell end table close square brackets space plus space open square brackets table row 1 cell 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 equals space open square brackets table row 7 cell space space space space space 12 end cell row 4 cell space space space space 7 end cell end table close square brackets space plus space open square brackets table row cell negative 8 end cell cell space space space minus 12 end cell row cell negative 4 end cell cell space space space minus 8 end cell end table close square brackets space plus 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 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 7 minus 8 plus 1 end cell cell space space space space space space space 12 minus 12 plus 0 end cell row cell 4 minus 4 plus 0 end cell cell space space space space space space space space space space space space 7 minus 8 plus 1 end cell end table close square brackets space equals space open square brackets table row 0 cell space space space 0 end cell row 0 cell space space space 0 end cell end table close square brackets space equals space straight O therefore space space space space straight A squared minus 4 straight A plus straight I space equals space straight O space space space space space rightwards double arrow space space space straight A space equals open square brackets table row 2 cell space space space 3 end cell row 1 cell space space 2 end cell end table close square brackets space satisfies space straight A squared minus 4 straight A plus straight I space equals space straight O Now comma space straight A squared minus 4 straight A plus straight I space equals space straight O space space space space space space space rightwards double arrow space space space straight I space equals space minus straight A squared plus 4 straight A rightwards double arrow space space space space space space space space space space space straight A to the power of negative 1 end exponent space equals space minus straight A space plus space 4 straight I space equals space minus open square brackets table row 2 cell space space space space space 3 end cell row 1 cell space space space space space space 2 end cell end table close square brackets space plus space 4 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 space space space space space space

equals space open square brackets table row cell negative 2 end cell cell space space space space minus 3 end cell row cell negative 1 end cell cell space space space space minus 2 end cell end table close square brackets plus space open square brackets table row 4 cell space space space space 0 end cell row 0 cell space space space space 4 end cell end table close square brackets space equals space open square brackets table row cell negative 2 plus 4 end cell cell space space space space space minus 3 plus 0 end cell row cell negative 1 plus 0 end cell cell space space space space minus 2 plus 4 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 open square brackets table row cell negative 2 end cell cell space space space space space minus 3 end cell row cell negative 1 end cell cell space space space space space space space 2 end cell end table close square brackets

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Determinants

Show that the matrix $straight A space equals space open square brackets table row 2 cell space space space space space 3 end cell row 1 cell space space space space 2 end cell end table close square brackets$ satisfies the equation A² – 4A + I = O and 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

Show that the matrix satisfies the equation A2 – 4A + I = O and hence find A–1.

Some More Questions From Determinants Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Show that the matrix $straight A space equals space open square brackets table row 2 cell space space space space space 3 end cell row 1 cell space space space space 2 end cell end table close square brackets$ satisfies the equation A² – 4A + I = O and hence find A^–1.