"If $A = \left( {\matrix{ 2 & 2 \cr 9 & 4 \cr } } \right)$ and $I = \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right)$ then 10A–1 is equal to :"

Question

Accepted Answer

"According to Cayley Hamilton equation
|A – $\lambda$I| = 0

$\Rightarrow$ $$\left| {\matrix{ {2 - \lambda } & 2 \cr 9 & {4 - \lambda } \cr } } \right|$$ = 0

$\Rightarrow$ (2 – $\lambda$)(4 – $\lambda$) – 18 = 0

$\Rightarrow$ 8 – 2$\lambda$ – 4$\lambda$ + $\lambda$² – 18 = 0

$\Rightarrow$ $\lambda$² – 6$\lambda$ – 10 = 0

$\therefore$ A² – 6A– 10 = 0

$\Rightarrow$ A^–1(A²) – 6A^–1A – 10A^–1 = 0

$\Rightarrow$ A – 6I – 10A^–1 = 0

$\Rightarrow$ 10A^–1 = A – 6I"

If $A = \left( {\matrix{ 2 & 2 \cr 9 & 4 \cr } } \right)$ and $I = \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right)$ then 10A^–1 is equal to :

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If $A = \left( {\matrix{ 2 & 2 \cr 9 & 4 \cr } } \right)$ and $I = \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right)$ then 10A–1 is equal to :

Solution

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If $A = \left( {\matrix{ 2 & 2 \cr 9 & 4 \cr } } \right)$ and $I = \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right)$ then 10A^–1 is equal to :