Question

# Find the adjoint of $${ \left( \dfrac { { 7 }^{ -4 } }{ { 4 }^{ -2 } } \right) }^{ \dfrac { 1 }{ 4 } }\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$$

Solution

## Let  $$A={ \left( { \dfrac { { { 7^{ -4 } } } }{ { { 4^{ -2 } } } } } \right) ^{ \dfrac { 1 }{ 4 } } }\left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 2 \\ 3 & 0 & { -2 } \\ 1 & 0 & 3 \end{array} } \right]$$ $$==\left( { \dfrac { { \dfrac { 1 }{ 7 } } }{ { \dfrac { 1 }{ 2 } } } } \right) \left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 2 \\ 3 & 0 & { -2 } \\ 1 & 0 & 3 \end{array} } \right]$$ $$=\left( { \dfrac { 2 }{ 7 } } \right) \left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 2 \\ 3 & 0 & { -2 } \\ 1 & 0 & 3 \end{array} } \right]$$ $$adjA=\dfrac{2}{7}adj\left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 2 \\ 3 & 0 & { -2 } \\ 1 & 0 & 3 \end{array} } \right]$$ $$=\left( { \dfrac { 2 }{ 7 } } \right) { \left[ { \begin{array} { *{ 20 }{ c } }0 & { -11 } & 0 \\ 3 & 1 & { -1 } \\ 2 & 8 & 3 \end{array} } \right] ^{ \prime } }$$ $$=\left( { \dfrac { 2 }{ 7 } } \right) \left[ { \begin{array} { *{ 20 }{ c } }0 & 3 & 2 \\ { -11 } & 1 & 8 \\ 0 & { -1 } & 3 \end{array} } \right]$$ $$=\left[ { \begin{array} { *{ 20 }{ c } }0 & { \dfrac { 6 }{ 7 } } & { \dfrac { 1 }{ 7 } } \\ { \dfrac { { -22 } }{ 7 } } & { \dfrac { 2 }{ 7 } } & { \dfrac { { 16 } }{ 7 } } \\ 0 & { \dfrac { { -2 } }{ 7 } } & { \dfrac { 6 }{ 7 } } \end{array} } \right]$$ Mathematics

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