Question

M= -2   2        I2= 1   0
      -1  -1              0   1
determinati nr reale a si b astfel incat M*M*M=aM+bI2

Answer 1

$\displaystyle \mathtt{M=\left(\begin{array}{ccc}\mathtt{-2}&\mathtt2\\\mathtt{-1}&\mathtt{-1}\\\end{array}\right),~I_2=\left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\\\end{array}\right)}\\ \\ \mathtt{M\cdot M\cdot M=aM+bI_2}$

$\displaystyle \mathtt{M \cdot M=\left(\begin{array}{ccc}\mathtt{-2}&\mathtt2\\\mathtt{-1}&\mathtt{-1}\\\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt{-2}&\mathtt2\\\mathtt{-1}&\mathtt{-1}\\\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt{(-2)\cdot(-2)+2\cdot(-1)}&\mathtt{(-2)\cdot2+2\cdot(-1)}\\\mathtt{(-1)\cdot(-2)+(-1)\cdot(-1)}&\mathtt{(-1)\cdot2+(-1)\cdot(-1)}\\\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{4-2}&\mathtt{(-4)-2}\\\mathtt{2+1}&\mathtt{(-2)+1}\\\end{array}\right) =\left(\begin{array}{ccc}\mathtt{2}&\mathtt{-6}\\\mathtt3&\mathtt{-1}\\\end{array}\right)}$

$\displaystyle \mathtt{M\cdot M \cdot M=\left(\begin{array}{ccc}\mathtt{2}&\mathtt{-6}\\\mathtt3&\mathtt{-1}\\\end{array}\right)\cdot\left(\begin{array}{ccc}\mathtt{-2}&\mathtt2\\\mathtt{-1}&\mathtt{-1}\\\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt{2\cdot(-2)+(-6)\cdot(-1)}&\mathtt{2\cdot2+(-6)\cdot(-1)}\\\mathtt{3\cdot(-2)+(-1)\cdot(-1)}&\mathtt{3\cdot2+(-1)\cdot(-1)}\\\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{(-4)+6}&\mathtt{4+6}\\\mathtt{(-6)+1}&\mathtt{6+1}\\\end{array}\right)=\left(\begin{array}{ccc}\mathtt{2}&\mathtt{10}\\\mathtt{-5}&\mathtt{7}\\\end{array}\right)}$

$\displaystyle \mathtt{\left(\begin{array}{ccc}\mathtt{2}&\mathtt{10}\\\mathtt{-5}&\mathtt{7}\\\end{array}\right)=a\cdot\left(\begin{array}{ccc}\mathtt{-2}&\mathtt2\\\mathtt{-1}&\mathtt{-1}\\\end{array}\right)+b\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\\\end{array}\right)}$

$\displaystyle \mathtt{\left(\begin{array}{ccc}\mathtt{2}&\mathtt{10}\\\mathtt{-5}&\mathtt{7}\\\end{array}\right)=\left(\begin{array}{ccc}\mathtt{-2a}&\mathtt{2a}\\\mathtt{-a}&\mathtt{-a}\\\end{array}\right)+\left(\begin{array}{ccc}\mathtt{b}&\mathtt{0}\\\mathtt{0}&\mathtt{b}\\\end{array}\right)}$

$\displaystyle \mathtt{\left(\begin{array}{ccc}\mathtt{2}&\mathtt{10}\\\mathtt{-5}&\mathtt{7}\\\end{array}\right)=\left(\begin{array}{ccc}\mathtt{-2a+b}&\mathtt{2a}\\\mathtt{-a}&\mathtt{-a+b}\\\end{array}\right) }$

$\displaystyle \mathtt{ \left \{ {{2=-2a+b} \atop {7=-a+b}} \right. \Rightarrow \left \{ {{2a-b=-2}|\cdot (-1) \atop {a-b=-7}} \right. \Rightarrow \left \{ {{-2a+b=2} \atop {a-b=-7}} \right. } \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-------}\\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-a~~/~~=-5\Rightarrow \boxed{\mathtt{a=5}}}\\ \\ \mathtt{7=-a+b\Rightarrow 7=-5+b\Rightarrow \boxed{\mathtt{b=12}}}$