Question

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Answer 1

Avem:
$A=\left(\begin{array}{ccc}1&2&1\\0&1&1\\0&m&1\end{array}\right)$ 
$B=\left(\begin{array}{ccc}1&0&0\\0&2&m\\0&1&m\end{array}\right)$

a). Pentru ca matricea A să fie inversabilă, trebuie ca determinantul să fie diferit de 0.

det(A) = (1+0+0-0-m-0)
           = 1-m
[tex]1-m \neq 0 -m \neq -1 m \neq 1[/tex]

Deci, m∈R/{1}

b). $A*B= \left(\begin{array}{ccc}1&2&1\\0&1&1\\0&m&1\end{array}\right)* \left(\begin{array}{ccc}1&0&0\\0&2&m\\0&1&m\end{array}\right) = \left(\begin{array}{ccc}1&5&3m\\0&3&2m\\0&2m+1&m^2+m\end{array}\right)$

S = 1+5+3m+0+3+2m+0+2m+1+m²+m
   = m²+8m+10
m²+8m+10=3
m²+8m+7=0
Δ=b²-4ac
Δ=64-28
Δ=36
[tex]m_{1}=\frac{-b- \sqrt{delta}}{2a}=\frac{-8-6}{2}=-7 m_{2}=\frac{-b+ \sqrt{delta}}{2a}=\frac{-8+6}{2}=-1 [/tex]

c). Dacă m=0, atunci $A=\left(\begin{array}{ccc}1&2&1\\0&1&1\\0&0&1\end{array}\right)$

det(A) = 1-m = 1-0 = 1

$A^t= \left(\begin{array}{ccc}1&0&0\\2&1&0\\1&1&1\end{array}\right)$

[tex]A_{11}=(-1)^2* \left|\begin{array}{cc}1&0\\1&1\end{array}\right| = 1 A_{12}=(-1)^3* \left|\begin{array}{cc}2&0\\1&1\end{array}\right|=-2 A_{13}=(-1)^4* \left|\begin{array}{cc}2&1\\1&1\end{array}\right|=1 A_{21}=(-1)^3* \left|\begin{array}{cc}0&0\\1&1\end{array}\right|=0 A_{22}=(-1)^4* \left|\begin{array}{cc}1&0\\1&1\end{array}\right|=1 A_{23}=(-1)^5* \left|\begin{array}{cc}1&0\\1&1\end{array}\right|=-1 A_{31}=(-1)^4* \left|\begin{array}{cc}0&0\\1&0\end{array}\right|= 0[/tex]
[tex]A_{32}=(-1)^5* \left|\begin{array}{cc}1&0\\2&0\end{array}\right|= 0 A_{33}=(-1)^6* \left|\begin{array}{cc}1&0\\2&1\end{array}\right|= 1[/tex]

$A^x= \left(\begin{array}{ccc}1&-2&1\\0&1&-1\\0&0&1\end{array}\right)$

Inversa matricei A este A⁻¹, care se află astfel:

[tex]A^{-1}=\frac1{det(A)}*A^x [/tex]

Determinantul matricei A este 1, deci A⁻¹ va fi egal cu Aˣ :

$A^{-1}=A^x=\left(\begin{array}{ccc}1&-2&1\\0&1&-1\\0&0&1\end{array}\right)$

Answer 2

$\displaystyle \mathtt{A=\left(\begin{array}{ccc}\mathtt1&\mathtt2&\mathtt1\\\mathtt0&\mathtt1&\mathtt1\\\mathtt0&\mathtt m&\mathtt1\end{array}\right),~B=\left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt0\\\mathtt0&\mathtt2&\mathtt m\\\mathtt0&\mathtt1&\mathtt m\end{array}\right),~m \in \mathbb{R}}$

$\displaystyle \mathtt{a)~det(A) \not = 0}\\ \\ \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt2&\mathtt1\\\mathtt0&\mathtt1&\mathtt1\\\mathtt0&\mathtt m&\mathtt1\end{array}\right|=1 \cdot 1 \cdot 1+1 \cdot 0 \cdot m+2 \cdot 1 \cdot 0-1 \cdot 1 \cdot 0-}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-2 \cdot 0 \cdot 1-1 \cdot 1 \cdot m=1-m}\\ \\ \mathtt{1-m \not =0\Rightarrow -m \not =0-1\Rightarrow -m \not = -1 \Rightarrow m\not=1\Rightarrow m \in \mathbb{R}-\{1\}}$

$\displaystyle \mathtt{b)~A \cdot B=\left(\begin{array}{ccc}\mathtt1&\mathtt2&\mathtt1\\\mathtt0&\mathtt1&\mathtt1\\\mathtt0&\mathtt m&\mathtt1\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt0\\\mathtt0&\mathtt2&\mathtt m\\\mathtt0&\mathtt1&\mathtt m\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{1 \cdot 1+2 \cdot 0 +1 \cdot 0}&\mathtt{1 \cdot 0+2 \cdot 2+1 \cdot 1}&\mathtt{1 \cdot 0+2 \cdot m+1 \cdot m}\\\mathtt{0 \cdot 1+1\cdot0+1\cdot0}&\mathtt{0 \cdot 0+1 \cdot2+1\cdot1}&\mathtt{0 \cdot0+1\cdot m+1\cdot m}\\\mathtt{0\cdot1+m\cdot0+1\cdot0}&\mathtt{0\cdot0+m\cdot2+1\cdot1}&\mathtt {0\cdot0+m\cdot m+1\cdot m}\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{1+0+0}&\mathtt{0+4+1}&\mathtt{0+2m+m}\\\mathtt{0+0+0}&\mathtt{0+2+1}&\mathtt{0+m+m}\\\mathtt{0+0+0}&\mathtt{0+2m+1}&\mathtt{0+m^2+m}\end{array}\right)=\left(\begin{array}{ccc}\mathtt1&\mathtt5&\mathtt{3m}\\\mathtt0&\mathtt3&\mathtt{2m}\\\mathtt0&\mathtt{2m+1}&\mathtt{m^2+m}\end{array}\right)}$

$\displaystyle \mathtt{1+5+3m+0+3+2m+0+2m+1+m^2+m=3}\\ \\ \mathtt{m^2+8m+10-3=0}\\ \\ \mathtt{m^2+8m+7=0}\\ \\ \mathtt{\Delta=8^2-4 \cdot 1 \cdot 7=64-28=36\ \textgreater \ 0}\\ \\ \mathtt{m_1= \frac{-8- \sqrt{36} }{2 \cdot 1}= \frac{-8-6}{2}= \frac{-14}{2} =-7}\\ \\ \mathtt{m_1= \frac{-8+ \sqrt{36} }{2 \cdot 1}= \frac{-8+6}{2}= \frac{-2}{2} =-1}$

$\displaystyle \mathtt{c)~m=0\Rightarrow A=\left(\begin{array}{ccc}\mathtt1&\mathtt2&\mathtt1\\\mathtt0&\mathtt1&\mathtt1\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)~~~~~~~~~~~~~~~~~~~~~~~~~~~~A^{-1}= \frac{1}{det(A)}\cdot A^*}$

$\displaystyle \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt2&\mathtt1\\\mathtt0&\mathtt1&\mathtt1\\\mathtt0&\mathtt0&\mathtt1\end{array}\right|=1 \cdot 1 \cdot 1+1 \cdot 0 \cdot 0+2 \cdot 1 \cdot 0-1 \cdot 1 \cdot 0-}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-2 \cdot 0 \cdot 1 -1 \cdot 1 \cdot 0=1}$

$\displaystyle \mathtt{A=\left(\begin{array}{ccc}\mathtt1&\mathtt2&\mathtt1\\\mathtt0&\mathtt1&\mathtt1\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)\Rightarrow A^T=\left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt0\\\mathtt2&\mathtt1&\mathtt0\\\mathtt1&\mathtt1&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{D_{11}=(-1)^{1+1}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt1&\mathtt1\end{array}\right|=1 \cdot 1=1}\\ \\ \mathtt{D_{12}=(-1)^{1+2}\cdot \left|\begin{array}{ccc}\mathtt2&\mathtt0\\\mathtt2&\mathtt1\end{array}\right|=(-1) \cdot 2=-2}\\ \\ \mathtt{D_{13}=(-1)^{1+3}\cdot \left|\begin{array}{ccc}\mathtt2&\mathtt1\\\mathtt1&\mathtt1\end{array}\right|=1 \cdot 1=1}$
$\displaystyle \mathtt{D_{21}=(-1)^{2+1}\cdot \left|\begin{array}{ccc}\mathtt0&\mathtt0\\\mathtt1&\mathtt1\end{array}\right|=(-1) \cdot 0=0}\\ \\ \mathtt{D_{22}=(-1)^{2+2}\cdot\left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt1&\mathtt1\end{array}\right| =1 \cdot 1=1}\\ \\ \mathtt{D_{23}=(-1)^{2+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt1&\mathtt1\end{array}\right|=(-1) \cdot 1=-1}$
$\displaystyle \mathtt{D_{31}=(-1)^{3+1}\cdot \left|\begin{array}{ccc}\mathtt0&\mathtt0\\\mathtt1&\mathtt0\end{array}\right|=1\cdot0=0}\\ \\ \mathtt{D_{32}=(-1)^{3+2}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt2&\mathtt0\end{array}\right|=(-1) \cdot 0=0}\\ \\ \mathtt{D_{33}=(-1)^{3+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt2&\mathtt1\end{array}\right|=1 \cdot 1=1}$

$\displaystyle \mathtt{A^*= \left(\begin{array}{ccc}\mathtt1&\mathtt{-2}&\mathtt1\\\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)}\\ \\ \mathtt{A^{-1}= \frac{1}{1} \cdot \left(\begin{array}{ccc}\mathtt1&\mathtt{-2}&\mathtt1\\\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)= \left(\begin{array}{ccc}\mathtt1&\mathtt{-2}&\mathtt1\\\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{A^{-1}= \left(\begin{array}{ccc}\mathtt1&\mathtt{-2}&\mathtt1\\\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)}$