Calculează \( \displaystyle \Delta=\begin{vmatrix} 2&1\\ 5&3\end{vmatrix} \)

\( \displaystyle \Delta=1 \)

\( \displaystyle \Delta=2\cdot 3-1\cdot 5=6-5=1 \)

Exerciții

Calculează \( \displaystyle \Delta=\begin{vmatrix} 1&0&2\\ 0&3&1\\ 0&4&-1\end{vmatrix} \) dezvoltând după coloana 1

Calculează \( \displaystyle \Delta=\begin{vmatrix} 0&2&1\\ 1&-1&0\\ 3&0&2\end{vmatrix} \) dezvoltând după linia 1

Calculează \( \displaystyle \Delta=\begin{vmatrix} 1&2&3\\ 0&-1&4\\ 2&0&1\end{vmatrix} \) dezvoltând după linia 2

Determină \( \displaystyle x \) astfel încât \( \displaystyle \begin{vmatrix} x&1\\ 2&3\end{vmatrix}=0 \)

Răspunsuri

Rezolvări

Coloana 1 are doi de zero, deci \( \displaystyle \Delta=1\cdot(-1)^{1+1}\begin{vmatrix} 3&1\\ 4&-1\end{vmatrix}=3\cdot(-1)-1\cdot 4=-3-4=-7 \)

\( \displaystyle \Delta=0\cdot C_{11}+2\cdot C_{12}+1\cdot C_{13} \), \( \displaystyle C_{12}=(-1)^{3}\begin{vmatrix} 1&0\\ 3&2\end{vmatrix}=-(2)= -2 \), \( \displaystyle C_{13}=(-1)^{4}\begin{vmatrix} 1&-1\\ 3&0\end{vmatrix}=3 \), deci \( \displaystyle \Delta=2\cdot(-2)+1\cdot 3=-4+3=-1 \) apoi verificare: am greșit cofactorul, \( \displaystyle \begin{vmatrix} 1&0\\ 3&2\end{vmatrix}=2 \Rightarrow C_{12}=-2 \), \( \displaystyle \begin{vmatrix} 1&-1\\ 3&0\end{vmatrix}=3 \Rightarrow \Delta=-1 \)

Linia 2: \( \displaystyle \Delta=0\cdot C_{21}+(-1)\cdot C_{22}+4\cdot C_{23} \), \( \displaystyle M_{22}=\begin{vmatrix} 1&3\\ 2&1\end{vmatrix}=1-6=-5 \Rightarrow C_{22}=(-1)^{4}\cdot(-5)=-5 \), \( \displaystyle M_{23}=\begin{vmatrix} 1&2\\ 2&0\end{vmatrix}=-4 \Rightarrow C_{23}=(-1)^{5}\cdot(-4)=4 \), deci \( \displaystyle \Delta=(-1)\cdot(-5)+4\cdot 4=5+16=21 \)

\( \displaystyle 3x-2=0\Rightarrow x=\frac{2}{3} \)