Explicação passo-a-passo:
Dada a igualdade de matrizes:
\begin{gathered}\mathsf{\begin{pmatrix}a+b-1&0\\ a-3c&b\\ 2b&0\end{pmatrix}=\begin{pmatrix}0&0\\ 0&0\\ 0&0\end{pmatrix}}\end{gathered}
⎝
⎛
a+b−1
a−3c
2b
0
b
0
⎠
⎞
=
⎝
⎛
0
0
0
0
0
0
⎠
⎞
Considerando:
\begin{gathered}\mathsf{A=\begin{pmatrix}a+b-1&0\\ a-3c&b\\ 2b&0\end{pmatrix}\:\:\:e\:\:\:B=\begin{pmatrix}0&0\\ 0&0\\ 0&0\end{pmatrix}}\end{gathered}
A=
⎝
⎛
a+b−1
a−3c
2b
0
b
0
⎠
⎞
eB=
⎝
⎛
0
0
0
0
0
0
⎠
⎞
Onde, para que as matrizes sejam iguais elas devem ter:
๏ Elementos correspondentes iguais
๏ Mesma ordem (numero de linhas e colunas)
Logo:
๏ \mathbf{a_{11}=b_{11}}a
11
=b
11
\begin{gathered}\mathsf{a+b-1=0}\\\mathsf{a+b=1 \: \: (i)}\end{gathered}
a+b−1=0
a+b=1(i)
๏ \mathbf{a_{12}=b_{12}}a
12
=b
12
\mathsf{0=0}0=0
๏ \mathbf{a_{21}=b_{21}}a
21
=b
21
\begin{gathered}\mathsf{a-3c=0}\\\mathsf{a=3c \: \: (ii) }\end{gathered}
a−3c=0
a=3c(ii)
๏ \mathbf{a_{31}=b_{31}}a
31
=b
31
\mathsf{b=0 \: \: (iii)}b=0(iii)
๏ \mathbf{a_{32}=b_{32}}a
32
=b
32
\begin{gathered}\mathsf{2b=0}\\\mathsf{b=0}\end{gathered}
2b=0
b=0
Substituindo (iii) em (i):
\begin{gathered}\mathsf{a+0=1}\\\mathsf{a=1 \: \: (iv)}\end{gathered}
a+0=1
a=1(iv)
Substituindo (iv) em (ii):
\begin{gathered}\mathsf{1=3c}\\\\\mathsf{c=\dfrac{1}{3}}\end{gathered}
1=3c
c=
3
1
