Respostas
Resposta:
Temos o seguinte sobre matriz inversa:
\begin{gathered}[\begin{array}{ccc}1&2\\1&0\\\end{array}] \cdot [\begin{array}{ccc}a&b\\c&d\\\end{array}] = [\begin{array}{ccc}1&0\\0&1\\\end{array}] \\ \\ \\ \{ {{1a+2c=1} \atop {1a +0c=0}} . \to \{ {{a+2c=1} \atop {a =0}} . \\ \\ \{ {{1b + 2d=0} \atop {1b + 0d=1}} . \to \{ {{b + 2d=0} \atop {b =1}} .\end{gathered}
[
1
1
2
0
]⋅[
a
c
b
d
]=[
1
0
0
1
]
{
1a+0c=0
1a+2c=1
.→{
a=0
a+2c=1
.
{
1b+0d=1
1b+2d=0
.→{
b=1
b+2d=0
.
Logo temos:
\begin{gathered}a + 2c = 1 \to 0 + 2c = 1 \to c = \frac{1}{2} \\ \\ b + 2d = 0 \to 1 + 2d = 0 \to d = - \frac{1}{2}\end{gathered}
a+2c=1→0+2c=1→c=
2
1
b+2d=0→1+2d=0→d=−
2
1
Logo a matriz inversa:
\begin{gathered}[\begin{array}{ccc}a&b\\c&d\\\end{array}] = [\begin{array}{ccc}0&1\\ \frac{1}{2} &- \frac{1}{2} \\\end{array}]\end{gathered}
[
a
c
b
d
]=[
0
2
1
1
−
2
1
]