Răspuns:
Explicație pas cu pas:
[tex]\left(\begin{array}{ccc}1&2\\3&5\\\end{array}\right) \cdot \left(\begin{array}{ccc}a&b\\c&d\\\end{array}\right)=\left(\begin{array}{ccc}1&0\\0&1\\\end{array}\right)[/tex]
[tex]\left(\begin{array}{ccc}a+2c&b+2d\\3a+5c&3b+5d\\\end{array}\right)=\left(\begin{array}{ccc}1&0\\0&1\\\end{array}\right)[/tex]
a+2c=1 |×3
3a+5c=0
3a+6c=3
3a+5c=0
c=3⇒a=1-6=-5
b+2d=0 |×3
3b+5d=1
3b+6d=0
3b+5d=1
d=-1⇒b=2
[tex]X=\left(\begin{array}{ccc}-5&2\\3&-1\\\end{array}\right)[/tex]
