Recent content by jubej

hello thnx for the reply Defennder. but i still get it wrong and i don't know why. 1) \left[\begin{array}{cc}a-b & b+c \\ 3d+c & 2a-4d\end{array}\right] I treat it like you suggested a-b=8 b+c=1 3d+c=7 2a+4d=6 and then i did something like this: \left[\begin{array}{cc}1,-1,0,0,8...

Im going crazy here please help me, i have tried to solve this but i don't know where to start. 1) Solve the following matrix equeations for a,b,c and d. \left[\begin{array}{cc}a-b & b+c \\ 3d+c & 2a-4d\end{array}\right] = \left[\begin{array}{cc}8 & 1 \\ 7 & 6\end{array}\right] this...

that would be maybe z² = - (2z - 3)/(2z - 3) z² = -1 its that right?

i can't factor out 2² in 2) because : 2z³ - 3z² + 2z -3 = 0 has 2z so its not and z² type. what to do?

there are 2 separate problems. the first one its so hard i don't where to start. or how to solve it and the second one ill start to factor out term z² and see what happens thnx.

hello! I am a studen from sweden and i having some troble solves this kind of problems. i have tried but the book its really bad to expleain. and i really have to know this for the coming exan, so if anyone could help me i really be so happy :) here are some stuff i have hard time solving...