摘要： Find the inverse of the following matrices: (i) (ii) (iii) (iv) This leads us into the realm of simultaneous equations: Consider the problem of solving the set of simultaneous equations: This can be represented as The purpose of solving these equations is to find the values of x and y. Therefore we need to remove the matrix at the front of the column matrix. This can be done by multiplying by the inverse of . Using the method above or by using your calculator find the inverse which is . Therefore the solution is x = -2 and y = 3. Type in the matrix you are working with: Use the ANS feature to find the inverse: Multiply the inverse by the answer matrix. This of course means we can solve very complex sets of simultaneous equations with ease. Exercise 4:
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以下是用Scilab编写的程序，输出a，b的含义是________________．
a=input(“please give the first number”)
b=input(“please give the second number”)
while a＜＞b
if a＞=b
a=a－b；
else
b=b－a；
end
end
print(％io(2)，a，b)
查看习题详情和答案>>以下是用
Scilab编写的程序，输出a，b的含义是________________．a=input(
“please give the first number”)b=input(
“please give the second number”)while
a＜＞bif a＞=b
a=a－b；
else
b=b－a；
end
end
print(
％io(2)，a，b) 查看习题详情和答案>>过点A（a，a）可作圆x2+y2-2ax+a2+2a-3=0的两条切线，则实数a的取值范围为（ ）
A、a＜-3或1＜a＜
| ||
B、1＜a＜
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| C、a＜-3 | ||
D、-3＜a＜1或a＞
|