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高二数学 | 高级数学
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2.2 Inverse Matrices逆矩阵


[1]

Given A = \left(\begin{matrix}7&0\\2&3\\\end{matrix}\right) and B = \left(\begin{matrix}35&7\\10&11\\\end{matrix}\right), if AX = B, find X.

已知A = \left(\begin{matrix}7&0\\2&3\\\end{matrix}\right) 和 B = \left(\begin{matrix}35&7\\10&11\\\end{matrix}\right), 若 AX = B, 求 X.

Answer:
\left(\begin{matrix}5&1\\0&3\\\end{matrix}\right)
[2]

A = \left(\begin{matrix}3&-2\\0&2\\\end{matrix}\right) , B = \left(\begin{matrix}6&0\\0&6\\\end{matrix}\right)

[a]

Given that A^2 = kA–BI, where k is a constant, find the values of k.

已知A^2 = kA–BI, 其中k 是常数, 求 k的值.

[b]

If AX = I, find matrix X

若AX = I, 求 X矩阵。

Answer:
(a) 5

(b) \left(\begin{matrix}{\frac{1}{3}}&{\frac{1}{3}}\\0&{\frac{1}{2}}\\\end{matrix}\right)

[3]

Given A= \left(\begin{matrix}3&k-2\\k&5\\\end{matrix}\right), if the inverse of A does not exist, find the value of k.

已知A= \left(\begin{matrix}3&k-2\\k&5\\\end{matrix}\right) , 若 A 的逆矩阵不存在, 求k 的值.

Answer:
–3, 5

[4]

A=\left(\begin{matrix}x^2&x-\frac{1}{2}\\-2&3\\\end{matrix}\right)

[a] if the inverse of A does not exist, find the value of x, 若A 的逆矩阵不存在, 求x 的值.

[b] if A^{-1}=\frac{1}{2}\left(\begin{matrix}6&1-2x\\4&2x^2\\\end{matrix}\right), find the value of x.

Answer:

(a) \frac{1}{3}, -1

(b) \frac{-1\pm\sqrt7}{3}

[5]

Find the inverse of the following matrices

求下列各矩阵的逆矩阵

[a] \left(\begin{matrix}1&0&3\\3&1&9\\-2&2&-4\\\end{matrix}\right)

[b] \left(\begin{matrix}1&-1&5\\3&1&4\\-2&5&2\\\end{matrix}\right)

Answer:
(a) \left(\begin{matrix}-11&3&-{\frac{3}{2}}\\-3&1&0\\4&-1&{\frac{1}{2}}\\\end{matrix}\right)

(b) \left(\begin{matrix}-{\frac{2}{9}}&{\frac{1}{3}}&-{\frac{1}{9}}\\-{\frac{14}{81}}&{\frac{4}{27}}&{\frac{11}{81}}\\{\frac{17}{81}}&-\frac{1}{27}&{\frac{4}{81}}\\\end{matrix}\right)