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受欢迎的代数 >

solvefor A,ABXA-1B-1=I+A

解答

求解

A, A BX A - 1 B - 1 = I + A

解答

A = \frac{1 + 1 - 4 BX ( - B - 1 - I )}{2 BX}, A = \frac{1 - 1 - 4 BX ( - B - 1 - I )}{2 BX}; BX \neq = 0

求解步骤

A BX A - 1 \cdot B - 1 = I + A

展开

A BX A - 1 \cdot B - 1 : A^{2} BX - B - 1

A^{2} BX - B - 1 = I + A

将

A

para o lado esquerdo

A^{2} BX - B - 1 - A = I

将

I

para o lado esquerdo

A^{2} BX - B - 1 - A - I = 0

改写成标准形式

a x^{2} + b x + c = 0

BX A^{2} - A - B - 1 - I = 0

使用求根公式求解

A_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 BX ( - B - 1 - I )}{2 BX}

化简

(- 1)^{2} - 4 BX (- B - 1 - I) : 1 - 4 BX (- B - 1 - I)

A_{1, 2} = \frac{- ( - 1 ) \pm 1 - 4 BX ( - B - 1 - I )}{2 BX}; BX \neq = 0

将解分隔开

A_{1} = \frac{- ( - 1 ) + 1 - 4 BX ( - B - 1 - I )}{2 BX}, A_{2} = \frac{- ( - 1 ) - 1 - 4 BX ( - B - 1 - I )}{2 BX}

A = \frac{- ( - 1 ) + 1 - 4 BX ( - B - 1 - I )}{2 BX} : \frac{1 + 1 - 4 BX ( - B - 1 - I )}{2 BX}

A = \frac{- ( - 1 ) - 1 - 4 BX ( - B - 1 - I )}{2 BX} : \frac{1 - 1 - 4 BX ( - B - 1 - I )}{2 BX}

二次方程组的解是:

A = \frac{1 + 1 - 4 BX ( - B - 1 - I )}{2 BX}, A = \frac{1 - 1 - 4 BX ( - B - 1 - I )}{2 BX}; BX \neq = 0

作图

流行的例子

(x+3)^2-(x-3)^2=x^2+32

(x + 3)^{2} - (x - 3)^{2} = x^{2} + 32

solvefor y,x^3+xy^2-y^2=0

so l v e f or y, x^{3} + x y^{2} - y^{2} = 0

-5q^2+9q+13=-6q^2

- 5 q^{2} + 9 q + 13 = - 6 q^{2}

solvefor y,x^3+x^2+2xy^2+2y^2-4x-4=0

so l v e f or y, x^{3} + x^{2} + 2 x y^{2} + 2 y^{2} - 4 x - 4 = 0

\frac{1}{16} = N^{2}