heureka

2. -2 A^-2(A^-2+A^-3)^-1+2A³(A²+A³)-1+ 3/4 A(A-1+I)^-1+ 3/4 A³(A+I)^-1

\(\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} = \ ?\\ \hline \end{array} \)

1.
\(\small{ \begin{array}{|rcll|} \hline &&\mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\\\ && & (A^{-2}+A^{-3})^{-1} = [(A^{-1}+I)A^{-2}]^{-1} \\ && & (A^{2}+A^{3})^{-1} = [(A+I)A^{2}]^{-1} \\\\ &=& -2 A^{-2}[(A^{-1}+I)A^{-2}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2} = A^{-1}A^{-1} = (AA)^{-1} = (A^2)^{-1} \\ &=& -2 A^{-2}[(A^{-1}+I)(A^2)^{-1}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A^{-1}+I)(A^2)^{-1}]^{-1} = [A^2(A^{-1}+I)^{-1}] \\ &=& -2 A^{-2}[A^2(A^{-1}+I)^{-1}]+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 A^{-2}A^2(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2}A^2 = A^{-1}A^{-1}AA = A^{-1}(A^{-1}A)A = A^{-1}(I)A = A^{-1}A = I \\ &=& -2 I(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A+I)A^{2}]^{-1} = [(A^2)^{-1}(A+I)^{-1} ] \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A^2)^{-1}(A+I)^{-1} ] \quad & | \quad (A^2)^{-1} = A^{-2} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3A^{-2}(A+I)^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2AA^2A^{-2}(A+I)^{-1} \quad & | \quad A^2A^{-2} = I \\ &=& -2 (A^{-1}+I)^{-1}+2AI(A+I)^{-1} \\ &=&\color{red}{ -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} } \quad & | \quad A(A+I)^{-1} = [(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2[(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(AA^{-1}+IA^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(I+A^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(A^{-1}+I )^{-1} \\ &=& \mathbf{0} \\\\ && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } = 0 \\ \hline \end{array} }\)

2.

\(\begin{array}{|rcrcl|} \hline && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\ &=& \color{red}{-2 (A^{-1}+I)^{-1}+2A(A+I)^{-1}} & =& 0 \\ && -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} & =& 0 \\ && 2 (A^{-1}+I)^{-1} &=& 2A(A+I)^{-1} \\ && \mathbf{(A^{-1}+I)^{-1}} &\mathbf{=}& \mathbf{ A(A+I)^{-1}} \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& 0 + \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A\underbrace{(A^{-1}+I)^{-1}}_{=A(A+I)^{-1}}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 AA(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A^2(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 ( A^2 + A^3)(A+I)^{-1} \\ &=& \frac34 A^2(I+A)(A+I)^{-1} \\ &=& \frac34 A^2\underbrace{(I+A)(I+A)^{-1}}_{=I} \\ &=& \frac34 A^2 I \\ &\mathbf{=}& \mathbf{\frac34 A^2 } \\ \hline \end{array} \)

1. A³(A³+A²)^-1+A²(A²+A)^-1+2(A³+A²)^-1A⁴

Question see also:

1. A³(A³+A²)^-1+A²(A²+A)^-1+2(A³+A²)^-1A⁴

Question see also:

Matrix Problem: A \(\mathbf{\in \mu}\) (2x2).

Find the possible conditions and possible cases in the real numbers for which A² = 0

In linear algebra, a nilpotent matrix is a square matrix A such that

\({\displaystyle A^{k}=0\,}\)

for some positive integer k.

If A is a nilpotent matrix then det(A) = 0

\(\begin{array}{|rcll|} \hline det(A) &=& \begin{vmatrix} a & b \\ c&d \end{vmatrix} = ad-bc = 0 \\\\ \Rightarrow \qquad \mathbf{bc} & \mathbf{=} & \mathbf{ad} \\ \hline \end{array}\)

\(A^2 = 0 \\ \begin{array}{|rcll|} \hline A^2 &=& \begin{pmatrix} a & b \\ c&d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c&d \end{pmatrix} \\\\ &=& \begin{pmatrix} a^2+bc & b\cdot (a+d) \\ c\cdot (a+d) & d^2+bc \end{pmatrix} \quad & | \quad \mathbf{bc=ad} \\\\ &=& \begin{pmatrix} a^2+ad & b\cdot (a+d) \\ c\cdot (a+d) & d^2+ad \end{pmatrix} \\\\ &=& \begin{pmatrix} a\cdot (a+d) & b\cdot (a+d) \\ c\cdot (a+d) & d\cdot(a+d) \end{pmatrix} \\\\ A^2 &=& \underbrace{(a+d)}_{=0}\cdot \underbrace{\begin{pmatrix} a & b \\ c & d \end{pmatrix}}_{A\ne 0} = 0\\\\ a+d &=& 0 \\ \mathbf{a} &\mathbf{=}& \mathbf{-d} \\ \hline \end{array}\)

if \(A\ne 0 \) and \(A^2 = 0\), then \( a = -d\) and \(det(A) = 0\).

\(\text{Example } 1:\\ \begin{array}{|rcll|} \hline A &=& \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \\ det(A) &=& \begin{vmatrix} 0 & 1 \\ 0&0 \end{vmatrix} = 0\cdot 0 - 0 \cdot 1 = 0 \\ a &=& -d \\ 0 &=& 0\ \checkmark \\\\ A^2 &=& \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} = 0\ \checkmark \\ \hline \end{array}\)

\(\text{Example } 2:\\ \begin{array}{|rcll|} \hline A &=& \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} \\ det(A) &=& \begin{vmatrix} 4 & 8 \\ -2 & -4 \end{vmatrix} = 4\cdot(-4) - (-2) \cdot 8 = -16+16 = 0 \\ a &=& -d \\ 4 &=& -(-4) \\ 4 &=&4\ \checkmark \\\\ A^2 &=& \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} \cdot \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} = 0\ \checkmark \\ \hline \end{array}\)

Solve for the positive value of x such that  \sqrt[3]{x^2 - 4x + 4} = 16

(\(\sqrt[3]{x^2 - 4x + 4}\) ) .

Formula:

\(\begin{array}{|rcll|} \hline && \sqrt[n]{a^m} \\ &=& (a^m)^\frac{1}{n} \\ &=& a^{m\cdot \frac{1}{n}} \\ &=& a^{\frac{m}{n}} \\ &=& a^{\frac{1}{n}\cdot m } \\ &=& (a^{\frac{1}{n}})^m \\ &=& (\sqrt[n]{a})^m \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sqrt[3]{x^2 - 4x + 4} &=& 16 \quad & | \quad x^2 - 4x + 4 = (x-2)^2 \\ \sqrt[3]{(x-2)^2} &=& 16 \quad & | \quad \sqrt[n]{a^m} = (\sqrt[n]{a})^m \\ \left(~ \sqrt[3]{x-2} ~\right)^2 &=& 16 \quad & | \quad \sqrt{} \\ \sqrt[3]{x-2} &=& 4 \quad & | \quad cube \\ x-2 &=& 4^3 \\ x-2 &=& 64 \\ x &=& 64+2 \\ \mathbf{x} & \mathbf{=} & \mathbf{66} \\ \hline \end{array}\)

x|x|=2x+1 

The lines are absolute value

\(\begin{array}{|rcll|} \hline x|x| &=& 2x+1 \quad & | \quad \text{square both sides} \\ x^2\cdot |x|^2 &=& (2x+1)^2 \quad & | \quad |x|^2 = x^2 \\ x^2\cdot x^2 &=& (2x+1)^2 \\ x^4 &=& 4x^2+4x+1 \\ \mathbf{x^4 - 4x^2+4x+1} & \mathbf{=} & \mathbf{0} \\ \hline \end{array} \)

solutions of \(x^4 - 4x^2+4x+1 = 0\)

Rhombus  ABCD has perimeter 148, and one of its diagonals has length 24.

How long is the other diagonal?

Let e one of its diagonal = 24

Let f the other diagonal = ?

Let a be the side of the rhombus. All sides are equal.

\(\mathbf{a= \ ? }\\ \begin{array}{|rcll|} \hline 4a &=& \text{perimeter} \\ a &=& \frac{ \text{perimeter} } {4} \\ a &=& \frac{ 148 } {4} \\ \mathbf{a} & \mathbf{=} & \mathbf{37} \\ \hline \end{array}\)

\(\mathbf{f= \ ? }\\ \begin{array}{|rcll|} \hline \mathbf{4a^2} & \mathbf{=} & \mathbf{e^2+f^2 } \\\\ 4\cdot 37^2 &=& 24^2+f^2 \\ 4\cdot 1369 &=& 576 + f^2 \\ 5476 &=& 576 + f^2 \quad & | \quad -576 \\ 5476-576 &=& f^2 \\ 4900 &=& f^2 \\ 70 &=& f \\\\ \mathbf{f} & \mathbf{=} & \mathbf{70} \\ \hline \end{array}\)

The other diagonal is 70

What is a:   sqrt(4+sqrt(16+16a))+sqrt(1+sqrt(1+a))=6

(\(\sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}}=6.\))

\(\begin{array}{|rcll|} \hline \sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ \sqrt{4+\sqrt{16(1+a)}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ \sqrt{4+4\sqrt{1+a}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ \sqrt{4(1+\sqrt{1+a})}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ 2\sqrt{1+\sqrt{1+a}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ 3\sqrt{1+\sqrt{1+a}} &=& 6 \quad & | \quad : 3 \\ \sqrt{1+\sqrt{1+a}} &=& 2 \quad & | \quad \text{square both sides} \\ 1+\sqrt{1+a} &=& 4 \quad & | \quad -1 \\ \sqrt{1+a} &=& 3 \quad & | \quad \text{square both sides} \\ 1+a &=& 9 \quad & | \quad -1 \\ \mathbf{ a } & \mathbf{=} & \mathbf{8} \\ \hline \end{array}\)

Proof:

\(\begin{array}{rcll} \sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}} &\overset{?}{=}& 6 \qquad a = 8\\ \sqrt{4+\sqrt{16+16\cdot 8}}+\sqrt{1+\sqrt{1+8}} & \overset{?}{=} & 6 \\ \sqrt{4+\sqrt{144}}+\sqrt{1+3} & \overset{?}{=} & 6 \\ \sqrt{4+12}+\sqrt{4} & \overset{?}{=} & 6 \\ \sqrt{16}+2 & \overset{?}{=} & 6 \\ 4+2 & \overset{?}{=} & 6 \\ 6 & \overset{!}{=} & 6 \quad \checkmark\\ \end{array}\)

Two altitudes of a triangle have lengths 12 and 14. What is the longest possible length of the third altitude,

if it is a positive integer?

Let h = the length of the third altitude.

Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h

Let A = Area of the triangle.

Let 2A = a*12
Let 2A = b*14
Let 2A = c*h

1.

\(\begin{array}{|rcll|} \hline a &=& \frac{2A}{12} \\ b &=& \frac{2A}{14} \\ c &=& \frac{2A}{h} \\ \hline \end{array}\)

2.

A triangle exists with side lengths a, b, and c
if and only if they satisfy the three triangle inequalities:

\(\begin{array}{|lrcll|} \hline (1) & a & < & b + c \\ (2) & b & < & c + a \\ (3) & c & < & a + b \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & \frac{2A}{12} & < & \frac{2A}{14} + \frac{2A}{h} \quad & | \quad :2A \\ & \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\\\ (2) & \frac{2A}{14} & < & \frac{2A}{h} + \frac{2A}{12} \quad & | \quad :2A \\ & \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\\\ (3) & \frac{2A}{h} & < & \frac{2A}{12} + \frac{2A}{14} \quad & | \quad :2A \\ & \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \hline \end{array}\)

(1):

\( \begin{array}{|rcll|} \hline \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\ \frac{1}{12}-\frac{1}{14} & < & \frac{1}{h} \\ \frac{12-14}{12\cdot 14} & < & \frac{1}{h} \\ \frac{2}{168} & < & \frac{1}{h} \\ \frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ \mathbf{ h } & \mathbf{ < } & \mathbf{ 84 } \\ \hline \end{array} \)

(2):

\(\begin{array}{|rcll|} \hline \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\ \frac{1}{14} - \frac{1}{12} & < & \frac{1}{h} \\ \frac{12-14}{14\cdot 12} & < & \frac{1}{h} \\ -\frac{2}{168} & < & \frac{1}{h} \\ -\frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ -h & < & 84 \quad & | \quad \cdot (-1) \qquad switch "<" to ">" \\ \mathbf{ h } & \mathbf{ > } & \mathbf{ -84 } \\ \hline \end{array} \)

(3):

\(\begin{array}{|rcll|} \hline \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \frac{1}{h} & < & \frac{14+12}{12\cdot 14} \\ \frac{1}{h} & < & \frac{26}{168} \\ \frac{1}{h} & < & \frac{13}{84} \quad & | \quad \cdot \frac{84}{13}\ h \\ \frac{84}{13} & < & h \\ \mathbf{ 6.46153846154 } & \mathbf{ < } & \mathbf{ h } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 6.46153846154 < h < 84 \\ \hline \end{array}\)

The longest possible length of the third altitude, if it is a positive integer is 83

When changing the base of a logarithm, from 

logb(a) = log(a)/log(b)

Is it possible for the bases of the new logarithms to be any number? or only base 10?

You can use any number:

Change of Base Formula

The change of base formula for logarithms is:

\(\begin{array}{|rcll|} \hline \log_a{(x)} &=& \dfrac{ \log_b{(x)} } { \log_b{(a)} } \\ \hline \end{array}\)

Example:

\(\begin{array}{|rcll|} \hline \log_4{(16)} &=& \log_4{(4^2)} \\ &=& 2 \\\\ \log_4{(16)} &=& \frac{ \log_2{(16)} }{ \log_2{(4)} } \\ &=& \frac{ \log_2{(2^4)} }{ \log_2{(2^2)} } \\ &=& \frac{ 4\cdot \log_2{(2)} }{ 2\cdot \log_2{(2)} } \\ &=& \frac{ 4 }{ 2 } \\ &=& 2 \quad \checkmark \\ \hline \end{array} \)