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Sum of Opposite Angles of a Cyclic Quadrilateral

If Show that ...

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If Show that AB ≠ BA

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Solution

Hence,

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a = \frac{- 1}{7}

b = \frac{3}{7},

a b = b a .

= [\begin{matrix} a & b - b & a \end{matrix}]

B = [\begin{matrix} c & d - d & c \end{matrix}]

show that A product B commute

i. e. AB = BA also verify that

| A B | = | A | . | B |

A = [\begin{matrix} 4 & 1 3 & 2 \end{matrix}]

B = [\begin{matrix} 1 & 3 4 & 2 \end{matrix}]

A B \neq B A

a > b

a^{b} b^{b} > a^{b} b^{a}

log \frac{b}{a} < log \frac{1 + b}{1 + a}

A = [\begin{matrix} 1 & - 2 & 3 3 & 2 & - 1 \end{matrix}]

B = ⎡ ⎢ ⎣ \begin{matrix} 2 & 3 - 1 & 2 4 & - 5 \end{matrix} ⎤ ⎥ ⎦

. Find AB and BA and show that

A B \neq B A .

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Sum of Opposite Angles of a Cyclic Quadrilateral

Standard IX Mathematics

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