if-matrix-A-2-1-3-0-1-then-matrix-A-

Question Number 119124 by abdullahquwatan last updated on 22/Oct/20

if matrix A^{2} = (\begin{matrix} 1 3 \\ 0 1 \end{matrix}) then matrix A = \dots

Answered by benjo_mathlover last updated on 22/Oct/20

\Leftrightarrow A = (\begin{matrix} 1 \frac{3}{2} \\ 0 1 \end{matrix})

A^{2} = (\begin{matrix} 1 \frac{3}{2} \\ 0 1 \end{matrix}) (\begin{matrix} 1 \frac{3}{2} \\ 0 1 \end{matrix}) = (\begin{matrix} 1 \frac{6}{2} \\ 0 1 \end{matrix})

Commented by abdullahquwatan last updated on 22/Oct/20

thank you sir

Answered by mindispower last updated on 23/Oct/20

d e t (A^{2}) = 1, i f d e t A = 1

d e t (A^{2} - {a I}_{2}) = 0 \Rightarrow a = 1

i f d e t (A - I_{2}) = 0

A = (\begin{matrix} a b \\ c d \end{matrix})

X^{2} - (a + d) X + d e t (A) = P (X)

P (1) = 1 - (a + d) + 1 = 0 \Rightarrow a + d = 2

A^{2} = (\begin{matrix} a^{2} + b c b a + b d \\ a c + c d b c + d^{2} \end{matrix})

a c + c d = c (a + d) = 0 \Rightarrow c = 0

b a + b d = b (a + d) = 2 b = 3 \Rightarrow b = \frac{3}{2}

b c + d^{2} = 1 \Rightarrow d = \underline{+} 1

a = \underline{+} 1, a + d = 2 \Rightarrow a = d = 1

A = (\begin{matrix} 1 \frac{3}{2} \\ 0 1 \end{matrix})