Question

Puteti va rog frumos sa mi faceti exercitiul astea pe o foaie

Answer 1

$A=\left(\begin{array}{ccc}a-1&1\\1&a-1\\\end{array}\right)$

a)

$A(2)\cdot A(-3)=\left(\begin{array}{ccc}1&1\\1&1\\\end{array}\right)\cdot \left(\begin{array}{ccc}-4&1\\1&-4\\\end{array}\right)\\\\=\left(\begin{array}{ccc}-3&-3\\-3&-3\\\end{array}\right)$

b)

$A\cdot A=\left(\begin{array}{ccc}a-1&1\\1&a-1\\\end{array}\right)\cdot \left(\begin{array}{ccc}a-1&1\\1&a-1\\\end{array}\right)=\left(\begin{array}{ccc}(a-1)^2+1&2(a-1)\\2(a-1)&(a-1)^2+1\\\end{array}\right)\\\\=\left(\begin{array}{ccc}a(a-2)+2&2(a-1)\\2(a-1)&a(a-2)+2\\\end{array}\right)$

$(2a-2)A-(a^2-2a)I_2=\left(\begin{array}{ccc}2(a-1)^2&2a-2\\2a-2&2(a-1)^2\\\end{array}\right)-\left(\begin{array}{ccc}a^2-2a&0\\0&a^2-2a\\\end{array}\right)=\\\\=\left(\begin{array}{ccc}a(a-2)+2&2(a-1)\\2(a-1)&a(a-2)+2\\\end{array}\right)$

Se observa ca sunt egale

c)

$A^2=2A\\\\\left(\begin{array}{ccc}a(a-2)+2&2(a-1)\\2(a-1)&a(a-2)+2\\\end{array}\right)=\left(\begin{array}{ccc}2(a-1)&2\\2&2(a-1)\\\end{array}\right)$

Se egaleaza membrii si obtinem:

2(a-1)=2

a-1=1

a=2

Un alt exercitiu gasesti aici: https://brainly.ro/tema/2721055

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