Consider A=⎡⎢⎣a210b00−3c⎤⎥⎦ where a,b,c are the roots of the equation x3−3x2+2x−1=0. If matrix B is such that AB=BA,|A+B−2I|≠0 and A2−B2=4I−4B, where I is the 3×3 identity matrix, then the value of √det(B) is
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Solution
|A|=∣∣
∣∣a210b00−3c∣∣
∣∣=abc=1
where x3−3x2+2x−1=(x−a)(x−b)(x−c)⋯(1)
Given, A2−B2=4I−4B ⇒A2=B2−4B+4I⇒A2−(B−2I)2=O⇒(A+B−2I)(A−B+2I)=O(∵AB=BA)
Since |A+B−2I|≠0, ∴A−B+2I=0⇒B=A+2I=∣∣
∣∣a+2210b+200−3c+2∣∣
∣∣
|B|=(a+2)(b+2)(c+2)
Putting x=−2 in equation (1), we get −8−12−4−1≡(−2−a)(−2−b)(−2−c)⇒25=(a+2)(b+2)(c+2)∴|B|=25∴√|B|=5