Let \(A=\left[\begin{matrix}2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b\end{matrix}\right]\). If \({A}^{3}=4{A}^{2}-A-21I\), whe…

Given \(A^3-4A^2+A+21I=O\)

Therefore every eigenvalue of \(A\) satisfies

\(x^3-4x^2+x+21=0\)

\(=(x-3)(x^2-x-7)\)

Since the factors are distinct, the eigenvalues of \(A\) are

\(3,\ \dfrac{1+\sqrt{29}}{2},\ \dfrac{1-\sqrt{29}}{2}\)

The trace of \(A\) is

\(\operatorname{tr}(A)=2+3+b=5+b\)

Also, the sum of eigenvalues is

\(3+\dfrac{1+\sqrt{29}}{2}+\dfrac{1-\sqrt{29}}{2}=4\)

\(5+b=4\)

\(\Rightarrow b=-1\)

\(\det(A)=2(3b-5)-ab\)

Putting \(b=-1\),

\(\det(A)=-16+a\)

The product of eigenvalues is

\(3\cdot\dfrac{1+\sqrt{29}}{2}\cdot\dfrac{1-\sqrt{29}}{2}\)

\(=3\cdot\dfrac{1-29}{4}\)

\(=-21\)

\(-16+a=-21\)

\(\Rightarrow a=-5\)

\(2a+3b=2(-5)+3(-1)\)

\(=-10-3\)

\(=-13\)