Question

The eigenvalues of a \(3\times3\) real matrix \(A\) are \(0\), \(-2\), \(-2\) and it satisfies the equation \[ A^4+4A^3=KA^2, \] then the value of \(K\) is

• A. \(-3\)
• B. \(-4\)
• C. \(3\)
• D. \(4\)

Hint:

If a matrix satisfies \[ P(A)=0, \] then every eigenvalue \(\lambda\) satisfies \[ \boxed{P(\lambda)=0.} \]

Answer 1

The Correct Option is B

Step 1: Use the polynomial satisfied by the eigenvalues. If \[ A^4+4A^3=KA^2, \] then every eigenvalue \(\lambda\) of \(A\) satisfies \[ \lambda^4+4\lambda^3=K\lambda^2. \]

Step 2: Substitute a non-zero eigenvalue. Since one eigenvalue is \[ \lambda=-2, \] we obtain \[ (-2)^4+4(-2)^3 = K(-2)^2. \] That is, \[ 16-32=4K, \] \[ -16=4K, \] \[ K=-4. \] Hence, \[ \boxed{K=-4.} \] Therefore, \[ \boxed{(B)} \] is the correct answer.