Question

Consider the matrix \(A = \begin{bmatrix} 2 & 2 & 3 \\ 2 & 5 & 6 \\ 3 & 4 & 10 \end{bmatrix}\). Which ONE of the following options CORRECTLY lists the eigenvalues of \(A\)?

• A. \(1;\ 8-\sqrt{37};\ -8+\sqrt{37}\)
• B. \(2;\ 8-\sqrt{37};\ -8+\sqrt{37}\)
• C. \(1;\ 8+\sqrt{37};\ 8-\sqrt{37}\)
• D. \(2;\ 8+\sqrt{37};\ 8+\sqrt{37}\)

Hint:

Expand det(A - lambda I) into a cubic, find one integer root by inspection, then factor out a quadratic and use the quadratic formula for the remaining two eigenvalues.

Answer 1

The Correct Option is C

Step 1: Set up the characteristic equation. The eigenvalues of \(A\) satisfy \(\det(A-\lambda I)=0\), i.e. \(\det\begin{bmatrix} 2-\lambda & 2 & 3 \\ 2 & 5-\lambda & 6 \\ 3 & 4 & 10-\lambda \end{bmatrix}=0\).

Step 2: Expand the determinant along the first row. \(\det = (2-\lambda)\big[(5-\lambda)(10-\lambda)-24\big] - 2\big[2(10-\lambda)-18\big] + 3\big[8-3(5-\lambda)\big]\).

Step 3: Simplify each bracket. \((5-\lambda)(10-\lambda)-24 = \lambda^2-15\lambda+50-24=\lambda^2-15\lambda+26\). Also \(2(10-\lambda)-18 = 2-2\lambda\), and \(8-3(5-\lambda) = -7+3\lambda\).

Step 4: Substitute back and expand. \((2-\lambda)(\lambda^2-15\lambda+26) = -\lambda^3+17\lambda^2-56\lambda+52\). Adding \(-2(2-2\lambda) = -4+4\lambda\) and \(3(-7+3\lambda)=-21+9\lambda\), the determinant becomes \(-\lambda^3+17\lambda^2-43\lambda+27\).

Step 5: Solve the characteristic equation. Setting the determinant to zero gives \(\lambda^3-17\lambda^2+43\lambda-27=0\). Testing \(\lambda=1\): \(1-17+43-27=0\), so \(\lambda=1\) is a root. Dividing out \((\lambda-1)\) gives \(\lambda^2-16\lambda+27=0\).

Step 6: Solve the remaining quadratic. By the quadratic formula, \(\lambda = \dfrac{16\pm\sqrt{256-108}}{2} = \dfrac{16\pm\sqrt{148}}{2} = 8\pm\sqrt{37}\).

Step 7: Conclude. The three eigenvalues are \(1,\ 8+\sqrt{37},\ 8-\sqrt{37}\), which matches option (C). Options (A), (B) and (D) either have the wrong sign pattern for the surd terms or the wrong first eigenvalue, so they are incorrect.