Find the eigenvalues for [latex]A=\left[\begin{array}{rr}3 & 3 \\6 & -4\end{array}\right][/latex].

Our strategy is to determine the values of λ that satisfy det(A − λI[latex]_2[/latex]) = 0. We have [latex]A-\lambda I_2=\left[\begin{array}{rr}3 & 3 \\6 & -4\end{array}\right]-\left[\begin{array}{ll}\lambda & 0 \\0 & \lambda\end{array}\right]=\left[\begin{array}{cc}(3-\lambda) & 3 \\6 & (-4-\lambda)\end{array}\right][/latex] Next, we compute the determinant, [latex]\operatorname{det}\left(A-\lambda I_2\right)=(3-\lambda)(-4-\lambda)-18=\lambda^2+\lambda-30[/latex] Setting det(A − λI ) = 0, we have [latex]\lambda^2+\lambda-30=0 \quad \Longrightarrow \quad(\lambda-5)(\lambda+6)=0 \quad \Longrightarrow \quad \lambda=5 \quad \text{or}\quad \lambda=-6[/latex] Thus the eigenvalues for A are λ = 5 and λ = −6.

Question 6.1.3: Find the eigenvalues for A =[3 3 6 -4].

Linear Algebra with Applications

Find the eigenvalues for $A=\left[\begin{array}{rr}3 & 3 \\6 & -4\end{array}\right]$ .

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Our strategy is to determine the values of λ that satisfy det(A − λI $_2$ ) = 0. We have

$A-\lambda I_2=\left[\begin{array}{rr}3 & 3 \\6 & -4\end{array}\right]-\left[\begin{array}{ll}\lambda & 0 \\0 & \lambda\end{array}\right]=\left[\begin{array}{cc}(3-\lambda) & 3 \\6 & (-4-\lambda)\end{array}\right]$

Next, we compute the determinant,

$\operatorname{det}\left(A-\lambda I_2\right)=(3-\lambda)(-4-\lambda)-18=\lambda^2+\lambda-30$

Setting det(A − λI ) = 0, we have

$\lambda^2+\lambda-30=0 \quad \Longrightarrow \quad(\lambda-5)(\lambda+6)=0 \quad \Longrightarrow \quad \lambda=5 \quad \text{or}\quad \lambda=-6$

Thus the eigenvalues for A are λ = 5 and λ = −6.