Question

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Find a formula for the eigenvalues of $A$ in terms of $a$, $b$, $c$ and $d$. a. $= \lambda^2 - d\lambda - a\lambda + ad - bc$ b. $= (\lambda - a)(\lambda - d) - bc$ c. $= \frac{(a + d) \pm \sqrt{(a - d)^2 + 4bc}}{2}$ d. $= \frac{(a + d) \pm \sqrt{a^2 + d^2 - 4ad - 4bc}}{2}$ e. $= \frac{(a + d) \pm \sqrt{a^2 - 2ad + d^2 + 4bc}}{2}$ f. $p(\lambda) = \begin{vmatrix} \lambda - a & -b \\ -c & \lambda - d \end{vmatrix}$ g. $p(\lambda) = 0 \rightarrow$ h. $\lambda = \frac{(a + d) \pm \sqrt{-(a + d)^2 - 4(ad - bc)}}{2}$ i. $= \lambda^2 - (a + d)\lambda + ad - bc$ j. $= \frac{(a + d) \pm \sqrt{a^2 + d^2 - 4ad + 4bc}}{2}$

          Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Find a formula for the eigenvalues of $A$ in terms of $a$, $b$, $c$ and $d$.
a. $= \lambda^2 - d\lambda - a\lambda + ad - bc$
b. $= (\lambda - a)(\lambda - d) - bc$
c. $= \frac{(a + d) \pm \sqrt{(a - d)^2 + 4bc}}{2}$
d. $= \frac{(a + d) \pm \sqrt{a^2 + d^2 - 4ad - 4bc}}{2}$
e. $= \frac{(a + d) \pm \sqrt{a^2 - 2ad + d^2 + 4bc}}{2}$
f. $p(\lambda) = \begin{vmatrix} \lambda - a & -b \\ -c & \lambda - d \end{vmatrix}$ 
g. $p(\lambda) = 0 \rightarrow$
h. $\lambda = \frac{(a + d) \pm \sqrt{-(a + d)^2 - 4(ad - bc)}}{2}$
i. $= \lambda^2 - (a + d)\lambda + ad - bc$
j. $= \frac{(a + d) \pm \sqrt{a^2 + d^2 - 4ad + 4bc}}{2}$

Let A =
< b m a t r i x >. Find a formula for the eigenvalues of A in terms of a, b, c and d.
a. = λ^2 - dλ - aλ + ad - bc
b. = (λ - a)(λ - d) - bc
c. = ((a + d) ±√((a - d)^2 + 4bc))/(2)
d. = ((a + d) ±√(a^2 + d^2 - 4ad - 4bc))/(2)
e. = ((a + d) ±√(a^2 - 2ad + d^2 + 4bc))/(2)
f. p(λ) = λ - a -b
-c λ - d
g. p(λ) = 0 →
h. λ = ((a + d) ±√(-(a + d)^2 - 4(ad - bc)))/(2)
i. = λ^2 - (a + d)λ + ad - bc
j. = ((a + d) ±√(a^2 + d^2 - 4ad + 4bc))/(2)

Added by William G.

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