EQUATION AX = λX.Calculation:\(M = \LEFT[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ 1&{ - 2}&1\\ 0&{ - 1}&1 \end{array}} \right]\)\(\left| {M - λ I} \right| = \left| {\begin{array}{*{20}{c}} {1 - λ }&{ - 1}&0\\ 1&{ - 2 - λ }&1\\ 0&{ - 1}&{1 - λ } \end{array}} \right| = 0\)(1 – λ) [(-2 – λ) (1 – λ) + 1] + 1 (1 – λ) = 0(1 – λ)[(-2 – λ) (1 – λ) + 1 + 1] = 0 (1 – λ)[- 2 + 2λ - λ + λ2 + 2] = 0⇒ λ (λ + 1) (1 - λ) = 0⇒ λ = 0, ±1To cross check:DET: 1 (-2 + 1) + 1(1 - 0) + 0 = -1 + 1 = 0Product of eigen values = 0 Eigen vector corresponding to the Eigen value λ = 0 isMX = λX\(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ 1&{ - 2}&1\\ 0&{ - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\)From the above matrix representation,x – y = 0 ⇒ x = YX – 2y + z = 0 ⇒ y = zTherefore, x = y = zLet, x = y = z = kNow, the Eigen vector is \(\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} k\\ k\\ k \end{array}} \right]\)For k = 1, the Eigen vector becomes \(\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\\ 1\\ 1 \end{array}} \right]\)