Question

Let${\text{A = }}\left( {\begin{array}{*{20}{c}}
  6&2&2 \\
  { - 2}&{ - 3}&{ - 1} \\
  2&{ - 1}&3
\end{array}} \right)$and\[f(x) = \mid xI - A\mid \], then$f(A)$equals
$(a){\text{ 0}}$
$(b){\text{ A}}$
$(c){\text{ 2A}}$
$(d){\text{ - A}}$

Answer 1

Hint: We are asked to solve the equation in the above question by using the given matrix. Simply, substitute the given matrix and the identity matrix in the given equation\[f(x) = \mid xI - A\mid \] and evaluate it further.

Complete step-by-step answer:
We have the given matrix as

${\text{A = }}\left( {\begin{array}{*{20}{c}}

  6&2&2 \\

  { - 2}&{ - 3}&{ - 1} \\

  2&{ - 1}&3

\end{array}} \right)$

Now, we have the given equation as\[f(x) = \mid xI - A\mid \], therefore after substituting, the values in this equation, we get

\[ \Rightarrow \left| {xI - A} \right|{\text{ = }}\left| {x\left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}

  6&2&2 \\

  { - 2}&{ - 3}&{ - 1} \\

  2&{ - 1}&3

\end{array}} \right)} \right|\]

$ \Rightarrow \left| {xI - A} \right|{\text{ = }}\left( {\begin{array}{*{20}{c}}

  {x - 6}&{ - 2}&{ - 2} \\

  2&{x + 3}&1 \\

  { - 2}&1&{x - 3}

\end{array}} \right)$

After applying ${R_3} \to {R_2} + {R_3}$, we get

$ \Rightarrow \left| {xI - A} \right|{\text{ = }}\left( {\begin{array}{*{20}{c}}

  {x - 6}&{ - 2}&{ - 2} \\

  2&{x + 3}&1 \\

  0&{x + 4}&{x - 2}

\end{array}} \right)$

$ = (x - 6)[(x - 2)(x + 3) - 1(x + 4)] - 2[ - 2(x - 2) + 2(x + 4)]$

$ = (x - 6)(({x^2} + 3x - 2x - 6) - x - 4) - 2( - 2x + 4 + 2x + 8)$

$ = (x - 6)({x^2} - 10) - 2(12)$

$ = {x^3} - 10x - 6{x^2} + 60 - 24$

$ = {x^3} - 10x - 6{x^2} + 36$

$f(x) = \left| {xI - A} \right| = {x^3} - 6{x^2} - 10x + 36$ … (1)

Now, after substituting the value of the given matrix in equation (1), we get,

$f(A) = {A^3} - 6{A^2} - 10A + 36I$

$ = \left( {\begin{array}{*{20}{c}}

  {240}&{44}&{116} \\

  {68}&{ - 30}&{ - 34} \\

  {140}&{14}&{78}

\end{array}} \right) - 6\left( {\begin{array}{*{20}{c}}

  {36}&4&{16} \\

  { - 8}&6&{ - 4} \\

  {20}&4&{14}

\end{array}} \right) - 10\left( {\begin{array}{*{20}{c}}

  6&2&2 \\

  { - 2}&{ - 3}&{ - 1} \\

  2&{ - 1}&3

\end{array}} \right) + 36\left( {\begin{array}{*{20}{c}}

  1&0&0 \\

  0&1&0 \\

  0&0&1

\end{array}} \right)$

$ = \left( {\begin{array}{*{20}{c}}

  {240}&{44}&{116} \\

  {68}&{ - 30}&{ - 34} \\

  {140}&{14}&{78}

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

  { - 216}&{ - 24}&{ - 96} \\

  {48}&{ - 36}&{24} \\

  { - 120}&{ - 24}&{ - 84}

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

  { - 60}&{ - 20}&{ - 20} \\

  {20}&{30}&{10} \\

  { - 20}&{10}&{ - 30}

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

  {36}&0&0 \\

  0&{36}&0 \\

  0&0&{36}

\end{array}} \right)$

$ = \left( {\begin{array}{*{20}{c}}

  {240}&{44}&{116} \\

  {68}&{ - 30}&{ - 34} \\

  {140}&{14}&{78}

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

  { - 216}&{ - 24}&{ - 96} \\

  {48}&{ - 36}&{24} \\

  { - 120}&{ - 24}&{ - 84}

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

  { - 60 + 36}&{ - 20}&{ - 20} \\

  {20}&{30 + 36}&{10} \\

  { - 20}&{10}&{ - 30 + 36}

\end{array}} \right)$

$ = \left( {\begin{array}{*{20}{c}}

  {240}&{44}&{116} \\

  {68}&{ - 30}&{ - 34} \\

  {140}&{14}&{78}

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

  { - 24 - 216}&{ - 20 - 24}&{ - 20 - 96} \\

  {20 + 48}&{66 - 36}&{10 + 24} \\

  { - 20 - 120}&{10 - 24}&{6 - 84}

\end{array}} \right)$

$ = \left( {\begin{array}{*{20}{c}}

  {240}&{44}&{116} \\

  {68}&{ - 30}&{ - 34} \\

  {140}&{14}&{78}

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

  { - 240}&{ - 44}&{ - 116} \\

  { - 68}&{30}&{34} \\

  { - 140}&{ - 14}&{ - 78}

\end{array}} \right)$

$ = \left( {\begin{array}{*{20}{c}}

  0&0&0 \\

  0&0&0 \\

  0&0&0

\end{array}} \right)$

$ = 0$

Hence, the required solution is the option$(a){\text{ 0}}$.

Note: When we face such a type of problem, the key point is to have an adequate knowledge of the matrices and its various properties like addition, multiplication, subtraction, etc. Substitute the matrices in the given expression and perform the required operations like addition, multiplication, etc. to obtain the required solution.