Question

If $A = \left[ {\begin{array}{*{20}{c}}
  1&4&4 \\
  4&1&4 \\
  4&4&1
\end{array}} \right]$, then ${A^2} - 6A = $
$\left( a \right)27{I_3}$
$\left( b \right)5{I_3}$
$\left( c \right)20{I_3}$
$\left( d \right)30{I_3}$

Answer 1

Hint: In this particular question use the concept of matrix multiplication which is given as, $\left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {{a^2} + bd + cg}&{ab + be + ch}&{ac + bf + ci} \\
  {da + ed + fg}&{db + {e^2} + fh}&{dc + ef + fi} \\
  {ga + hd + ig}&{gb + he + ih}&{gc + hf + {i^2}}
\end{array}} \right]$, then use the concept of matrix subtraction, so use these concepts to reach the solution of the question.

Complete step by step answer:
$A = \left[ {\begin{array}{*{20}{c}}
  1&4&4 \\
  4&1&4 \\
  4&4&1
\end{array}} \right]$
Now we have to find out the value of ${A^2} - 6A = $
So, first find out the value of ${A^2}$
So use the concept of matrix multiplication so we have,
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  1&4&4 \\
  4&1&4 \\
  4&4&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&4&4 \\
  4&1&4 \\
  4&4&1
\end{array}} \right]$
Now multiply the matrices we have,
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {1 \times 1 + 4 \times 4 + 4 \times 4}&{1 \times 4 + 4 \times 1 + 4 \times 4}&{1 \times 4 + 4 \times 4 + 4 \times 1} \\
  {4 \times 1 + 1 \times 4 + 4 \times 4}&{4 \times 4 + 1 \times 1 + 4 \times 4}&{4 \times 4 + 1 \times 4 + 4 \times 1} \\
  {4 \times 1 + 4 \times 4 + 1 \times 4}&{4 \times 4 + 4 \times 1 + 1 \times 4}&{4 \times 4 + 4 \times 4 + 1 \times 1}
\end{array}} \right]$
Now simplify it we have,
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {33}&{24}&{24} \\
  {24}&{33}&{24} \\
  {24}&{24}&{33}
\end{array}} \right]$................ (1)
Now calculate the value of 6A, so we have,
$ \Rightarrow 6A = 6\left[ {\begin{array}{*{20}{c}}
  1&4&4 \\
  4&1&4 \\
  4&4&1
\end{array}} \right]$
Now when 6 multiplied inside the matrix it should be multiplied in all of the elements so we have,
\[ \Rightarrow 6A = \left[ {\begin{array}{*{20}{c}}
  6&{24}&{24} \\
  {24}&6&{24} \\
  {24}&{24}&6
\end{array}} \right]\].................. (2)
Now subtract (2) from (1) we have,
$ \Rightarrow {A^2} - 6A = \left[ {\begin{array}{*{20}{c}}
  {33}&{24}&{24} \\
  {24}&{33}&{24} \\
  {24}&{24}&{33}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  6&{24}&{24} \\
  {24}&6&{24} \\
  {24}&{24}&6
\end{array}} \right]$
$ \Rightarrow {A^2} - 6A = \left[ {\begin{array}{*{20}{c}}
  {33 - 6}&{24 - 24}&{24 - 24} \\
  {24 - 24}&{33 - 6}&{24 - 24} \\
  {24 - 24}&{24 - 24}&{33 - 6}
\end{array}} \right]$
$ \Rightarrow {A^2} - 6A = \left[ {\begin{array}{*{20}{c}}
  {27}&0&0 \\
  0&{27}&0 \\
  0&0&{27}
\end{array}} \right]$
Now take 27 common from the matrix we have,
$ \Rightarrow {A^2} - 6A = 27\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
Now as we know that $\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$ is known as identity matrix which is denoted by ${I_3}$, where subscript 3 denotes $\left( {3 \times 3} \right)$ matrix so we have,
$ \Rightarrow {A^2} - 6A = 27{I_3}$
So this is the required answer.

So, the correct answer is “Option A”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall how to multiply, add and subtract the matrices which is all stated above, and also recall what is identity matrix, so first multiply the matrix with itself as above, the multiply the given matrix with digit 6 as above, then subtract them we will get the required answer.