Question

If $ A = \left[ {\begin{array}{*{20}{c}}
  1&0&2 \\
  0&2&1 \\
  2&0&3
\end{array}} \right] $ and $ {A^3} - 6{A^2} + 7A + K{I_3} = 0 $ find k.

Answer 1

Hint: Matrix is the set of the arranged numbers. We get its multiplication by multiplying corresponding rows into the columns. First we will find the square of the matrix and then the square matrix is multiplied with the matrix for the cube of matrix and at last will substitute values in the given equation.

Complete step-by-step answer:
First find the square of the matrix.
 $ {A^2} = A \times A $
Place value in the above equation –
 $ {A^2} = \left[ {\begin{array}{*{20}{c}}
  1&0&2 \\
  0&2&1 \\
  2&0&3
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  1&0&2 \\
  0&2&1 \\
  2&0&3
\end{array}} \right] $
Multiply the above matrices and open the brackets with multiplying subsequent rows with columns.
 $ {A^2} = \left[ {\begin{array}{*{20}{c}}
  {1(1) + 0(0) + 2(2)}&{0(1) + 2(0) + 0(2)}&{2(1) + 1(0) + 3(2)} \\
  {1(0) + 0(2) + 2(1)}&{0(0) + 2(2) + 0(1)}&{2(0) + 1(2) + 3(1)} \\
  {1(2) + 0(0) + 2(3)}&{0(2) + 2(0) + 0(3)}&{2(2) + 1(0) + 3(3)}
\end{array}} \right] $
Simplify the above equation –
 $ {A^2} = \left[ {\begin{array}{*{20}{c}}
  {1 + 0 + 4}&{0 + 0 + 0}&{2 + 0 + 6} \\
  {0 + 0 + 2}&{0 + 4 + 0}&{0 + 2 + 3} \\
  {2 + 0 + 6}&{0 + 0 + 0}&{4 + 0 + 9}
\end{array}} \right] $
Simplify –
 $ {A^2} = \left[ {\begin{array}{*{20}{c}}
  5&0&8 \\
  2&4&5 \\
  8&0&{13}
\end{array}} \right] $ ..... (A)
Now, the cube of the matrix is equal to product of the square matrix with matrix.
 $ {A^3} = {A^2} \cdot A $
Place the values in the above equation –
 $ {A^3} = \left[ {\begin{array}{*{20}{c}}
  5&0&8 \\
  2&4&5 \\
  8&0&{13}
\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}
  1&0&2 \\
  0&2&1 \\
  2&0&3
\end{array}} \right] $
Multiply and simplify –
 $ \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  {21}&0&{34} \\
  {12}&8&{23} \\
  {34}&0&{55}
\end{array}} \right] $ ..... (B)
Now, the given equation –
 $ {A^3} - 6{A^2} + 7A + K{I_3} = 0 $
Place values in the above equation –
 $ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {21}&0&{34} \\
  {12}&8&{23} \\
  {34}&0&{55}
\end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}}
  5&0&8 \\
  2&4&5 \\
  8&0&{13}
\end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}}
  1&0&2 \\
  0&2&1 \\
  2&0&3
\end{array}} \right] + k\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = 0 $
Multiply the numbers outside with the respective matrices.
 $ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {21}&0&{34} \\
  {12}&8&{23} \\
  {34}&0&{55}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {30}&0&{48} \\
  {12}&{24}&{30} \\
  {48}&0&{48}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  7&0&{14} \\
  0&{14}&7 \\
  {14}&0&{21}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  k&0&0 \\
  0&k&0 \\
  0&0&k
\end{array}} \right] = 0 $
Open the brackets, and apply operations accordingly
 $ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {21 - 30 + 7 + k}&{0 + 0 + 0 + 0}&{34 - 48 + 14 + 0} \\
  {12 - 12 + 0 + 0}&{8 - 24 + 14 + k}&{23 - 30 + 7 + 0} \\
  {34 - 48 + 14 + 0}&{0 + 0 + 0 + 0}&{55 - 48 + 21 + k}
\end{array}} \right] = 0 $
Simplify the above matrix
 $ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  { - 2 + k}&0&0 \\
  0&{ - 2 + k}&0 \\
  0&0&{ - 2 + k}
\end{array}} \right] = 0 $
The above equation implies –
 $ - 2 + k = 0 $
Make the required unknown “K” the subject –
 $ \Rightarrow k = 2 $ is the required answer.
So, the correct answer is “2”.

Note: Matrix was introduced by the British Mathematician Arthur Cayley in the year 1858. We get the transpose of the square matrix by changing the rows of the original matrix changes to columns and columns are changed to rows. It is denoted by $ {A^T} $ . The determinant of the matrix is a special number which can be calculated from the square matrix. Be careful while multiplying the matrices, as a single mistake will lead to the wrong required answer. Follow step by step approach, it's lengthy but easy.