Question

If \[A = \left[ {\begin{array}{*{20}{c}}
  2&0&1 \\
  2&1&3 \\
  1&{ - 1}&0
\end{array}} \right]\] ,find\[\,{A^2} - 5A + 4I\] and hence find the matrix X such that \[{A^2} - 5A + 4I + X = 0\].

Answer 1

Hint: As you can see, this numerical is based on matrix. So, what is a matrix? In mathematics, matrix is defined as a set of numbers arranged in rows and columns, so as to form a rectangular array. In this numerical we need to find \[{A^2} - 5A + 4I\] , for that we will find \[{A^2} - 5A\,\,and\,\,4I\] separately and add them and we will get required answer.

Complete step by solution:
Given data: Matrix multiplication shares some properties with usual multiplication. However, matrix multiplication is not defined if the number of columns of the first factor differs from the number of rows of the second factor, and it is non-commutative, even when the product remains definite after changing the order of the factors
We are given the matrix A
\[A = \left[ {\begin{array}{*{20}{c}}
  2&0&1 \\
  2&1&3 \\
  1&{ - 1}&0
\end{array}} \right]\]
We need to find \[{A^2} - 5A + 4I\] , also find X when \[{A^2} - 5A + 4I + X = 0\] .
We will find \[{A^2}, - 5A,4I\] separately
We know,
\[
  {A^2} = A \cdot A \\
   \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  2&0&1 \\
  2&1&3 \\
  1&{ - 1}&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  2&0&1 \\
  2&1&3 \\
  1&{ - 1}&0
\end{array}} \right] \\
 \]
Multiplying the two matrices, we get
\[ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  5&{ - 1}&2 \\
  9&{ - 2}&5 \\
  0&{ - 1}&{ - 2}
\end{array}} \right]\]
Now,
\[
   \Rightarrow - 5A = - 5\left[ {\begin{array}{*{20}{c}}
  2&0&1 \\
  2&1&3 \\
  1&{ - 1}&0
\end{array}} \right] \\
   \Rightarrow - 5A = \left[ {\begin{array}{*{20}{c}}
  { - 10}&0&{ - 5} \\
  { - 10}&{ - 5}&{ - 15} \\
  { - 5}&5&0
\end{array}} \right] \\
 \]
Also,
\[
   \Rightarrow 4I = 4\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] \\
   \Rightarrow 4I = \left[ {\begin{array}{*{20}{c}}
  4&0&0 \\
  0&4&0 \\
  0&0&4
\end{array}} \right] \\
 \]
Now, we will add \[{A^2}, - 5A,4I\] together
\[
   \Rightarrow {A^2} - 5A + 4I = \left[ {\begin{array}{*{20}{c}}
  5&{ - 1}&2 \\
  9&{ - 2}&5 \\
  0&{ - 1}&{ - 2}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  { - 10}&0&{ - 5} \\
  { - 10}&{ - 5}&{ - 15} \\
  { - 5}&5&0
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  4&0&0 \\
  0&4&0 \\
  0&0&4
\end{array}} \right] \\
   \Rightarrow {A^2} - 5A + 4I = \left[ {\begin{array}{*{20}{c}}
  {5 - 10 + 4}&{ - 1 + 0 + 0}&{2 - 5 + 0} \\
  {9 - 10 - 0}&{ - 2 - 5 + 4}&{5 - 15 + 0} \\
  {0 - 5 + 0}&{ - 1 + 5 + 0}&{ - 2 + 0 + 4}
\end{array}} \right] \\
   \Rightarrow {A^2} - 5A + 4I = \left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 1}&{ - 3} \\
  { - 1}&{ - 3}&{ - 10} \\
  { - 5}&4&2
\end{array}} \right] \\
 \]
i.e.: the required matrix is given by
\[ \Rightarrow {A^2} - 5A + 4I = \left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 1}&{ - 3} \\
  { - 1}&{ - 3}&{ - 10} \\
  { - 5}&4&2
\end{array}} \right]\]
Now again, to find the value of X when \[{A^2} - 5A + 4I + X = 0\]
Here,
\[
  {A^2} - 5A + 4I + X = 0 \\
   \Rightarrow X = 0 - ({A^2} - 5A + 4I) \\
   \Rightarrow X = - ({A^2} - 5A + 4I) \\
   \Rightarrow X = - \left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 1}&{ - 3} \\
  { - 1}&{ - 3}&{ - 10} \\
  { - 5}&4&2
\end{array}} \right] \\
   \Rightarrow X = \left[ {\begin{array}{*{20}{c}}
  1&1&3 \\
  1&3&{10} \\
  5&{ - 4}&{ - 2}
\end{array}} \right] \\
 \]
Hence the required X matrix is given by
\[X = \left[ {\begin{array}{*{20}{c}}
  1&1&3 \\
  1&3&{10} \\
  5&{ - 4}&{ - 2}
\end{array}} \right]\]


Note: Students often make mistakes in multiplying the matrices. You should always be careful while computing the values of multiplication of matrices. Also, do not be confused between matrices and determinants. The identity matrix is denoted by a boldface one 1. Identity matrix is a square matrix of order n, and also a special kind of diagonal matrix. It is called an identity matrix because multiplication with it leaves a matrix unchanged.