Question

If \[M = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right)\;\] and \[{M^2} - \lambda M - {I_2} = O\;\],where \[\lambda \] is constant, then \[\lambda \] equals
A) \[ - 2\]
B) \[2\]
C) \[ - 4\]
D) \[4\]

Answer 1

Hint: Add the equation \[{M^2} - \lambda M - {I_2} = O\;\]by \[\lambda M\]on both sides of the equation. Now square the matrix M and substitute the value of I. Now subtract both the terms and the answer has to be converted in terms of M. So, at last cancelling M from both sides we will get the value of \[\lambda \].

Complete step-by-step answer:
Given \[M = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right)\;\] and \[{M^2} - \lambda M - {I_2} = O\;\]
Where \[{I_2} = \left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\] (An identity matrix is a square matrix having 1s on the main diagonal, and 0s everywhere else.)
\[{M^2} - \lambda M - {I_2} = O\;\]
Add both sides by \[\lambda M\]
\[ \Rightarrow {M^2} - {I_2} = \lambda M\;\]
Now substituting moving LHS to RHS and RHS to LHS. We get
\[ \Rightarrow \lambda M = {M^2} - {I_2}\;\]
Substituting the value of matrix M and identity matrix \[{I_2}\]on RHS, we get
\[ \Rightarrow \lambda M = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right)\;\left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right)\; - \left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\;\]
Solving for \[{M^2}\], we get the value as
Now I need to multiply the ROWS of M by the COLUMNS of M. By this I mean that I first take the first row of M and the first column of M, and I multiply the first entries and then the second entries, and then I add the two products. The sum is one entry in the product matrix MM; in fact, being the product of row 1 and column 1, the result is the 1,1-entry of MM. Then I continue in like manner. For instance, the sum of the products from row 2 of M and column 1 of M is the 2,1-entry of MM.
 \[ \Rightarrow \lambda M = \left( {\begin{array}{*{20}{c}}
  {1 + 4}&{2 + 6} \\
  {2 + 6}&{4 + 9}
\end{array}} \right)\; - \left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\;\]
 \[ \Rightarrow \lambda M = \left( {\begin{array}{*{20}{c}}
  5&8 \\
  8&{13}
\end{array}} \right)\; - \left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\;\]
Now we subtract the two matrices.
\[ \Rightarrow \lambda M = \left( {\begin{array}{*{20}{c}}
  {5 - 1}&{8 - 0} \\
  {8 - 0}&{13 - 1}
\end{array}} \right)\;\]
\[ \Rightarrow \lambda M = \left( {\begin{array}{*{20}{c}}
  4&8 \\
  8&{12}
\end{array}} \right)\;\]
Taking 4 common from the matrix on the RHS side.
\[ \Rightarrow \lambda M = 4\left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right)\;\]
Replacing the value of \[M = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right)\;\]
\[ \Rightarrow \lambda M = 4M\]
Now cancelling M from both sides, we get
\[ \Rightarrow \lambda = 4\]

So, option (D) is the correct answer.

Note: We were able to calculate the value by using the simple algebra of linear equations. The basic properties of addition for real numbers also hold true for matrices such as commutative, associative, distributive etc. Whereas unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices.