Question

If $M = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right]$ and ${M^2} - \lambda M - I = 0$, then $\lambda $ equals
A. -2
B. 2
C. -4
D. 4

Answer 1

Hint: First, find the value of ${M^2}$ by multiplying the matrix $M$ with itself. Then find the value of $\lambda M$ by multiplying each entry/element by $\lambda $. Now substitute the values of ${M^2}$, $\lambda M$ and $I$ in the equation ${M^2} - \lambda M - I = 0$.

Formula Used: $\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  e&f \\
  g&h
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\left( a \right)\left( e \right) + \left( b \right)\left( g \right)}&{\left( a \right)\left( f \right) + \left( b \right)\left( h \right)} \\
  {\left( c \right)\left( e \right) + \left( d \right)\left( g \right)}&{\left( c \right)\left( f \right) + \left( d \right)\left( h \right)}
\end{array}} \right]$

Complete step by step Solution:
Let us first find the value of ${M^2}$ by multiplying the matrix with itself.
\[{M^2} = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\left( 1 \right)\left( 1 \right) + \left( 2 \right)\left( 2 \right)}&{\left( 1 \right)\left( 2 \right) + \left( 2 \right)\left( 3 \right)} \\
  {\left( 2 \right)\left( 1 \right) + \left( 3 \right)\left( 2 \right)}&{\left( 2 \right)\left( 2 \right) + \left( 3 \right)\left( 3 \right)}
\end{array}} \right]\]
${M^2} = \left[ {\begin{array}{*{20}{c}}
  5&8 \\
  8&{13}
\end{array}} \right]$
Let us now find the value of $\lambda M$ and $I$.
$\lambda M = \left[ {\begin{array}{*{20}{c}}
  \lambda &{2\lambda } \\
  {2\lambda }&{3\lambda }
\end{array}} \right]$
$I = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
It is given that ${M^2} - \lambda M - I = 0$. Therefore,
$\left[ {\begin{array}{*{20}{c}}
  5&8 \\
  8&{13}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  \lambda &{2\lambda } \\
  {2\lambda }&{3\lambda }
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = 0$
$\left[ {\begin{array}{*{20}{c}}
  {5 - \lambda - 1}&{8 - 2\lambda } \\
  {8 - 2\lambda }&{13 - 3\lambda - 1}
\end{array}} \right] = 0$
Solving any of the four elements, let’s say $8 - 2\lambda = 0$, we get $\lambda = 4$.

Therefore, the correct answer is option (D)

Note: A shorter and alternate way of doing this question: Write the equation ${M^2} - \lambda M - I = 0$ as ${M^2} - \lambda M = I$. Let us just solve for the first element in all the 3 matrices. The first element in\[{M^2}\] is \[\left( 1 \right)\left( 1 \right) + \left( 2 \right)\left( 2 \right) = 5\], the first element in $\lambda M$ is $\lambda $ and the first element in $I$ is $1$. Therefore, $5 - \lambda = 1$ and from this we get $\lambda = 4$.