Question

If \[A = \left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  4&{ - 2}
\end{array}} \right]\], find the value of \[K\] such that \[{A^2} = KA - 2I\], where \[I\] is the identity element.

Answer 1

Hint: Here, in the question, we have been given a matrix \[A\] and a scalar \[K\] and also the relation between \[A\] and \[K\]. And we are asked to find the value of \[K\] using the given information. We will first multiply the matrix \[A\] with itself to get another matrix as \[{A^2}\] and then use the relation to get the desired answer.

Complete step-by-step answer:
Given \[A = \left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  4&{ - 2}
\end{array}} \right]\]
\[{A^2} = KA - 2I\]
Now we know \[{A^2} = A \times A\]
\[\therefore {A^2} = \left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  4&{ - 2}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  4&{ - 2}
\end{array}} \right]\]
Multiplying the matrices, we get
\[
  \therefore {A^2} = \left[ {\begin{array}{*{20}{c}}
  {9 + \left( { - 8} \right)}&{ - 6 + 4} \\
  {12 + \left( { - 8} \right)}&{ - 8 + 4}
\end{array}} \right] \\
   \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  4&{ - 4}
\end{array}} \right] \\
 \]
Now we have \[{A^2} = KA - 2I\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  4&{ - 4}
\end{array}} \right] = K\left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  4&{ - 2}
\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\]
Simplifying the above matrices, we get,
\[\left[ {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  4&{ - 4}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {3K}&{ - 2K} \\
  {4K}&{ - 2K}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  2&0 \\
  0&2
\end{array}} \right]\]
Simplifying it further, we obtain,
\[\left[ {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  4&{ - 4}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {3K - 2}&{ - 2K - 0} \\
  {4K - 0}&{ - 2K - 2}
\end{array}} \right]\]
Now, equating the corresponding elements of matrices, we get,
\[
  3K - 2 = 1 \\
   \Rightarrow K = \dfrac{3}{3} \\
   \Rightarrow K = 1 \\
 \]
\[
   - 2K = - 2 \\
   \Rightarrow K = 1 \\
 \]
\[
  4K = 4 \\
   \Rightarrow K = 1 \\
 \]
\[
   - 2K - 2 = - 4 \\
   \Rightarrow 2K + 2 = 4 \\
   \Rightarrow K = \dfrac{2}{2} \\
   \Rightarrow K = 1 \\
 \]
By comparing all the corresponding values, we get \[K = 1\].
\[K = 1\]
So, the correct answer is “Option B”.

Note: While solving such types of questions, it is important that we understand the application of mathematical operations to matrices. First, we should check the order of matrices.
A matrix having \[m\] rows and \[n\] columns is known as a matrix of order \[m \times n\] or simply \[m \times n\] matrices read as \[m\] by \[n\] matrix.
Two matrices can be added or subtracted (if they are of the same order) by adding or subtracting all the corresponding elements.
If \[A = {\left[ {{a_{ij}}} \right]_{m \times n}}\] is a matrix and \[k\] is any scalar, then \[kA\] is another matrix which is obtained by multiplying each element of \[A\] by a scalar \[k\].
Let \[A = {\left[ {{a_{ij}}} \right]_{m \times n}}\] and \[B = {\left[ {{b_{ij}}} \right]_{m \times n}}\], then the product of matrices \[A\] and \[B\] will be the matrix \[C\] of order \[m \times p\]. To find the \[{\left( {i,k} \right)^{th}}\] element \[{c_{ik}}\] of the matrix \[C\], we take \[{i^{th}}\] row of \[A\] and \[{k^{th}}\] column of \[B\], multiply them element-wise and take the sum of all these products.