Question

If A = \left( {\begin{array}{*{20}{l}}1&{ - 3}\\\2&K;\end{array}} \right) and ${A^2} - 4A + 10I = A$ then $K=$
A.1 or 4
B.4
C.$- 4$
D.0

Answer 1

Here we will first find the value of ${A^2}$ by using the multiplication rule of a matrix. Then we will find the value of $4A$ and $10I$ using scalar multiplication. Finally, we will substitute all the values in the given condition and get one matrix on both sides. We will compare each term of the matrix on both sides and form the equation. We will then solve the equation to get our answer.

Complete step-by-step answer:
We have our matrix as A = \left( {\begin{array}{*{20}{l}}1&{ - 3}\\\2&K;\end{array}} \right).
So, now we will find the value of each of the term of the condition separately
First, we will find the value of ${A^2}$ as:
${A^2} = A \times A$
Substituting A = \left( {\begin{array}{*{20}{l}}1&{ - 3}\\\2&K;\end{array}} \right) in the above equation, we get
\Rightarrow {A^2} = \left( {\begin{array}{*{20}{l}}1&{ - 3}\\\2&K;\end{array}} \right) \times \left( {\begin{array}{*{20}{l}}1&{ - 3}\\\2&K;\end{array}} \right)
Now by using matrix multiplication method, we get
{A^2} = \left( {\begin{array}{*{20}{l}}{1 \times 1 + \left( { - 3} \right) \times 2}&{1 \times \left( { - 3} \right) + \left( { - 3} \right) + K}\\\\{2 \times 1 + K \times 2}&{2 \times \left( { - 3} \right) + K \times K}\end{array}} \right)
Simplifying the equation, we get
{A^2} = \left( {\begin{array}{*{20}{l}}{ - 5}&{ - 3K - 3}\\\\{2K + 2}&{{K^2} - 6}\end{array}} \right)…………….$\left( 1 \right)$
Next, we will find the value of $- 4A$ .
4A = 4\left( {\begin{array}{*{20}{l}}1&{ - 3}\\\2&K;\end{array}} \right)
Using scalar multiplication, we get
\Rightarrow 4A = \left( {\begin{array}{*{20}{l}}4&{ - 12}\\\8&{4K}\end{array}} \right)…………..…….$\left( 2 \right)$
Next, we will find the value of $10I$.
We know that I = \left( {\begin{array}{*{20}{l}}1&0\\\0&1\end{array}} \right), so
10I = 10\left( {\begin{array}{*{20}{l}}1&0\\\0&1\end{array}} \right)
Using scaler multiplication, we get
\Rightarrow 10I = \left( {\begin{array}{*{20}{l}}{10}&0\\\0&{10}\end{array}} \right)……$\left( 3 \right)$
Now substituting all the value from equation $\left( 1 \right)$, $\left( 2 \right)$ and $\left( 3 \right)$ in equation ${A^2} - 4A + 10I = A$ we get
\left( {\begin{array}{*{20}{l}}{ - 5}&{ - 3K - 3}\\\\{2K + 2}&{{K^2} - 6}\end{array}} \right) - \left( {\begin{array}{*{20}{l}}4&{ - 12}\\\8&{4K}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}{10}&0\\\0&{10}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1&{ - 3}\\\2&K;\end{array}} \right)
Simplifying the equation, we get
\begin{array}{l} \Rightarrow \left( {\begin{array}{*{20}{l}}{ - 5 - 4 + 10}&{ - 3K - 3 + 12 + 0}\\\\{2K + 2 - 8 + 0}&{{K^2} - 6 - 4K + 10}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1&{ - 3}\\\2&K;\end{array}} \right)\\\ \Rightarrow \left( {\begin{array}{*{20}{l}}1&{ - 3K + 9}\\\\{2K - 6}&{{K^2} - 4K + 4}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1&{ - 3}\\\2&K;\end{array}} \right)\end{array}
Comparing both side we get the below equations,
$- 3K + 9 = \left( { - 3} \right)$…………$\left( 4 \right)$
$2K - 6 = 2$……….....$\left( 5 \right)$
${K^2} - 4K + 4 = K$……$\left( 6 \right)$
On solving equation $\left( 4 \right)$, we get
$\begin{array}{l} - 3K + 9 = \left( { - 3} \right)\\\ \Rightarrow - 3K = \left( { - 3} \right) - 9\end{array}$
Dividing both side by $- 3$, we get
$\Rightarrow K = \dfrac{{ - 12}}{{ - 3}} = 4$
On solving equation $\left( 5 \right)$, we get
$\begin{array}{l}2K - 6 = 2\\\ \Rightarrow 2K = 2 + 6\end{array}$
Dividing both side by 4, we get
$\Rightarrow K = \dfrac{8}{2} = 4$
On solving equation $\left( 6 \right)$ we get,
$\begin{array}{l}{K^2} - 4K + 4 = K\\\ \Rightarrow {K^2} - 4K - K + 4 = 0\end{array}$
Factoring out common terms, we get
$\Rightarrow K\left( {K - 4} \right) - \left( {K - 4} \right) = 0$
Again factoring out common terms, we get
$\Rightarrow \left( {K - 1} \right)\left( {K - 4} \right) = 0$
Using zero product property, we get
$\begin{array}{l} \Rightarrow K - 1 = 0\\\ \Rightarrow K = 1\end{array}$
Or
$\begin{array}{l} \Rightarrow K - 4 = 0\\\ \Rightarrow K = 4\end{array}$
As we can see that at $K = 4$ all the equations are being satisfied but on $K = 1$ only equation $\left( 6 \right)$is satisfying.
Therefore $K = 4$ is the correct answer.
Hence, option (B) is correct.

Note: One more way to find the value of $K$ is that we can find the Eigenvalue of the matrix. As we know that the Eigenvalue satisfies all the conditions of a matrix, we can substitute it in the condition and thus can find the value of $K$. As we can see that we got two values of $K$ but only one value satisfies all the equations so we can make a mistake by writing both 1 and 4 as our answer. So, it becomes important to cross check our answer.