Q73 / 180 · COMEDK 2016 · Mathematics
If $A=\left[\begin{array}{cc}1 & -3 \\ 2 & k\end{array}\right]$ and $A^{2}-4 A+10 I=A$, then $k=$
- Correct answerA1 or 4
- B4 and not 1
- C$-4$
- D0
Solution
We have, $A=\left[\begin{array}{cc}1 & -3 \\ 2 & k\end{array}\right]$
Now, $\quad A=\left[\begin{array}{cc}1 & -3 \\ 2 & k\end{array}\right]\left[\begin{array}{cc}1 & -3 \\ 2 & k\end{array}\right]$
$=\left[\begin{array}{cc}
1-6 & -3-3 k \\
2+2 k & -6+k^{2}
\end{array}\right]=\left[\begin{array}{cc}
-5 & -3-3 k \\
2+2 k & k^{2}-6
\end{array}\right]$
Now, $A^{2}-4 A+10 I=A$
$\begin{aligned} \Rightarrow\left[\begin{array}{cc}-5 & -3-3 k \\ 2+2 k & k^{2}-6\end{array}\right]-4\left[\begin{array}{cc}1 & -3 \\ 2 & k\end{array}\right] \\ &+10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & -3 \\ 2 & k\end{array}\right] \end{aligned}$
$\begin{aligned}
&\Rightarrow\left[\begin{array}{cc}
-5-4+10 & -3-3 k+12+0 \\
2+2 k-8+0 & k^{2}-6-4 k+10
\end{array}\right]=\left[\begin{array}{cc}
1 & -3 \\
2 & k
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{cc}
-1 & -3 k+9 \\
2 k-6 & k^{2}-4 k+4
\end{array}\right]=\left[\begin{array}{cc}
1 & -3 \\
2 & k
\end{array}\right]
\end{aligned}$
On comparing both sides, we get
$k^{2}-4 k+4=k$
$\begin{array}{ll}\Rightarrow & k^{2}-5 k+4=0 \\ \Rightarrow & k^{2}-4 k-k+4=0 \\ \Rightarrow & k(k-4)-1(k-4)=0 \\ \Rightarrow & (k-4)(k-1)=0 \\ \Rightarrow & k=4,1\end{array}$
Hence, the correct answer is option: 'A'.