Question

If $\left[\begin{array}{cc}{\lambda }^2-2\lambda +1& \lambda -2\\ 1-{\lambda }^2+3\lambda & 1-{\lambda }^2\end{array}\right]={A\lambda }^2+B\lambda +C$ where $A,B,C$ are matrices then $B+C=$

• A. $\left[\begin{array}{cc}-1& -1\\ 4& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}1& -1\\ 4& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 1\\ -4& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}-1& -1\\ -4& 1\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}-1& -1\\ 4& 1\end{array}\right]$

Let the matrix A,b and C are as

$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right),B=\left(\begin{array}{cc}w& x\\ y& z\end{array}\right)$ and $C=\left(\begin{array}{cc}m& n\\ o& p\end{array}\right)$

As given

$\left(\begin{array}{cc}{\lambda }^2-2\lambda +1& \lambda -2\\ 1-{\lambda }^2+4\lambda & 1-{\lambda }^2\end{array}\right)={\lambda }^2\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)+\lambda \left(\begin{array}{cc}w& x\\ y& z\end{array}\right)+\left(\begin{array}{cc}m& n\\ o& p\end{array}\right)$

$\left(\begin{array}{cc}{\lambda }^2-2\lambda +1& \lambda -2\\ 1-{\lambda }^2+4\lambda & 1-{\lambda }^2\end{array}\right)=\left(\begin{array}{cc}a{\lambda }^2+w\lambda +m& b{\lambda }^2+x\lambda +p\\ c{\lambda }^2+y\lambda +o& d{\lambda }^2+z\lambda +p\end{array}\right)$

On comparing both sides we get

$a=1,b=0,c=-1,d=-1,w=-2,x=1,y=3,z=0,m=1,n=-2,o=1,p=1$

$B+C=\left(\begin{array}{cc}w& x\\ y& z\end{array}\right)+\left(\begin{array}{cc}m& n\\ o& p\end{array}\right)=\left(\begin{array}{cc}-2& 1\\ 3& 0\end{array}\right)+\left(\begin{array}{cc}1& -2\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}-1& -1\\ 4& 1\end{array}\right)$