Let A, B and C be three 2 2 matrices with real entries such that B=(I+A)^-1 and A+C=I.

Question

Let $A, B$ and $C$ be three $2 	imes 2$ matrices with real entries such that $B=(I+A)^{-1}$ and $\mathrm{A}+\mathrm{C}=\mathrm{I}$. 
If $\mathrm{BC}=\left[\begin{array}{cc} 1 & -5 \ -1 & 2 \end{array}ight]$

Accepted Answer

We are given:

$B=(I+A{)}^{-1}\implies {B}^{-1}=I+A$

$A+C=I\implies A=I-C$

Substitute $A$ into the first equation:

${B}^{-1}=I+(I-C)$

${B}^{-1}=2I-C........$

pre-multiply $B:B\cdot {B}^{-1}=B(2I-C)I=2B-BC\implies BC=2B-I$ ........ (2)

post multiply in eqn (1): ${B}^{-1}B=(2I-C)\cdot BI=2B-CB\implies CB=2B-I$ .......(3)

from (2) \& (3), $BC=CB$

Set up the System of Equations Since $BC=CB$, we use the given matrix for $BC$ :

$CB=[\begin{gathered} 1 -5 \ -1 2 ]$

Now, substitute this into the vector equation :

$[ 1 -5 \ -1 2 \end{gathered}][\begin{gathered} {x}_{1} \ {x}_{2} \end{gathered}]=[\begin{gathered} 12 \ -6 \end{gathered}]$

This results in two linear equations :

${x}_{1}-5{x}_{2}=12$

$-{x}_{1}+2{x}_{2}=-6$

Solve for ${x}_{1}$ and ${x}_{2}$

Adding the two equations :

$({x}_{1}-{x}_{1})+(-5{x}_{2}+2{x}_{2})=12-6$

$\Rightarrow$ ${x}_{2}=-2$

Substitute ${x}_{2}=-2$ into the second equation :

$\begin{gathered} -{x}_{1}+2(-2)=-6 \ -{x}_{1}-4=-6\implies -{x}_{1}=-2\implies {x}_{1}=2 \end{gathered}$

Final result ${x}_{1}+{x}_{2}=2+(-2)=0$

Solutions · 1