Let A be a 3 [] 3 real matrix such that [A( pmatrix 1 1 0 pmatrix ) = ( pmatrix 1 1 0 pmat…

Question

Let A be a 3 $$\times$$ 3 real matrix such that

$$A\left( {\begin{pmatrix} 1  \
   1  \
   0  \ \end{pmatrix} } \right) = \left( {\begin{pmatrix} 1  \
   1  \
   0  \ \end{pmatrix} } \right);A\left( {\begin{pmatrix} 1  \
   0  \
   1  \ \end{pmatrix} } \right) = \left( {\begin{pmatrix} { - 1}  \
   0  \
   1  \ \end{pmatrix} } \right)$$ and $$A\left( {\begin{pmatrix} 0  \
   0  \
   1  \ \end{pmatrix} } \right) = \left( {\begin{pmatrix} 1  \
   1  \
   2  \ \end{pmatrix} } \right)$$.

If $X = {({x_1},{x_2},{x_3})^T}$ and I is an identity matrix of order 3, then the system $$(A - 2I)X = \left( {\begin{pmatrix} 4  \
   1  \
   1  \ \end{pmatrix} } \right)$$ has :

Accepted Answer

Let $$A = \left[ {\begin{pmatrix} a & b & c  \
   d & e & f  \
   g & h & i  \

\end{pmatrix} } \right]$$

$$A = \left[ {\begin{pmatrix} 1  \
   1  \
   0  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 1  \
   1  \
   0  \

\end{pmatrix} } \right] \Rightarrow \left[ {\begin{pmatrix} a & b & c  \
   d & e & f  \
   g & h & i  \

\end{pmatrix} } \right]\left[ {\begin{pmatrix} 1  \
   1  \
   0  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 1  \
   1  \
   0  \

\end{pmatrix} } \right] \Rightarrow \begin{pmatrix} {a + b = 1}  \
   {d + e = 1}  \
   {g + h = 0}  \

\end{pmatrix} $$

$$A = \left[ {\begin{pmatrix} 1  \
   0  \
   1  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} { - 1}  \
   0  \
   1  \

\end{pmatrix} } \right] \Rightarrow \left[ {\begin{pmatrix} a & b & c  \
   d & e & f  \
   g & h & i  \

\end{pmatrix} } \right]\left[ {\begin{pmatrix} 1  \
   0  \
   1  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} { - 1}  \
   0  \
   1  \

\end{pmatrix} } \right] \Rightarrow \begin{pmatrix} {a + c =  - 1}  \
   {d + f = 1}  \
   {g + i = 0}  \

\end{pmatrix} $$

$$A = \left[ {\begin{pmatrix} 0  \
   0  \
   1  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 1  \
   1  \
   2  \

\end{pmatrix} } \right] \Rightarrow \left[ {\begin{pmatrix} a & b & c  \
   d & e & f  \
   g & h & i  \

\end{pmatrix} } \right]\left[ {\begin{pmatrix} 0  \
   0  \
   1  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 1  \
   1  \
   2  \

\end{pmatrix} } \right] \Rightarrow \begin{pmatrix} {c = 1}  \
   {f = 1}  \
   {i = 2}  \

\end{pmatrix} $$

Solving will get

$$a =  - 2,\,b = 3,\,c = 1,\,d =  - 1,\,e = 2,\,f = 1,\,g =  - 1,\,h = 1,\,i = 2$$

$$A = \left[ {\begin{pmatrix} { - 2} & 3 & 1  \
   { - 1} & 2 & 1  \
   { - 1} & 1 & 2  \

\end{pmatrix} } \right] \Rightarrow A = 2I = \left[ {\begin{pmatrix} { - 4} & 3 & 1  \
   { - 1} & 0 & 1  \
   { - 1} & 1 & 0  \

\end{pmatrix} } \right]$$

$$(A - 2I)x = \left[ {\begin{pmatrix} 4  \
   1  \
   1  \

\end{pmatrix} } \right]$$

$$ \Rightarrow  - 4{x_1} + 3{x_2} + {x_3} = 4$$ ..... (i)

$- {x_1} + {x_3} = 1$ ...... (ii)

$- {x_1} + {x_2} = 1$ ...... (iii)

So 3(iii) + (ii) = (i)

$$\therefore$$ Infinite solution

Solutions · 1