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  • Let A be a  real matrix such that  <img alt="A\left(\begin{array}{l} 1 \\ 1 \\ 0

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

A\left(\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right)=\left(\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right) ; A\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)=\left(\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right) \text { and } A\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right)=\left(\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right). 

If \mathrm{X}=\left(x_{1}, x_{2}, x_{3}\right)^{\mathrm{T}} and \mathrm{I} is an identity matrix of order 3, then the system  (A-2 I) X=\left(\begin{array}{l} 4 \\ 1 \\ 1 \end{array}\right) has:

Option: 1

no solution


Option: 2

infinitely many solutions


Option: 3

unique solution


Option: 4

exactly two solutions


Answers (1)

best_answer

\mathrm{A= \begin{bmatrix} a &b &c \\ l & m &n \\ p & q & r \end{bmatrix}}

\mathrm{A\begin{bmatrix} 0\\0 \\ 1 \end{bmatrix}= {\begin{bmatrix} c\\n \\ r \end{bmatrix}= {\begin{bmatrix} 1\\1 \\ 2 \end{bmatrix}}

\mathrm{\Rightarrow c= 1,\; n= 1,\; r= 2}
\mathrm{A\begin{bmatrix} 1\\0 \\ 1 \end{bmatrix}= \begin{bmatrix} c+a\\ n+l \\r+p \end{bmatrix}=\begin{bmatrix} -1\\ 0 \\ 1 \end{bmatrix}}

\mathrm{\mathrm{\Rightarrow a= -2,\; l= -1,\; p= -1}}
\mathrm{A\begin{bmatrix} 1\\1 \\ 0 \end{bmatrix}= \begin{bmatrix} a+b\\ l+m \\ p+q \end{bmatrix}= \begin{bmatrix} 1\\ 1 \\ 0 \end{bmatrix}}

\mathrm{\Rightarrow b= 3,\; m= 2,\; q= 1}
\mathrm{A= \begin{bmatrix} -2 & 3 &1 \\ -1 & 2 & 1\\ -1&1 & 2 \end{bmatrix}\Rightarrow A-2I= \begin{bmatrix} -4 & 3& 1\\ -1& 0 &1 \\ -1&1 & 0 \end{bmatrix}}

\mathrm{\left | A-2I \right |= 0}

Now,
, \mathrm{\begin{bmatrix} -4 & 3& 1\\ -1& 0 &1 \\ -1&1 & 0 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \\ x_{3} \end{bmatrix}= \begin{bmatrix} 4\\1 \\ 1 \end{bmatrix}}

\mathrm{-4x_{1}+3x_{2}+x_{3}= 4\; ----(1)}
\mathrm{-x_{1}+x_{3}= 1\; ----(2)}
\mathrm{-x_{1}+x_{2}= 1\; ----(3)}

\mathrm{From\; eqn (1)-\left [ eqn(2)+3\times eqn(3) \right ]}
\mathrm{0= 0\Rightarrow infinite\: solution}

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