Matrices and Determinants 4 Question 25
26. $A=\begin{bmatrix}a & 0 & 1 \\ 1 & c & b \\ 1 & d & b\end{bmatrix}, B=\begin{bmatrix}a & 1 & 1 \\ 0 & d & c \\ f & g & h\end{bmatrix}, $
$U =\begin{bmatrix}f \\ g\\ h\end{bmatrix}, $ $V=\left|\begin{aligned} & a^{2} \\ & 0 \\ & 0\end{aligned}\right| $
If there is a vector matrix $X$, such that $A X=U$ has infinitely many solutions, then prove that $B X=V$ cannot have a unique solution. If $a f d \neq 0$. Then, prove that $B X=V$ has no solution.
$(2004,4 \mathrm{M})$
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Answer:
Correct Answer: 26. $-\sqrt{2} \leq \lambda \leq \sqrt{2}, \alpha=n \pi, n \pi+\frac{\pi}{4}$
Solution:
- Since, $A X=U$ has infinitely many solutions.
$ \begin{aligned} \Rightarrow|A|=0 \Rightarrow \begin{bmatrix} a & 0 & 1 \\ 1 & c & b \\ 1 & d & b \end{bmatrix} & =0 \\ \Rightarrow \quad a(b c-b d)+1(d-c) & =0 \Rightarrow \quad(d-c)(a b-1)=0 \\ \therefore \quad a b & =1 \text { or } d=c \end{aligned} $
$ \begin{aligned} & \text { Again, } \quad\left|A_{3}\right|=\begin{bmatrix} a & 0 & f \\ 1 & c & g \\ 1 & d & h \end{bmatrix}=0 \Rightarrow g=h \\ & \Rightarrow \quad\left|A_{2}\right|=\begin{bmatrix} a & f & 1 \\ 1 & g & b \\ 1 & h & b \end{bmatrix}=0 \Rightarrow g=h \\ & \text { and } \quad\left|A_{1}\right|=\begin{bmatrix} f & 0 & 1 \\ g & c & b \\ h & d & b \end{bmatrix} \Rightarrow 0 \Rightarrow g=h \\ & \therefore \quad g=h, c=d \text { and } a b=1 \\ & \text { Now, } \quad B X=V \\ & |B|=\begin{bmatrix} a & 1 & 1 \\ 0 & d & c \\ f & g & h \end{bmatrix}=0 \end{aligned} $
[since, $C_{2}$ and $C_{3}$ are equal]
$ \therefore \quad B X=V \text { has no solution. } $
$ \left|B_{1}\right|=\begin{bmatrix} a^{2} & 1 & 1 \\ 0 & d & c \\ 0 & g & h \end{bmatrix}=0 $
$ \left|B_{2}\right|=\begin{bmatrix} a & a^{2} & 1 \\ 0 & 0 & c \\ f & 0 & h \end{bmatrix}=a^{2} c f=a^{2} d f $
[since, $c=d$ and $g=h$ ]
$ \text { Since, } \quad a d f \neq 0 \Rightarrow\left|B_{2}\right| \neq 0 $
$BX =V$ has no solution.
