MTE-02 Assignment Solution: Linear Algebra [1 Jan 2019 to 31 Dec 2019]
1 (a) Which of the following are binary operations on \( \not\subset \)? Justify your answer.
(i) The operation \( \nabla \) defined by \( x \nabla y=|x y| \)
(ii) The operation \( \Delta \) defined by \( x \Delta y=\overline{x y} \) where the bar denotes complex conjugation.
Also, for those operations which are binary operations, check whether they are associative and commutative.
Answer:- (i) The operation \( \nabla \) defined by \( x \nabla y=|x y| \)
Let x, y \( \not\subset \) then \( x \nabla y=|x y| \)
\( \in \mathbb{R} \subset \not\subset \)
i.e, \( x \nabla y \in \not\subset \)
i.e \( \nabla \) is binary operation.
(ii) The operation \( \Delta \) defined by \( x \Delta y=\overrightarrow{x y} \)
Let,
\( \begin{array}{l} x, y \in \not \subset \\ x=a+i b \quad ; \quad a, b, c, d \in \mathbb{R} \\ y=c+i d \end{array} \)
then,
\( \begin{aligned} x \Delta y &=\overline{x y}=\overline{(a+i b)(c+i d)} \\ &=\overline{\left[a c+i a d+i b c+i^{2} b d\right]} \quad \because i^{2}=-1 \\ &=[a c-b d+i(a d+b c)\\ &=[(a c-b d)-i(a d+b c)] \in \not\subset \end{aligned} \)
i.e \( x \Delta y \in \not\subset \)
So, again \( \Delta \) is binary operation in \( \not\subset \)
Now we will check commutative and associative property in both cases,
(i)
\( \begin{aligned} x \nabla y=&|x y| \\ & x, y, z \in d \end{aligned} \)
Commutative:- To show:- \( x \nabla y=y \nabla x \)
L.H.S,
\( \begin{aligned} x \nabla y &=|x y| \\ &=|y x| \\ &=y \nabla x \\ &=R H S . \end{aligned} \)
\( i.e\: \nabla \text { is commutative in } \not\subset \)
Associative:- To show:- \( x \nabla(y \nabla z)=(x \nabla y) \nabla z \)
\( \begin{aligned} \operatorname{L.H.S,}\: x \nabla(y \nabla z) &=x \nabla(|y z|) \\ &=|| x|| y z|| \\ &=|| x|||y z| \mid \\ &=|x||y z| \\ &=|x| \cdot|y||z| \\ &=|x y||z| \\ &=(|x y|) \nabla \mid z\:|) \\ &=(x \nabla y) \nabla z \:\:\text { R.H.S } \end{aligned} \)
i.e \( \nabla \) is Associative.
(ii)
\( \begin{array}{l} x \Delta y=\overline{x y} \\ x, y \in \not\subset \end{array} \)
Commutative:- To show, \( x \Delta y=y \Delta x \)
\( \text { L.H.S }=x \Delta y=\overline{x y}=\overline{y x}=y \Delta x \)
i.e \( \Delta \text { is Commutative in } \not\subset. \)
Associative:- Let, \( x, y \geq \in\not\subset \) To show:- \( x \Delta(y \Delta z)=(x \Delta y) \Delta z \)
L.H.S,
\( \begin{aligned} x \Delta(y \Delta z) &=x \Delta \overline{y z} \\ &=\overline{x y z} \\ &=\overline{x y} \Delta z \\ &=(x \Delta y) \Delta z \\ &=R H S . \end{aligned} \)
i.e \( \triangle \) is Associative in \( \not\subset \)
1(b). Find the vector equation of the plane determined by the points \( (1, 1, -1),(1, 1, 1) \) and \( (0,1,1) \). Also find the point of intersection of the line \( \vec{r}=(1+3 t) \hat{i}+(2-t) \hat{j}+(1+t) \hat{k} \) and the plane.
Answer:- Given three points \( A(1, 1,-1), B(1,1,1) \:\:\&\:\: C(0,1,1) \)
\( \vec{a}=(\hat{i}+\hat{j}-\hat{k}), \vec{b}=(\hat{i}+\hat{j}+\hat{k}) \quad\&\:\: \vec{c}=(\hat{j}+\hat{k}) \)
Vector equation of plane passing through points is
\( \begin{array}{c} (\vec{r}-\vec{a}) \cdot[(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})]=0 \\ {[\vec{r}-(\hat{i}+\hat{j} -\hat{k})] \cdot[(\hat{i}+\hat{j}+\hat{k}-\hat{i}-\hat{j}+\hat{k}) \times(\hat{j}+\hat{k}-\hat{i}-\hat{j}+\hat{k})]=0} \\ (\vec{\jmath}-(\hat{i}+\hat{j}-\hat{k})] \cdot[(2 \hat{k}) \times(-\hat{i}+2 \hat{k})]=0 \\ (\vec{\jmath}-(\hat{i}+\hat{j}-\hat{k})] \cdot\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 2 \\ -1 & 0 & 2 \end{array}\right|=0 \\ \left(\vec{s}-(\hat{i}+\hat{j}-\hat{k}) \cdot\left[\begin{array}{cc} -\hat{j}(0+2)] \end{array}\right]=0\right. \end{array} \)
