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Basic matrix operations Quiz 2

mathematics

 Let $A$ be a matrix of quadratic form ${{\left( X-Y+2Z \right)}^{2}}$, then trace of $A$ is:

 Since the trace of a matrix is the sum of the diagonal elements so adding up the elements will give the trace of $A$.

  $\text{Here, }A={{\left( X-Y+2Z \right)}^{2}} \\ \text{Now, construct the matrix for the above Quadratic form} \\ \text{Then we get} \\ A=\begin{bmatrix}   X & Y & Z  \\ \end{bmatrix} \, \begin{bmatrix}   1 & -1 & 2  \\   -1 & 1 & -2  \\   2 & -2 & 4  \\ \end{bmatrix} \, \begin{bmatrix}   X  \\   Y  \\   Z  \\ \end{bmatrix}\\ \text{Now, }trace(A)=1+1+4=6$ 

 Find the matrix $A$ such that: $ \\ \begin{bmatrix}   2  \\   0  \\   -2  \\\end{bmatrix}\begin{bmatrix}   -1  \\   1  \\   4  \\ \end{bmatrix} X= \begin{bmatrix}   -1 & -8 & -10  \\   3 & 4 & 0  \\   10 & 20 & 10  \\ \end{bmatrix} $

 First find let $X$ to be a variable matrix then multiply as per the relation given, then from the linear equations we will get the every variable.

  $\text{Let, } X=\begin{bmatrix}   a  \\   d  \\ \end{bmatrix} \begin{bmatrix}   b  \\   e  \\ \end{bmatrix} \begin{bmatrix}   c  \\   f  \\ \end{bmatrix} \\ \begin{bmatrix}   2  \\   0  \\   -2  \\ \end{bmatrix} \begin{bmatrix}   -1  \\   1  \\   4  \\ \end{bmatrix} \begin{bmatrix}   a  \\   d  \\ \end{bmatrix}\begin{bmatrix}   b  \\   e  \\ \end{bmatrix}\begin{bmatrix}   c  \\   f  \\ \end{bmatrix} =\begin{bmatrix}   -1 & -8 & -10  \\   3 & 4 & 0  \\   10 & 20 & 10  \\ \end{bmatrix} \\ \Rightarrow 2a-d=-1,d=3-2a+4d=10,a=1 \\ 2b-e=-8,e=-4,-2b+4e=20,b=-2 \\ 2c-f=-10,f=0,-2c+4f=10,c=-5 \\ X= \begin{bmatrix}   1  \\   3  \\ \end{bmatrix} \begin{bmatrix}   -2  \\   4  \\ \end{bmatrix} \begin{bmatrix}   -5  \\   0  \\ \end{bmatrix}$ 

 If $P, Q$ and $R$ are the three matrices of order $4\times 3,4\times 3\text{ and }2\times 3$ respectively, find the order of the matrix $Z={{\left[ R{{\left( {{P}^{T}}Q \right)}^{-1}}{{X}^{T}} \right]}^{T}}$.

 From the rule that, when two matrices of order $\left( x,y \right)\text{ and }(a,b)$ are to be multiplied, the multiplication is possible when $y=a$ , and the resultant matrix is of the order  $\left( x\times a \right)$, we will get the required order of the matrix.

  $\text{When two matrices of order }\left( x,y \right)\text{ and }(a,b)\text{ are to be multiplied,} \text{ the multiplication is possible when }y=a,\text{ and the resultant matrix is of the order }\left( x\times a \right) \\ \text{We know that, }{{X}^{T}}=\text{ The matrix in which rows and columns are interchanged} \\ {{\left\{ \left( 2\times 3 \right)\times {{\left[ \left( 3\times 4 \right)\times \left( 4\times 3 \right) \right]}^{-1}}~\times \left( 3\times 2 \right) \right\}}^{T}}  \\  \Rightarrow {{\left\{ \left( 2\times 3 \right)\times \left( 3\times 3 \right)\times \left( 3\times 2 \right) \right\}}^{T}}  \\  \Rightarrow {{\left\{ \left( 2\times 3 \right)\times \left( 3\times 2 \right) \right\}}^{T}}  \\   \Rightarrow {{\left( 2\times 2 \right)}^{T}}  \\  \Rightarrow 2\times 2$ 

 $A=\begin{bmatrix}   1 & x  \\  -1 & -1  \\ \end{bmatrix}$ and $B=\begin{bmatrix}   1 & 2x  \\   1 & -1  \\ \end{bmatrix}$. If ${{\left( A+B \right)}^{2}}={{A}^{2}}+{{B}^{2}}$, find $x$.

 From the distributive law of matrices, we will get, $AB=-BA$, from here putting the values of both the matrices, we will get the required value of $x$.

  ${{\left( A+B \right)}^{2}}={{A}^{2}}+{{B}^{2}}  \\   \left( A+B \right)\left( A+B \right)={{A}^{2}}+{{B}^{2}}  \\   {{A}^{2}}+AB+BA+{{B}^{2}}={{A}^{2}}+{{B}^{2}}~\left[ \text{Matrix multiplication is distributive} \right]  \\   i.e.AB=-BA \\ AB=\begin{bmatrix}   1 & x  \\   -1 & -1  \\ \end{bmatrix}\begin{bmatrix}   1 & 2x  \\   1 & -1  \\ \end{bmatrix}=\begin{bmatrix}   1+x & x  \\   -2 & -2x+1  \\ \end{bmatrix} \\ BA=\begin{bmatrix}   1 & 2x  \\   1 & -1  \\ \end{bmatrix}\begin{bmatrix}   1 & x  \\   -1 & -1  \\ \end{bmatrix}=\begin{bmatrix}   1-2x & -x  \\   2 & x+1  \\ \end{bmatrix} \\ \text{Therefore, } \\ -BA=\begin{bmatrix}   2x-1 & x  \\   -2 & -x-1  \\ \end{bmatrix}\\ \Rightarrow \begin{bmatrix}   1+x & x  \\   -2 & -2x+1  \\ \end{bmatrix}=\begin{bmatrix}   2x-1 & x  \\   -2 & -x-1  \\ \end{bmatrix}\\ x=2$ 

 $A=\begin{bmatrix}   1 & -2 & 4  \\   4 & 4 & 3  \\   2 & -2 & \dfrac{8}{3}  \\ \end{bmatrix}$ and $B=k\begin{bmatrix}   1  \\   1  \\   1  \\ \end{bmatrix}$. If $\dfrac{1}{4}AC=B$, where C is a matrix, find the least possible positive integral value of k?

 From the given relation, $\dfrac{1}{4}AC=B$, then by considering two equations every time we will get 3 relations, from there by solving the simultaneous equations, we will get the answer.

  $\text{Since, }\dfrac{1}{4}AC=B\text{ and B is a }3\times 1\text{ matrix, }\dfrac{1}{4}AC\text{ is also a }3\times 1\text{ matrix, so C is also a }3\times 1\text{ matrix.} \\ \text{Let } C=\begin{bmatrix}   x  \\   y  \\   z  \\ \end{bmatrix}\\ \text{Now},\dfrac{1}{4}\begin{bmatrix}   1 & -2 & 4  \\   4 & 4 & 3  \\   2 & -2 & \dfrac{8}{3}  \\ \end{bmatrix}\begin{bmatrix}   x  \\   y  \\   z  \\ \end{bmatrix}=\begin{bmatrix}   k  \\   k  \\   k  \\ \end{bmatrix}\\ \Rightarrow \begin{bmatrix}   \dfrac{1}{4}x-\dfrac{1}{2}y+z  \\   x+y+\dfrac{3}{4}z  \\   \dfrac{1}{2}x-\dfrac{1}{2}y+\dfrac{2}{3}z  \\ \end{bmatrix}=\begin{bmatrix}   k  \\   k  \\   k  \\ \end{bmatrix}\\ \Rightarrow \dfrac{1}{4}x-\dfrac{1}{2}y+z=x+y+\dfrac{3}{4}z=\dfrac{1}{2}x-\dfrac{1}{2}y+\dfrac{2}{3}z=k-(1) \\ \text{Multiply all the three expressions by 12.} \\ \text{Then, }3x6y+12z=12x+12y+9z~=6x6y+8z \\ \text{Let us consider the first and the third expressions.} \\ 3x6y+12z=6x6y+8z  \\   \dfrac{x}{z}=\dfrac{4}{3}-----\left( 2 \right) \\ \text{Let }x=4a\text{ and }z=3a \\ \text{Substitute in first and second expressions;} \\ 3x6y+12z=12x+12y+9z~  \\   3\left( 4a \right)6y+12\left( 3a \right)=12\left( 4a \right)+12y+9\left( 3a \right)  \\   -18y=27a  \\   i.e.\dfrac{y}{a}=\dfrac{\left( -3 \right)}{2}--------\left( 3 \right)  \\ \text{From }\left( 2 \right)\text{ and }\left( 3 \right) \\ \text{If }a=2,\text{ then;} \\ x=4a=8  \\   y=-3  \\   z=3a=6  \\   i.e.x:y:z=8:-3:6  \\ \text{Let }x=8n, y=-3n\text{ and }z=6n \\ \text{From }\left( 1 \right) \\ \dfrac{1}{4}x-\dfrac{1}{2}y+z=k  \\   \dfrac{1}{4}\left( 8n \right)-\dfrac{1}{2}\left( -3n \right)+\left( 6n \right)=k  \\   2n+1.5n+6n=k  \\   9.5n=k  \\ \text{When }n=2, k=19\text{ and it is the least possible positive integral value of k.}$