Question

(a)

If $H=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 3& 1& 2\end{array}\right],G=\left[\begin{array}{ccc}2& 6& -10\\ 0& -4& 8\\ 2& -8& 4\end{array}\right]$

Find $3H-\frac{1}{2}G.$
$\left[3\right]$

(b) If $3A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ x& 2& y\end{array}\right]$ and $A\cdot A^T=I$, then find the value of $x+y$.
$\left[3\right]$

(c) If $A=\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right],B=\left[\begin{array}{cc}4& -2\\ -1& 3\end{array}\right]$ and $I$ is Identity matrix of same order and $A^t$ is the transpose of matrix $A$ find $A^t.B+BI$.
$\left[4\right]$

Answer 1

(a)

Given $H=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 3& 1& 2\end{array}\right],G=\left[\begin{array}{ccc}2& 6& -10\\ 0& -4& 8\\ 2& -8& 4\end{array}\right]$

$3H=\left[\begin{array}{ccc}3\times 1& 3\times 2& 3\times 3\\ 3\times 2& 3\times 3& 3\times 1\\ 3\times 3& 3\times 1& 3\times 2\end{array}\right]=\left[\begin{array}{ccc}3& 6& 9\\ 6& 9& 3\\ 9& 3& 6\end{array}\right]$
$\left[1\right]$

$\frac{1}{2}G=\left[\begin{array}{ccc}2÷2& 6÷2& -10÷2\\ 0÷2& -4÷2& 8÷2\\ 2÷2& -8÷2& 4÷2\end{array}\right]=\left[\begin{array}{ccc}1& 3& -5\\ 0& -2& 4\\ 1& -4& 2\end{array}\right]$
$\left[1\right]$

$3H-\frac{1}{2}G=\left[\begin{array}{ccc}2& 3& 14\\ 6& 11& -1\\ 8& 7& 4\end{array}\right]$
$\left[1\right]$

(b) $Giventhat,3A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ x& 2& y\end{array}\right]$
$\Rightarrow A=\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{x}{3}& \frac{2}{3}& \frac{y}{3}\end{array}\right]$
$\left[0.5\right]$
$\therefore A^T=\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{x}{3}\\ \frac{2}{3}& \frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{y}{3}\end{array}\right]$
$\left[0.5\right]$
$A{A}^T=I\Rightarrow \left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{x}{3}& \frac{2}{3}& \frac{y}{3}\end{array}\right]\cdot \left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{x}{3}\\ \frac{2}{3}& \frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{y}{3}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{ccc}\frac{1}{9}+\frac{4}{9}+\frac{4}{9}& \frac{2}{9}+\frac{2}{9}-\frac{4}{9}& \frac{x}{9}+\frac{4}{9}+\frac{2y}{9}\\ \frac{2}{9}+\frac{2}{9}-\frac{4}{9}& \frac{4}{9}+\frac{1}{9}+\frac{4}{9}& \frac{2x}{9}+\frac{2}{9}-\frac{2y}{9}\\ \frac{x}{9}+\frac{4}{9}+\frac{2y}{9}& \frac{2x}{9}+\frac{2}{9}-\frac{2y}{9}& \frac{x^2}{9}+\frac{4}{9}+\frac{y^2}{9}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\Rightarrow x^2+y^2+4=9\Rightarrow x^2+y^2=5...\left(1\right)$

$\Rightarrow 2x+2-2y=0\Rightarrow y=x+1...\left(2\right)$

$\Rightarrow x+4+2y=0\Rightarrow x+2y=-4...\left(3\right)$

Solving Equations (1) and (2)

$x^2+(x+1{)}^2=5\Rightarrow x^2+x^2+2x+1=5\Rightarrow x^2+x-2=0\Rightarrow x=-2,1\Rightarrow (x,y)\equiv (1,2),(-2,-1)$

Satisfying these in Equation (3)

$(-2)+2(-1)=-4\left(1\right)+2\left(2\right)\ne -4$
Hence, $(-2,-1)$ satisfies.

So $x+y=-3$.
$\left[2\right]$

(c) $A=\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]$
$\therefore A^t=\left[\begin{array}{cc}2& 1\\ 5& 3\end{array}\right]$
$\left[1\right]$
$A^t.B=\left[\begin{array}{cc}2& 1\\ 5& 3\end{array}\right]\left[\begin{array}{cc}4& -2\\ -1& 3\end{array}\right]$
$=\left[\begin{array}{cc}2\times 4+1\times (-1)& 2\times (-2)+1\times 3\\ 3\times 4+3\times (-1)& 5\times (-2)+3\times 3\end{array}\right]$
$=\left[\begin{array}{cc}7& -1\\ 17& -1\end{array}\right]$
$\left[1\right]$
$B.I=\left[\begin{array}{cc}4& -2\\ -1& 3\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}4& -2\\ -1& 3\end{array}\right]$
$\left[1\right]$
$\therefore A^t.B+BI=\left[\begin{array}{cc}7& -1\\ 17& -1\end{array}\right]+\left[\begin{array}{cc}4& -2\\ -1& 3\end{array}\right]$
$=\left[\begin{array}{cc}11& -3\\ 16& 2\end{array}\right]$
$\left[1\right]$