Question

If $5A=\left(\begin{array}{cc}3& -4\\ 4& x\end{array}\right)$ and $A{A}^T=A^{T}A=I$, then the value of $x$, is

• A. $3$
• B. $-3$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $3$

Given:

$5A=\left[\begin{array}{cc}3& -4\\ 4& x\end{array}\right]$

Dividing both sides by $5$

$A=\left[\begin{array}{cc}\frac{3}{5}& -\frac{4}{5}\\ \frac{4}{5}& \frac{x}{5}\end{array}\right]$

It is given that
$A{A}^T=A^{T}A=I$
where , $I=$ Identity matrix
Using property of transpose.
$A^T=\left[\begin{array}{cc}\frac{3}{5}& \frac{4}{5}\\ -\frac{4}{5}& \frac{x}{5}\end{array}\right]$
$A{A}^T=\left[\begin{array}{cc}\frac{3}{5}\times \frac{3}{5}+\frac{4}{5}\times \frac{4}{5}& \frac{3}{5}\times \frac{4}{5}-\frac{4}{5}\times \frac{x}{5}\\ \frac{4}{5}\times \frac{3}{5}-\frac{x}{5}\times \frac{4}{5}& \frac{4}{5}\times \frac{4}{5}+\frac{x}{5}\times \frac{x}{5}\end{array}\right]$
$\Rightarrow A{A}^T=\left[\begin{array}{cc}1& \frac{4(3-x)}{25}\\ \frac{4(3-x)}{25}& \frac{16+x^2}{25}\end{array}\right]$

But,

$A{A}^T=I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

So, comparing the corresponding terms,
$\frac{4}{25}(3-x)=0$
$\Rightarrow x=3$

Hence, option A.