Question

Given that $A=\left[\begin{array}{cc}3& 2\\ 5& 7\\ 8& 9\end{array}\right],B=\left[\begin{array}{cc}1& 9\\ 0& 3\\ 4& 10\end{array}\right]$ and $X=\left[\begin{array}{cc}6& 29\\ 5& 16\\ 20& 39\end{array}\right]$ and if $2A+6B=kX$, then the value of $k$ is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A)

$\because 2A+6B=kX$
$\therefore 2\left[\begin{array}{cc}3& 2\\ 5& 7\\ 8& 9\end{array}\right]+6\left[\begin{array}{cc}1& 9\\ 0& 3\\ 4& 10\end{array}\right]=k\left[\begin{array}{cc}6& 29\\ 5& 16\\ 20& 39\end{array}\right]$
or $\left[\begin{array}{cc}6& 4\\ 10& 14\\ 16& 18\end{array}\right]+\left[\begin{array}{cc}6& 54\\ 0& 18\\ 24& 60\end{array}\right]=k\left[\begin{array}{cc}6& 29\\ 5& 16\\ 20& 39\end{array}\right]$
or $\left[\begin{array}{cc}12& 58\\ 10& 32\\ 40& 78\end{array}\right]=k\left[\begin{array}{cc}6& 29\\ 5& 16\\ 20& 39\end{array}\right]$
or $2\left[\begin{array}{cc}6& 29\\ 5& 16\\ 20& 39\end{array}\right]=k\left[\begin{array}{cc}6& 29\\ 5& 16\\ 20& 39\end{array}\right]\therefore k=2$