Practicing Success
Practicing Success
CUET Preparation Today
If $A=\begin{bmatrix}n&0&0\\0&n&0\\0&0&n\end{bmatrix}$ and $B=\begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}$, then $AB$ is equal to |
$B$ $nB$ $B^n$ $A+B$ |
$nB$ |
We have, $A =nI_3$ $∴AB=(nI_3) B=n (I_3 B) =nB$. |
