# Find the value of λ, a non-zero scalar,

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Question:

Find the value of $\lambda$, a non-zero scalar, if $\lambda\left[\begin{array}{lll}1 & 0 & 2 \\ 3 & 4 & 5\end{array}\right]+2\left[\begin{array}{rrr}1 & 2 & 3 \\ -1 & -3 & 2\end{array}\right]=\left[\begin{array}{lll}4 & 4 & 10 \\ 4 & 2 & 14\end{array}\right]$

Solution:

Given : $\lambda\left[\begin{array}{lll}1 & 0 & 2 \\ 3 & 4 & 5\end{array}\right]+2\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & -3 & 2\end{array}\right]=\left[\begin{array}{lll}4 & 4 & 10 \\ 4 & 2 & 14\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}\lambda & 0 & 2 \lambda \\ 3 \lambda & 4 \lambda & 5 \lambda\end{array}\right]+\left[\begin{array}{ccc}2 & 4 & 6 \\ -2 & -6 & 4\end{array}\right]=\left[\begin{array}{ccc}4 & 4 & 10 \\ 4 & 2 & 14\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}\lambda+2 & 0+4 & 2 \lambda+6 \\ 3 \lambda-2 & 4 \lambda-6 & 5 \lambda+4\end{array}\right]=\left[\begin{array}{ccc}4 & 4 & 10 \\ 4 & 2 & 14\end{array}\right]$

$\Rightarrow \lambda+2=4$

$\Rightarrow \lambda=4-2$

$\therefore \lambda=2$