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# Let $A\;=\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}, B\;=\begin{bmatrix}1 & 3\\-2 & 5\end{bmatrix}, C\;=\begin{bmatrix}-2 & 5\\3 & 4\end{bmatrix}$. Find $3A-C$

$\begin{array}{1 1} \begin{bmatrix}8 & 7\\6 & 0\end{bmatrix} \\ \begin{bmatrix}8 & -7\\6 & 2\end{bmatrix} \\ \begin{bmatrix}8 & 7\\6 & 2\end{bmatrix} \\ \begin{bmatrix}8 & 7\\6 & 2\end{bmatrix} \end{array}$

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Toolbox:
• The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
• The difference $A-B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A - B)_{i,j} = A_{i,j} - B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Given $A\;=\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}$, $3A = \begin{bmatrix}6 & 12\\9 & 6\end{bmatrix}$.
Therefore $3A-C = \begin{bmatrix}6 & 12\\9 & 6\end{bmatrix} - \begin{bmatrix}-2 & 5\\3 & 4\end{bmatrix} = \begin{bmatrix}6 - (-2) & 12-5\\9-3 & 6-4\end{bmatrix} = \begin{bmatrix}8 & 7\\6 & 2\end{bmatrix}$
answered Feb 27, 2013