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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES2 Marks
Question
Let $\text{A}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix},\ \text{B}=\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}.$ Verify that AB = AC though B ≠ C, A ≠ O.
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Answer

Here,
$\text{A}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix},\ \text{B}=\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}$
Now,
$\text{A}\text{B}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix}\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$
$\Rightarrow\text{A}\text{B}=\begin{bmatrix}3+5-2&1+2+4\\9+15-6&3+6+12\end{bmatrix}$
$\Rightarrow\text{A}\text{B}=\begin{bmatrix}6&7\\18&21\end{bmatrix}$
$\text{AC}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix}\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}$
$\Rightarrow\text{AC}=\begin{bmatrix}4-3+5&2+5+0\\12-9+15&6+15+0\end{bmatrix}$
$\Rightarrow\text{AC}=\begin{bmatrix}6&7\\18&21\end{bmatrix}$
So, AB = AC though B ≠ C, A ≠ O.

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