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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES2 Marks
Question
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{AB})^{\text{T}}=\text{B}^{\text{T}}\text{A}^{\text{T}}.$
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Answer

We have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2

$\text{AB}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}4&0\\1&5\end{bmatrix}$

$=\begin{bmatrix}4+2&0+10\\-4+3&0+15\end{bmatrix}=\begin{bmatrix}6&10\\-1&15\end{bmatrix}$

$\therefore\ (\text{AB})^{\text{T}}=\begin{bmatrix}6&-1\\10&15\end{bmatrix}$

Now, $\text{B}^{\text{T}}\text{A}^{\text{T}}=\begin{bmatrix}4&1\\0&5\end{bmatrix}\begin{bmatrix}1&-1\\2&3\end{bmatrix}$

$=\begin{bmatrix}6&-1\\10&15\end{bmatrix}$

$=(\text{AB})^{\text{T}}$

Hence proved.

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