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Algebra of Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices3 Marks
Question
If $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{ B}=\text{diag}\begin{pmatrix}1&1&-4\end{pmatrix}$ and $\text{C}=\text{diag}\begin{pmatrix}-6&3&4\end{pmatrix},$ find.
$2\text{A}+3\text{B}-5\text{C}$
✓

Answer

 Given, $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{B}\begin{pmatrix}1&1&-4\end{pmatrix}$ and $\text{C}=\text{diag}\begin{pmatrix}-\text{b}&3&4\end{pmatrix}$

$2\text{A}+3\text{B}-5\text{C}$

$\Rightarrow2\text{A}+3\text{B}-5\text{C}=2\begin{bmatrix}2&0&0\\0&-5&0\\0&0&9\end{bmatrix}+3\begin{bmatrix}1&0&0\\0&1&0\\0&0&-4\end{bmatrix}-5\begin{bmatrix}-6&0&0\\0&3&0\\0&0&4\end{bmatrix}$

$\Rightarrow2\text{A}+3\text{B}-5\text{C}=\begin{bmatrix}4&0&0\\0&-10&0\\0&0&18\end{bmatrix}+\begin{bmatrix}3&0&0\\0&3&0\\0&0&-12\end{bmatrix}-\begin{bmatrix}-30&0&0\\0&15&0\\0&0&20\end{bmatrix}$

$\Rightarrow2\text{A}+3\text{B}-5\text{C}=\begin{bmatrix}4+3+30&0+0-0&0+0-0\\0+0-0&-10+3-15&0+0-0\\0+0-0&0+0-0&18-12-20\end{bmatrix}$

$\Rightarrow2\text{A}+3\text{B}-5\text{C}=\begin{bmatrix}37&0&0\\0&-22&0\\0&0&-14\end{bmatrix}=\text{diag}\begin{pmatrix}37&-22&-14\end{pmatrix}$ 

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