Find the matrix A such that 2&-1 1&0 -3&4 A= -1&-8&-10 1&-2&-10 9&22&15 .

We have, 2&-1 1&0 -3&4 _ 3 2 A= -1&-8&-10 1&-2&-10 9&22&15 _ 3 3 From the given equation, it is clear that order of A should be 2 × 3. Let A= a&b&c &e&f 2&-1 1&0 -3&4 a&b&c &e&f = -1&-8&-10 1&-2&-10 9&22&15 2a-d&2b-e&2c-f &b&c -3a+4d&-3b+4e&-3c+4f = -1&-8&-10 1&-2&-5 9&22&15 By equality of matrices, we geta = 1, b = -2, c = -5 and 2a - d = -1 ⇒ d = 2a + 1 =3; 2b - e = -8 ⇒ e = 2(-2) + 8 = 4 2c - f = -10 ⇒ f = 2c + 10 = 0 A= 1&-2&-5 3&4&0

Find the matrix A such that 2&-1 1&0 -3&4 A= -1&-8&-10 1&-2&-10 9…

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Question

Find the matrix A such that $\begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}\text{A}=\begin{bmatrix}-1&-8&-10\\1&-2&-10\\9&22&15\end{bmatrix}.$

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Answer

We have, $\begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}_{3\times2}\text{A}=\begin{bmatrix}-1&-8&-10\\1&-2&-10\\9&22&15\end{bmatrix}_{3\times3}$ From the given equation, it is clear that order of A should be 2 × 3. Let $\text{A}=\begin{bmatrix}\text{a}&\text{b}&\text{c}\\\text{d}&\text{e}&\text{f}\end{bmatrix}$ $\therefore\ \begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}\begin{bmatrix}\text{a}&\text{b}&\text{c}\\\text{d}&\text{e}&\text{f}\end{bmatrix}=\begin{bmatrix}-1&-8&-10\\1&-2&-10\\9&22&15\end{bmatrix}$ $\Rightarrow\ \begin{bmatrix}2\text{a}-\text{d}&2\text{b}-\text{e}&2\text{c}-\text{f}\\\text{a}&\text{b}&\text{c}\\-3\text{a}+4\text{d}&-3\text{b}+4\text{e}&-3\text{c}+4\text{f}\end{bmatrix}=\begin{bmatrix}-1&-8&-10\\1&-2&-5\\9&22&15\end{bmatrix}$ By equality of matrices, we geta = 1, b = -2, c = -5
and 2a - d = -1 ⇒ d = 2a + 1 =3;
2b - e = -8 ⇒ e = 2(-2) + 8 = 4
2c - f = -10 ⇒ f = 2c + 10 = 0
$\therefore\ \text{A}=\begin{bmatrix}1&-2&-5\\3&4&0\end{bmatrix}$