Question

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\begin{gathered}\sf\left[\begin{array}{cc}2&-1\\ 1&0\\ - 3&4\end{array}\right]\end{gathered}A = \begin{gathered}\sf\left[\begin{array}{ccc} - 1&-8& - 10\\1& - 2&-5\\9&22&15\end{array}\right]\end{gathered}$

Since, we have to find unknown matrix A, so let first find the order of matrix A.

Let assume that,

$\rm :\longmapsto\:B = \begin{gathered}\sf\left[\begin{array}{ccc} - 1&-8& - 10\\1& - 2&-5\\9&22&15\end{array}\right]\end{gathered}$

and

$\rm :\longmapsto\:C = \begin{gathered}\sf\left[\begin{array}{cc}2&-1\\1&0\\ - 3&4\end{array}\right]\end{gathered}$

Now,

$\bf\implies \:BA = C$

Now,

Order of matrix B is 3 × 2 and order of matrix C is 3 × 3.

Let assume that order of matrix A is x × y.

Now, BA = C

It means BA is defined when 2 = x and resultant of BA = C

It means, order of BA = 3 × 3, so it implies y = 3.

Hence, order of matrix A is 2 × 3.

So, Let assume that,

$\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}a&b&c\\d&e&f\end{array}\right]\end{gathered}$

Now, we have

$\rm :\longmapsto\:BA = C$

On substituting the values, we get

$\rm :\longmapsto\:\begin{gathered}\sf\left[\begin{array}{cc}2&-1\\ 1&0\\ - 3&4\end{array}\right]\end{gathered}\begin{gathered}\sf \left[\begin{array}{ccc}a&b&c\\d&e&f\end{array}\right]\end{gathered} = \begin{gathered}\sf\left[\begin{array}{ccc} - 1&-8& - 10\\1& - 2&-5\\9&22&15\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf\left[\begin{array}{ccc}2a - d&2b - e&2c - f\\a&b&c\\ - 3a + 4d&-3b + 4e& - 3c + 4f\end{array}\right]\end{gathered} = \begin{gathered}\sf\left[\begin{array}{ccc} - 1&-8& - 10\\1& - 2&-5\\9&22&15\end{array}\right]\end{gathered}$

On comparing, we get

$\rm :\longmapsto\:a = 1$

$\rm :\longmapsto\:b = - 2$

$\rm :\longmapsto\:c = - 5$

$\rm :\longmapsto\:2a - d = - 1 \implies\:2 - d = - 1\implies\:d = 3$

$\rm :\longmapsto\:2b - e = - 8 \implies\: - 4 - e= - 8\implies\:e = 4$

$\rm :\longmapsto\:2c - f = - 10 \implies\: - 10 - f= - 10\implies\:f = 0$

Hence,

$\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}1& - 2& - 5\\3&4&0\end{array}\right]\end{gathered}$