If A= [ lll 1 & 0 & 1 0 & 2 & 3 1 & 2 & 1 ] and B= [ lll 1 & 2 & 3 1 & 1 & 5 2 & 4 & 7 ], then find a matrix X such that X A=B.

Consider XA = B By C_3-C_1, we get, X [ lll 1 & 0 & 0 0 & 2 & 3 1 & 2 & 0 ]= [ lll 1 & 2 & 2 1 & 1 & 4 2 & 4 & 5 ] By (1 2 ) C_2, we get, X [ lll 1 & 0 & 0 0 & 1 & 3 1 & 1 & 0 ]= [ rrr 1 & 1 & 2 1 & 1 / 2 & 4 2 & 2 & 5 ] By C_3-3 C_2, we get, X [ rrr 1 & 0 & 0 0 & 1 & 0 1 & 1 & -3 ]= [ rrr 1 & 1 & -1 1 & 1 / 2 & 5 / 2 2 & 2 & -1 ] By (-1 3 ) C_3, we get, X [ lll 1 & 0 & 0 0 & 1 & 0 1 & 1 & 1 ]= [ rrr 1 & 1 & 1 / 3 1 & 1 / 2 & -5 / 6 2 & 2 & 1 / 3 ] By C_1-C_3 and C_2-C_3, we get, X [ lll 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 ]= [ rrr 2 / 3 & 2 / 3 & 1 / 3 11 / 6 & 4 / 3 & -5 / 6 5 / 3 & 5 / 3 & 1 / 3 ] X=1 6 [ rrr 4 & 4 & 2 11 & 8 & -5 10 & 10 & 2 ]

If A= [ lll 1 & 0 & 1 0 & 2 & 3 1 & 2 & 1 ] and B= [ lll 1 & 2 & …

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Question

If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]$, then find a matrix $X$ such that $X A=B$.

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Answer

Consider $XA = B$ By $C_3-C_1,$ we get,
$X\left[\begin{array}{lll}1 & 0 & 0 \\0 & 2 & 3 \\1 & 2 & 0\end{array}\right]=\left[\begin{array}{lll}1 & 2 & 2 \\1 & 1 & 4 \\2 & 4 & 5\end{array}\right]$
By $\left(\frac{1}{2}\right) C_2$, we get,
$X\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 3 \\1 & 1 & 0\end{array}\right]=\left[\begin{array}{rrr}1 & 1 & 2 \\1 & 1 / 2 & 4 \\2 & 2 & 5\end{array}\right]$
By $\mathrm{C}_3-3 \mathrm{C}_2$, we get,
$X\left[\begin{array}{rrr}1 & 0 & 0 \\0 & 1 & 0 \\1 & 1 & -3\end{array}\right]=\left[\begin{array}{rrr}1 & 1 & -1 \\1 & 1 / 2 & 5 / 2 \\2 & 2 & -1\end{array}\right]$
By $\left(-\frac{1}{3}\right) C_3$, we get,
$X\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\1 & 1 & 1\end{array}\right]=\left[\begin{array}{rrr}1 & 1 & 1 / 3 \\1 & 1 / 2 & -5 / 6 \\2 & 2 & 1 / 3\end{array}\right]$
By $C_1-C_3$ and $C_2-C_3$, we get,
$X\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{rrr}2 / 3 & 2 / 3 & 1 / 3 \\ 11 / 6 & 4 / 3 & -5 / 6 \\ 5 / 3 & 5 / 3 & 1 / 3\end{array}\right] $
$ \therefore X=\frac{1}{6}\left[\begin{array}{rrr}4 & 4 & 2 \\ 11 & 8 & -5 \\ 10 & 10 & 2\end{array}\right] $

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