Question
Question: Find the matrix A, such that \[\left[ {\begin{array}{*{20}{c}} 4 \\\ 1 \\\ 3 \end{a...
Find the matrix A, such that \left[ {\begin{array}{*{20}{c}} 4 \\\ 1 \\\ 3 \end{array}} \right]A = \left[ {\begin{array}{*{20}{c}} { - 4}&8&4 \\\ { - 1}&2&1 \\\ { - 3}&6&3 \end{array}} \right].
Solution
We will first assume that the matrix A is \left[ {\begin{array}{*{20}{c}} x&y;&z; \end{array}} \right]. Then we will use this value of A in the given equation and simplify. Then we will use that the corresponding elements of two equal matrices are equal to find the required value.
Complete step by step answer:
We are given that
4 \\\ 1 \\\ 3 \end{array}} \right]A = \left[ {\begin{array}{*{20}{c}} { - 4}&8&4 \\\ { - 1}&2&1 \\\ { - 3}&6&3 \end{array}} \right]{\text{ ......eq.(1)}}$$ Let us assume that the matrix A is $$\left[ {\begin{array}{*{20}{c}} x&y;&z; \end{array}} \right]$$. Using the value of A in the equation (1), we get\Rightarrow \left[ {\begin{array}{{20}{c}}
4 \\
1 \\
3
\end{array}} \right]\left[ {\begin{array}{{20}{c}}
x&y;&z;
\end{array}} \right] = \left[ {\begin{array}{{20}{c}}
{ - 4}&8&4 \\
{ - 1}&2&1 \\
{ - 3}&6&3
\end{array}} \right] \\
\Rightarrow \left[ {\begin{array}{{20}{c}}
{4x}&{4y}&{4z} \\
x&y;&z; \\
{3x}&{3y}&{3z}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ - 4}&8&4 \\
{ - 1}&2&1 \\
{ - 3}&6&3
\end{array}} \right] \\
\Rightarrow \dfrac{{4x}}{4} = \dfrac{{ - 4}}{4} \\
\Rightarrow x = - 1 \\
\Rightarrow \dfrac{{4y}}{4} = \dfrac{8}{4} \\
\Rightarrow y = 2 \\
\Rightarrow \dfrac{{4z}}{4} = \dfrac{4}{4} \\
\Rightarrow z = 1 \\