If \[A=\left[ \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix} 2 & 2 & 4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \\ \end{matrix} \right],\] find AB. Use this to solve the system of equations |
\[x-y=3,\] \[2x+3y+4z=17\] and \[y+2z=7.\] |
OR |
By using elementary row operations, find the inverse of the matrix \[A=\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \\ \end{matrix} \right].\] |
Answer:
We have,
\[AB=\left[ \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \\ \end{matrix} \right]\,\,\left[ \begin{matrix} 2 & 2 & -\,4 \\ -\,4 & 2 & -\,4 \\ 2 & -1 & 5 \\ \end{matrix} \right]\]
\[=\left[ \begin{matrix} 2+4+0 & 2-2+0 & -\,4+4+0 \\ 4-12+8 & 4+6-4 & -\,8-12+20 \\ 0-4+4 & 0+2-2 & 0-4+10 \\ \end{matrix} \right]\,\]
\[=\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right]\,\,=6\,\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=6l\]
\[\therefore \] AB = 6 / ?(i)
Now given system of equations can be written in matrix form as \[AX=C\] \[\Rightarrow \] \[X={{A}^{-1}}C\] ? (ii)
Where,\[A=\left[ \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \\ \end{matrix} \right],\] \[X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]\] and \[C=\left[ \begin{matrix} 3 \\ 17 \\ 7 \\ \end{matrix} \right]\]
Now, from Eq, (i) we have
AB = 6 /
\[\Rightarrow \] \[{{A}^{-1}}(AB)={{A}^{-1}}(6l)\]
[Pre-multiply both sides by \[{{A}^{-1}}\]]
\[\Rightarrow \] \[({{A}^{-1}}A)B=6({{A}^{-1}}l)\]
\[\Rightarrow \] \[lB=6{{A}^{-1}}\]
\[[\because \,A{{A}^{-1}}={{A}^{-1}}A=l\,\,\text{and}\,\,Al=A-lA]\]
\[\Rightarrow \] \[B=6{{A}^{-1}}\]
\[\Rightarrow \] \[{{A}^{-1}}=\frac{1}{6}B\]
\[\Rightarrow \] \[{{A}^{-1}}=\frac{1}{6}\,\,\left[ \begin{matrix} 2 & 2 & -\,4 \\ -\,4 & 2 & -\,4 \\ 2 & -1 & 5 \\ \end{matrix} \right]\]
Now, from Eq. (ii) we have
\[X={{A}^{-1}}C\]
\[\Rightarrow \] \[\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\frac{1}{6}\,\,\left[ \begin{matrix} 6 & +\,34 & -\,28 \\ -\,12 & +\,34 & -\,28 \\ 6 & -\,17 & +\,35 \\ \end{matrix} \right]\]
\[=\frac{1}{6}\,\,\left[ \begin{matrix} 12 \\ -6 \\ 24 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 \\ -1 \\ 4 \\ \end{matrix} \right]\]
On comparing corresponding elements, we get
\[x=2,\] \[y=-\,1\] and z = 4
OR
We have,
\[\left[ \begin{matrix} 1 & 3 & -\,2 \\ -\,3 & 0 & -\,5 \\ 2 & 5 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\cdot A\]
\[\Rightarrow \] \[Z=22(0)+18(20)=360\]
\[\left[ \begin{align} & {{R}_{2}}\to {{R}_{2}}+3{{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-2{{R}_{1}} \\ \end{align} \right]\]
\[\Rightarrow \] \[\left[ \begin{matrix} 1 & 3 & -\,2 \\ 0 & -\,1 & 4 \\ 0 & 9 & -\,11 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ -\,2 & 0 & 1 \\ 3 & 1 & 0 \\ \end{matrix} \right]\cdot A\] \[\,\left[ {{R}_{2}}\leftrightarrow {{R}_{1}} \right]\]
\[\Rightarrow \] \[\left[ \begin{matrix} 1 & 0 & 10 \\ 0 & -\,1 & 4 \\ 0 & 0 & 25 \\ \end{matrix} \right]=\left[ \begin{matrix} -\,5 & 0 & 3 \\ -\,2 & 0 & 1 \\ -\,15 & 1 & 9 \\ \end{matrix} \right]\cdot A\]
\[\,\,\left[ \begin{matrix} {{R}_{1}}\to {{R}_{1}}+3{{R}_{2}} \\ {{R}_{3}}\to {{R}_{3}}+9{{R}_{2}} \\ \end{matrix} \right]\]
\[\Rightarrow \]\[\left[ \begin{matrix} 1 & 0 & 10 \\ 0 & 1 & -\,4 \\ 0 & 0 & 25 \\ \end{matrix} \right]=\left[ \begin{matrix} -\,5 & 0 & 3 \\ 2 & 0 & -\,1 \\ -\,15 & 1 & 9 \\ \end{matrix} \right]\cdot A\] \[\left[ {{R}_{2}}\to (-\,1){{R}_{2}} \right]\]
\[\Rightarrow \] \[\left[ \begin{matrix} 1 & 0 & 10 \\ 0 & 1 & -\,4 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -\,5 & 0 & 3 \\ 2 & 0 & -\,1 \\ -\,3/5 & 1/25 & 9/25 \\ \end{matrix} \right]\cdot A\]
\[\,\left[ {{R}_{3}}\to \frac{1}{25}{{R}_{3}} \right]\]
\[\Rightarrow \] \[\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & \frac{-\,2}{5} & \frac{-\,3}{5} \\ \frac{-\,2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{-\,3}{5} & \frac{1}{25} & \frac{9}{25} \\ \end{matrix} \right]\cdot A\]
\[\left[ \begin{matrix} {{R}_{1}}\to {{R}_{1}}\to 10{{R}_{3}} \\ {{R}_{2}}\to {{R}_{2}}\to 4{{R}_{3}} \\ \end{matrix} \right]\]
Hence, \[{{A}^{-1}}=\left[ \begin{matrix} 1 & \frac{-\,2}{5} & \frac{-\,3}{5} \\ \frac{-\,2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{-\,3}{5} & \frac{1}{25} & \frac{9}{25} \\ \end{matrix} \right]\]
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