Exercise 5.4
Question 6
\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]
Sol :
A=\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]
A=IA
\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A
R1↔R2
\left[\begin{array}{ll}5 & 7 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] A
R1→R1-2R2
\left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ 1 & 0\end{array}\right]A
R2→R2-2R1
\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ 5 & -2\end{array}\right]A
R1→R1-R2
\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-7 & 3 \\ 5 & -2\end{array}\right]A
A^{-1}=\left[\begin{array}{rr}-7 & 3 \\ 5 & -2\end{array}\right]
\left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]
Sol :
A=\left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]
A=IA
\left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A
R1→R1-R2
\left[\begin{array}{ll}1 & 1 \\ 3 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] A
R2→R2-3R1
\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -3 & 4\end{array}\right] A
R1→R1-R2
\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right]A
A^{-1}=\left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right]
\left[\begin{array}{rr}3 & -1 \\ -4 & 2\end{array}\right]
Sol :
A=\left[\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right]
A=IA
\left[\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A
R1→R1-R2
\left[\begin{array}{cc}-1 & 1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] A
R1→-1R1
\left[\begin{array}{cc}1 & -1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 0 & 1\end{array}\right] A
R2→R2+4R1
\left[\begin{array}{cc}1 & -1 \\ 0 & -2\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ -4 & -3\end{array}\right]A
R_{2} \rightarrow-\frac{1}{2} R_{2}
\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 2 & \frac{3}{2}\end{array}\right]A
R1→R1-R2
\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & \frac{1}{2} \\ 2 & \frac{3}{2}\end{array}\right]A
A^{-1}=\left[\begin{array}{ll}1 & \frac{1}{2} \\ 2 & \frac{3}{2}\end{array}\right]
Sol :
A=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]
A=IA
\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]A
R1↔R2
\left[\begin{array}{ll}4 & 2 \\ 2 & 1\end{array}\right]=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] A
R1→R1-2R2
\left[\begin{array}{cc}0 & 0 \\ 2 & 1\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ 1 & 0\end{array}\right] A
∴ A-1 का अस्तित्व नही हैं
Sol :
A=\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]
A=IA
\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]A
R2→R2-2R1
\left[\begin{array}{cc}1 & -1 \\ 0 & 5\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] A
R_{2} \rightarrow \frac{1}{5} R_{2}
\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]A
R1→R1+R2
\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]A
A^{-1}=\left[\begin{array}{cc}\frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]
Sol :
A=\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]
A=IA
\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A
R1↔R2
\left[\begin{array}{ll}5 & 7 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] A
R1→R1-2R2
\left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ 1 & 0\end{array}\right]A
R2→R2-2R1
\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ 5 & -2\end{array}\right]A
R1→R1-R2
\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-7 & 3 \\ 5 & -2\end{array}\right]A
A^{-1}=\left[\begin{array}{rr}-7 & 3 \\ 5 & -2\end{array}\right]
Question 7
Sol :
A=\left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]
A=IA
\left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A
R1→R1-R2
\left[\begin{array}{ll}1 & 1 \\ 3 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] A
R2→R2-3R1
\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -3 & 4\end{array}\right] A
R1→R1-R2
\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right]A
A^{-1}=\left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right]
Question 8
Sol :
A=\left[\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right]
A=IA
\left[\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A
R1→R1-R2
\left[\begin{array}{cc}-1 & 1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] A
R1→-1R1
\left[\begin{array}{cc}1 & -1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 0 & 1\end{array}\right] A
R2→R2+4R1
\left[\begin{array}{cc}1 & -1 \\ 0 & -2\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ -4 & -3\end{array}\right]A
R_{2} \rightarrow-\frac{1}{2} R_{2}
\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 2 & \frac{3}{2}\end{array}\right]A
R1→R1-R2
\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & \frac{1}{2} \\ 2 & \frac{3}{2}\end{array}\right]A
A^{-1}=\left[\begin{array}{ll}1 & \frac{1}{2} \\ 2 & \frac{3}{2}\end{array}\right]
Question 9
\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]Sol :
A=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]
A=IA
\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]A
R1↔R2
\left[\begin{array}{ll}4 & 2 \\ 2 & 1\end{array}\right]=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] A
R1→R1-2R2
\left[\begin{array}{cc}0 & 0 \\ 2 & 1\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ 1 & 0\end{array}\right] A
∴ A-1 का अस्तित्व नही हैं
Question 10
\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]Sol :
A=\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]
A=IA
\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]A
R2→R2-2R1
\left[\begin{array}{cc}1 & -1 \\ 0 & 5\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] A
R_{2} \rightarrow \frac{1}{5} R_{2}
\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]A
R1→R1+R2
\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]A
A^{-1}=\left[\begin{array}{cc}\frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]
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