Question#1: Find the determinant of the following matrices.
\(i\)
\[A=\left[ \begin{matrix}
-1 & 1 \\
2 & 0 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| A \right|&=\left| \begin{matrix}
-1 & 1 \\
2 & 0 \\
\end{matrix} \right| \\
& =\left( -1 \right)\left( 0 \right)-\left(1 \right)\left( 2 \right) \\
& =0-2 \\
& =-2 \\
\end{align}\]
\(ii\)
\[B=\left[ \begin{matrix}
1 & 3 \\
2 & -2 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| B \right|&=\left| \begin{matrix}
1 & 3 \\
2 & -2 \\
\end{matrix} \right| \\
& =\left( 1 \right)\left( -2 \right)-\left(3 \right)\left( 2 \right) \\
& =-2-6 \\
& =-8 \\
\end{align}\]
\(iii\)
\[C=\left[ \begin{matrix}
3 & 2 \\
3 & 2 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| C \right|&=\left| \begin{matrix}
3 & 2 \\
3 & 2 \\
\end{matrix} \right| \\
& =\left( 3 \right)\left( 2 \right)-\left(3 \right)\left( 2 \right) \\
& =6-6 \\
& =0 \\
\end{align}\]
\(iv\)
\[D=\left[ \begin{matrix}
3 & 2 \\
1 & 4 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| D \right|&=\left| \begin{matrix}
3 & 2 \\
1 & 4 \\
\end{matrix} \right| \\
& =\left( 3 \right)\left( 4 \right)-\left(2 \right)\left( 1 \right) \\
& =12-2 \\
& =10 \\
\end{align}\]
Question#2: Find which of the following matrices are singular or non-singular.
\(i\)
\[A=\left[ \begin{matrix}
3 & 6 \\
2 & 4 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| A \right|&=\left| \begin{matrix}
3 & 6 \\
2 & 4 \\
\end{matrix} \right| \\
& =\left( 3 \right)\left( 4 \right)-\left(6 \right)\left( 2 \right) \\
& =12-12 \\
& =0 \\
A&\textbf{ Matrix is Singular}
\end{align}\]
\(ii\)
\[B=\left[ \begin{matrix}
4 & 1 \\
3 & 2 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| B \right|&=\left| \begin{matrix}
4 & 1 \\
3 & 2 \\
\end{matrix} \right| \\
& =\left( 4 \right)\left( 2 \right)-\left(1 \right)\left( 3 \right) \\
& =8-3 \\
& =5 \\
B&\textbf{ Matrix is Non Singular}
\end{align}\]
\(iii\)
\[C=\left[ \begin{matrix}
7 & -9 \\
3 & 5 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| C \right|&=\left| \begin{matrix}
7 & -9 \\
3 & 5 \\
\end{matrix} \right| \\
& =\left( 7 \right)\left( 5 \right)-\left(-9 \right)\left( 3 \right) \\
& =35+27 \\
& =62 \\
C&\textbf{ Matrix is Non Singular}
\end{align}\]
\(iv\)
\[D=\left[ \begin{matrix}
5 & -10 \\
-2 & 4 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| D \right|&=\left| \begin{matrix}
5 & -10 \\
-2 & 4 \\
\end{matrix} \right| \\
& =\left( 5 \right)\left( 4 \right)-\left(-10 \right)\left( -2 \right) \\
& =20-20 \\
& =0 \\
D&\textbf{ Matrix is Singular}
\end{align}\]
Question#3: Find the multiplicative inverse (if it exists) of each.
\(i\)
\[A=\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| A \right|&=\left| \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right| \\
& =\left( -1 \right)\left( 0 \right)-\left(3 \right)\left( 2 \right) \\
& =0-6 \\
& =-6 \\
&\textbf{ Inverse is possible}\\
Adj\left( A \right) &=\left[ \begin{matrix}
0 & -3 \\
-2 & -1 \\
\end{matrix} \right] \\
{{A}^{-1}}&=\frac{Adj\left( A \right)}{\left| A \right|} \\
& =\frac{\left[ \begin{matrix}
0 & -3 \\
-2 & -1 \\
\end{matrix} \right]}{-6} \\
& =\left[ \begin{matrix}
\frac{0}{-6} & \frac{-3}{-6} \\
\frac{-2}{-6} & \frac{-1}{-6} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & \frac{1}{2} \\
\frac{1}{3} & \frac{1}{6} \\
\end{matrix} \right] \\
\end{align}\]
\(ii\)
\[B=\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| B \right|&=\left| \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right| \\
& =\left( 1 \right)\left( -5 \right)-\left(2 \right)\left( -3 \right) \\
& =-5+6 \\
& =1 \\
&\textbf{ Inverse is possible}\\
Adj\left( B \right) &=\left[ \begin{matrix}
-5 & -2 \\
3 & 1 \\
\end{matrix} \right] \\
{{B}^{-1}}&=\frac{Adj\left( B \right)}{\left| B \right|} \\
& =\frac{\left[ \begin{matrix}
-5 & -2 \\
3 & 1 \\
\end{matrix} \right]}{1} \\
& =\left[ \begin{matrix}
-5 & -2 \\
3 & 1 \\
\end{matrix} \right] \\
\end{align}\]
\(iii\)
\[C=\left[ \begin{matrix}
-2 & 6 \\
3 & -9 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| C \right|&=\left| \begin{matrix}
-2 & 6 \\
3 & -9 \\
\end{matrix} \right| \\
& =\left( -2 \right)\left( -9 \right)-\left(6 \right)\left( 3 \right) \\
& =18-18 \\
& =0 \\
&\textbf{ Inverse is not possible}\\
\end{align}\]
\(iv\)
\[D=\left[ \begin{matrix}
\frac{1}{2} & \frac{3}{4} \\
1 & 2 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left| D \right|&=\left| \begin{matrix}
\frac{1}{2} & \frac{3}{4} \\
1 & 2 \\
\end{matrix} \right| \\
& =\left( \frac{1}{2} \right)\left( 2 \right)-\left(\frac{3}{4} \right)\left( 1 \right) \\
& =1-\frac{3}{4} \\
& =\frac{4-3}{4} \\
& =\frac{1}{4} \\
&\textbf{ Inverse is possible}\\
Adj\left( D \right) &=\left[ \begin{matrix}
2 & -\frac{3}{4} \\
-1 & \frac{1}{2} \\
\end{matrix} \right] \\
{{D}^{-1}}&=\frac{Adj\left( D \right)}{\left| D \right|} \\
& =\frac{\left[ \begin{matrix}
2 & -\frac{3}{4} \\
-1 & \frac{1}{2} \\
\end{matrix} \right]}{\frac{1}{4}} \\
& =4\left[ \begin{matrix}
2 & -\frac{3}{4} \\
-1 & \frac{1}{2} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2\times 4 & -\frac{3}{4}\times 4 \\
-1\times 4 & \frac{1}{2}\times 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
8 & -3 \\
-4 & 2 \\
\end{matrix} \right] \\
\end{align}\]
Question#4: If \(A=\left[ \begin{matrix}1 & 2 \\4 & 6 \\ \end{matrix}\right]\) and \(B=\left[ \begin{matrix}3 & -1 \\2 & -2 \\ \end{matrix}\right]\) then prove that.
\(i\)
\[A\left( Adj\left( A \right) \right)=\left( Adj\left( A \right) \right)A=\left( \det A \right)I\]
Solution:
\[\begin{align}
Adj\left( A \right)&=\left[ \begin{matrix}
6 & -2 \\
-4 & 1 \\
\end{matrix} \right] \\
A\left( Adj\left( A \right) \right)&=\left[ \begin{matrix}
1 & 2 \\
4 & 6 \\
\end{matrix} \right]\left[ \begin{matrix}
6 & -2 \\
-4 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
6-8 & -2+2 \\
24-24 & -8+2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2 & 0 \\
0 & -2 \\
\end{matrix} \right]…(i) \\
\left( Adj\left( A \right) \right)A&=\left[ \begin{matrix}
6 & -2 \\
-4 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 \\
4 & 6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
6-8 & 12-12 \\
-4+4 & -8+6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2 & 0 \\
0 & -2 \\
\end{matrix} \right]…(ii) \\
\det A&=\left| \begin{matrix}
1 & 2 \\
4 & 6 \\
\end{matrix} \right| \\
& =\left( 1 \right)\left( 6 \right)-\left( 2 \right)\left( 4 \right) \\
& =6-8 \\
& =-2 \\
\left( \det A \right)I&=-2\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2 & 0 \\
0 & -2 \\
\end{matrix} \right]…(\text{iii}) \\
& \text{From (i), (ii), and (iii), we have} \\
& A\left( Adj\left( A \right) \right)=\left( Adj\left( A \right) \right)A=\left( \det A \right)I \\
& \text{ } \\
\end{align}\]
\(ii\)
\[B{{B}^{-1}}=I={{B}^{-1}}B\]
Solution:
\[\begin{align}
Adj\left( B \right) &=\left[ \begin{matrix}
-2 & 1 \\
-2 & 3 \\
\end{matrix} \right] \\
\left| B \right|&=\left| \begin{matrix}
3 & -1 \\
2 & -2 \\
\end{matrix} \right| \\
& =\left( 3 \right)\left( -2 \right)-\left( -1 \right)\left( 2 \right) \\
& =-6+2 \\
& =-4 \\
{{B}^{-1}} & =\frac{1}{\left| B \right|}\times Adj\left( B \right) \\
& =\frac{1}{-4}\left[ \begin{matrix}
2 & 1 \\
-2 & 3 \\
\end{matrix} \right] \\
B{{B}^{-1}}&=\left[ \begin{matrix}
3 & -1 \\
2 & -2 \\
\end{matrix} \right]\times \frac{1}{-4}\left[ \begin{matrix}
-2 & 1 \\
-2 & 3 \\
\end{matrix} \right] \\
& =\frac{1}{-4}\left[ \begin{matrix}
3 & -1 \\
2 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
-2 & 1 \\
-2 & 3 \\
\end{matrix} \right] \\
& =\frac{1}{-4}\left[ \begin{matrix}
-6+2 & 3-3 \\
-4+4 & 2-6 \\
\end{matrix} \right] \\
& =\frac{1}{-4}\left[ \begin{matrix}
-4 & 0 \\
0 & -4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{-4}{-4} & \frac{0}{-4} \\
\frac{0}{-4} & \frac{-4}{-4} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
& =I…(i) \\
{{B}^{-1}}B&=\frac{1}{-4}\left[ \begin{matrix}
-2 & 1 \\
-2 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
3 & -1 \\
2 & -2 \\
\end{matrix} \right] \\
& =\frac{1}{-4}\left[ \begin{matrix}
-2 & 1 \\
-2 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
3 & -1 \\
2 & -2 \\
\end{matrix} \right] \\
& =\frac{1}{-4}\left[ \begin{matrix}
-6+2 & 2-2 \\
-6+6 & 2-6 \\
\end{matrix} \right] \\
& =\frac{1}{-4}\left[ \begin{matrix}
-4 & 0 \\
0 & -4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{-4}{-4} & \frac{0}{-4} \\
\frac{0}{-4} & \frac{-4}{-4} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
& =I…(ii) \\
& \text{From (i) and (ii) we have} \\
B{{B}^{-1}} & =I={{B}^{-1}}B \\
\end{align}\]
Question#5: Determine whether the given matrices are multiplicative inverse of each other.
\(i\)
\[\left[ \begin{matrix}
3 & 5 \\
4 & 7 \\
\end{matrix} \right]\text{ and }\left[ \begin{matrix}
7 & -5 \\
-4 & 3 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
3 & 5 \\
4 & 7 \\
\end{matrix} \right]\left[ \begin{matrix}
7 & -5 \\
-4 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
21-20 & -15+15 \\
28-28 & -20+21 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
\end{align}\]
Given matrices are the multiplicative inverse of each other since their product equal to an identity matrix
\(ii\)
\[\left[ \begin{matrix}
1 & 2 \\
2 & 3 \\
\end{matrix} \right]\text{ and }\left[ \begin{matrix}
-3 & 2 \\
2 & -1 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
1 & 2 \\
2 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
-3 & 2 \\
2 & -1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-3+4 & 2-2 \\
-6+6 & 4-3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
\end{align}\]
Given matrices are the multiplicative inverse of each other since their product equal to an identity matrix
Question#6: If \(A=\left[ \begin{matrix}4 & 0 \\-1 & 2 \\ \end{matrix}\right]\) and \(B=\left[ \begin{matrix}-4 & -2 \\1 & -1 \\ \end{matrix}\right]\) and \(D=\left[ \begin{matrix}3 & 1 \\-2 & 2 \\ \end{matrix}\right]\) then prove that.
\(i\)
\[{{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}\]
Solution:
\[\begin{align}
AB&=\left[ \begin{matrix}
4 & 0 \\
-1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
-4 & -2 \\
1 & -1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-16+0 & -8-0 \\
4+2 & 2-2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-16 & -8 \\
6 & 0 \\
\end{matrix} \right] \\
\left| AB \right|&=\left| \begin{matrix}
-16 & -8 \\
6 & 0 \\
\end{matrix} \right| \\
& =\left( -16 \right)\left( 0 \right)-\left( -8 \right)\left( 6 \right) \\
& =0+48 \\
& =48 \\
Adj\left( AB \right) &=\left[ \begin{matrix}
0 & 8 \\
-6 & -16 \\
\end{matrix} \right] \\
{{\left( AB \right)}^{-1}} &=\frac{1}{\left| AB \right|}\times Adj\left( AB \right) \\
& =\frac{1}{48}\left[ \begin{matrix}
0 & 8 \\
-6 & -16 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{0}{48} & \frac{8}{48} \\
\frac{-6}{48} & \frac{-16}{48} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & \frac{1}{6} \\
-\frac{1}{8} & -\frac{1}{3} \\
\end{matrix} \right]…(i) \\
B& =\left[ \begin{matrix}
-4 & -2 \\
1 & -1 \\
\end{matrix} \right] \\
\left| B \right|& =\left| \begin{matrix}
-4 & -2 \\
1 & -1 \\
\end{matrix} \right| \\
& =\left( -4 \right)\left( -1 \right)-\left( -2 \right)\left( 1 \right) \\
& =4+2 \\
& =6 \\
Adj\left( B \right) &=\left[ \begin{matrix}
-1 & 2 \\
-1 & -4 \\
\end{matrix} \right] \\
{{B}^{-1}}&=\frac{1}{\left| B \right|}\times Adj\left( B \right) \\
& =\frac{1}{6}\left[ \begin{matrix}
-1 & 2 \\
-1 & -4 \\
\end{matrix} \right] \\
A&=\left[ \begin{matrix}
4 & 0 \\
-1 & 2 \\
\end{matrix} \right] \\
\left| A \right|&=\left| \begin{matrix}
4 & 0 \\
-1 & 2 \\
\end{matrix} \right| \\
& =\left( 4 \right)\left( 2 \right)-\left( 0 \right)\left( -1 \right) \\
& =8-0 \\
& =8 \\
Adj\left( A \right) &=\left[ \begin{matrix}
2 & 0 \\
1 & 4 \\
\end{matrix} \right] \\
{{A}^{-1}}&=\frac{1}{\left| A \right|}\times Adj\left( A \right) \\
& =\frac{1}{8}\left[ \begin{matrix}
2 & 0 \\
1 & 4 \\
\end{matrix} \right] \\
{{B}^{-1}}{{A}^{-1}}&=\frac{1}{6}\left[ \begin{matrix}
-1 & 2 \\
-1 & -4 \\
\end{matrix} \right]\times \frac{1}{8}\left[ \begin{matrix}
2 & 0 \\
1 & 4 \\
\end{matrix} \right] \\
& =\frac{1}{6\times 8}\left[ \begin{matrix}
-1 & 2 \\
-1 & -4 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & 0 \\
1 & 4 \\
\end{matrix} \right] \\
& =\frac{1}{48}\left[ \begin{matrix}
-2+2 & 0+8 \\
-2-4 & 0-16 \\
\end{matrix} \right] \\
& =\frac{1}{48}\left[ \begin{matrix}
0 & 8 \\
-6 & -16 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{0}{48} & \frac{8}{48} \\
\frac{-6}{48} & \frac{-16}{48} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & \frac{1}{6} \\
-\frac{1}{8} & -\frac{1}{3} \\
\end{matrix} \right]…(ii) \\
& \text{Frome (i) and (ii), we have} \\
{{\left( AB \right)}^{-1}} &={{B}^{-1}}{{A}^{-1}} \\
\end{align}\]
\(ii\)
\[{{\left( DA \right)}^{-1}}={{A}^{-1}}{{D}^{-1}}\]
Solution:
\[\begin{align}
DA&=\left[ \begin{matrix}
3 & 1 \\
-2 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
4 & 0 \\
-1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
12-1 & 0+2 \\
-8-2 & 0+4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
11 & 2 \\
-10 & 4 \\
\end{matrix} \right] \\
\left| DA \right|&=\left| \begin{matrix}
11 & 2 \\
-10 & 4 \\
\end{matrix} \right| \\
& =\left( 11 \right)\left( 4 \right)-\left( 2\right)\left( -10 \right) \\
& =44+20 \\
& =64 \\
Adj\left( DA \right) &=\left[ \begin{matrix}
4 & -2 \\
10 & 11 \\
\end{matrix} \right] \\
{{\left( DA \right)}^{-1}} &=\frac{1}{\left| DA \right|}\times Adj\left( DA \right) \\
& =\frac{1}{64}\left[ \begin{matrix}
4 & -2 \\
10 & 11 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{4}{64} & \frac{-2}{64} \\
\frac{10}{64} & \frac{11}{64} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{1}{16} & -\frac{1}{32} \\
\frac{5}{32} & \frac{11}{64} \\
\end{matrix} \right]…(i) \\
A& =\left[ \begin{matrix}
4 & 0 \\
-1 & 2 \\
\end{matrix} \right] \\
\left| A \right|& =\left| \begin{matrix}
4 & 0 \\
-1 & 2 \\\end{matrix} \right| \\
& =\left( 4 \right)\left( 2 \right)-\left( 0 \right)\left( -1 \right) \\
& =8+0 \\
& =8 \\
Adj\left( A \right) &=\left[ \begin{matrix}
2 & 0 \\
1 & 4 \\
\end{matrix} \right] \\
{{A}^{-1}}&=\frac{1}{\left| A \right|}\times Adj\left( A \right) \\
& =\frac{1}{8}\left[ \begin{matrix}
2 & 0 \\
1 & 4 \\
\end{matrix} \right] \\
D&=\left[ \begin{matrix}
3 & 1 \\
-2 & 2 \\
\end{matrix} \right] \\
\left| D \right|&=\left| \begin{matrix}
3 & 1 \\
-2 & 2 \\
\end{matrix} \right| \\
& =\left( 3 \right)\left( 2 \right)-\left( 1 \right)\left( -2 \right) \\
& =6+2 \\
& =8 \\
Adj\left( D \right) &=\left[ \begin{matrix}
2 & -1 \\
2 & 3 \\
\end{matrix} \right] \\
{{D}^{-1}}&=\frac{1}{\left| D \right|}\times Adj\left( D \right) \\
& =\frac{1}{8}\left[ \begin{matrix}
2 & -1 \\
2 & 3 \\
\end{matrix} \right] \\
{{A}^{-1}}{{D}^{-1}}&=\frac{1}{8}\left[ \begin{matrix}
2 & 0 \\
1 & 4 \\
\end{matrix} \right]\times \frac{1}{8}\left[ \begin{matrix}
2 & -1 \\
2 & 3 \\
\end{matrix} \right] \\
& =\frac{1}{8\times 8}\left[ \begin{matrix}
2 & 0 \\
1 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & -1 \\
2 & 3 \\
\end{matrix} \right] \\
& =\frac{1}{64}\left[ \begin{matrix}
4+0 & -2+0 \\
2+8 & -1+12 \\
\end{matrix} \right] \\
& =\frac{1}{64}\left[ \begin{matrix}
4 & -2 \\
10 & 11 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{4}{64} & \frac{-2}{64} \\
\frac{10}{64} & \frac{11}{64} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{1}{16} & -\frac{1}{32} \\
\frac{5}{32} & \frac{11}{64} \\
\end{matrix} \right]…(ii) \\
& \text{Frome (i) and (ii), we have} \\
{{\left( DA \right)}^{-1}} &={{A}^{-1}}{{D}^{-1}} \\
\end{align}\]
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