Question#1: Which of the following are conformable for addition?
Solution:
\[\begin{align}
& A=\left[ \begin{matrix}
2 & 1 \\
-1 & 3 \\
\end{matrix} \right],B=\left[ \begin{matrix}
3 \\
1 \\
\end{matrix} \right],C=\left[ \begin{matrix}
1 & 0 \\
2 & -1 \\
1 & -2 \\
\end{matrix} \right] \\
& D=\left[ \begin{matrix}
2+1 \\
3 \\
\end{matrix} \right],E=\left[ \begin{matrix}
-1 & 0 \\
1 & 2 \\
\end{matrix} \right],F=\left[ \begin{matrix}
3 & 2 \\
1+1 & -4 \\
3+2 & 2+1 \\
\end{matrix} \right] \\
& \text{A and E, B and D, C and F are conformable for addition}\text{.} \\
\end{align}\]
Question#2: Find the additive inverse of the following matrices.
\(i\)
\[A=\left[ \begin{matrix}
2 & 4 \\
-2 & 1 \\
\end{matrix} \right]\]
Solution:
\[-A=\left[ \begin{matrix}
-2 & -4 \\
2 & -1 \\
\end{matrix} \right]\]
\(ii\)
\[B=\left[ \begin{matrix}
1 & 0 & -1 \\
2 & -1 & 3 \\
3 & -2 & 1 \\
\end{matrix} \right]\]
Solution:
\[-B=\left[ \begin{matrix}
-1 & 0 & 1 \\
-2 & 1 & -3 \\
-3 & 2 & -1 \\
\end{matrix} \right]\]
\(iii\)
\[C=\left[ \begin{matrix}
4 \\
-2 \\
\end{matrix} \right]\]
Solution:
\[C=\left[ \begin{matrix}
4 \\
-2 \\
\end{matrix} \right]\]
\(iv\)
\[D=\left[ \begin{matrix}
1 & 0 \\
-3 & -2 \\
2 & 1 \\
\end{matrix} \right]\]
Solution:
\[-D=\left[ \begin{matrix}
-1 & 0 \\
3 & 2 \\
-2 & -1 \\
\end{matrix} \right]\]
\(v\)
\[E=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]\]
Solution:
\[-E=\left[ \begin{matrix}
-1 & 0 \\
0 & -1 \\
\end{matrix} \right]\]
\(vi\)
\[F=\left[ \begin{matrix}
\sqrt{3} & 1 \\
-1 & \sqrt{2} \\
\end{matrix} \right]\]
Solution:
\[-F=\left[ \begin{matrix}
-\sqrt{3} & -1 \\
1 & -\sqrt{2} \\
\end{matrix} \right]\]
Question#3: If \(A=\left[ \begin{matrix}
-1 & 2 \\
2 & 1 \\
\end{matrix} \right]\), \(B=\left[ \begin{matrix}
1 \\
-1 \\
\end{matrix} \right]\), \(C=\left[ \begin{matrix}
1 & -1 & 2 \\
\end{matrix} \right]\), \(D=\left[ \begin{matrix}
1 & 2 & 3 \\
-1 & 0 & 2 \\
\end{matrix} \right]\), then find.
\(i\)
\[A+\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =A+\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 & 2 \\
2 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1+1 & 2+1 \\
2+1 & 1+1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & 3 \\
3 & 2 \\
\end{matrix} \right] \\
\end{align}\]
\(ii\)
\[B+\left[ \begin{matrix}
-2 \\
3 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =B+\left[ \begin{matrix}
-2 \\
3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 \\
-1 \\
\end{matrix} \right]+\left[ \begin{matrix}
-2 \\
3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-2 \\
-1+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 \\
2 \\
\end{matrix} \right] \\
\end{align}\]
\(iii\)
\[C+\left[ \begin{matrix}
-2 & 1 & 3 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =C+\left[ \begin{matrix}
-2 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & -1 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
-2 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-2 & -1+1 & 2+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 & 0 & 5 \\
\end{matrix} \right] \\
\end{align}\]
\(iv\)
\[D+\left[ \begin{matrix}
0 & 1 & 0 \\
2 & 0 & 1 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =D+\left[ \begin{matrix}
0 & 1 & 0 \\
2 & 0 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 2 & 3 \\
-1 & 0 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
0 & 1 & 0 \\
2 & 0 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+0 & 2+1 & 3+0 \\
-1+2 & 0+0 & 2+1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 3 & 3 \\
1 & 0 & 3 \\
\end{matrix} \right] \\
\end{align}\]
\(v\)
\[2A\]
Solution:
\[\begin{align}
& =2A \\
& =2\left[ \begin{matrix}
-1 & 2 \\
2 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1\times 2 & 2\times 2 \\
2\times 2 & 1\times 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2 & 4 \\
4 & 2 \\
\end{matrix} \right] \\
\end{align}\]
\(vi\)
\[\left( -1 \right)B\]
Solution:
\[\begin{align}
& =\left( -1 \right)B \\
& =\left( -1 \right)\left[ \begin{matrix}
1 \\
-1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 \\
1 \\
\end{matrix} \right] \\
\end{align}\]
\(vii\)
\[\left( -2 \right)C\]
Solution:
\[\begin{align}
& =\left( -2 \right)C \\
& =\left( -2 \right)\left[ \begin{matrix}
1 & -1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\times -2 & -1\times -2 & 2 \\
\end{matrix}\times -2 \right] \\
& =\left[ \begin{matrix}
-2 & 2 & -4 \\
\end{matrix} \right] \\
\end{align}\]
\(viii\)
\[3D\]
Solution:
\[\begin{align}
& =3D \\
& =3\left[ \begin{matrix}
1 & 2 & 3 \\
-1 & 0 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\times 3 & 2\times 3 & 3\times 3 \\
-1\times 3 & 0\times 3 & 2\times 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 6 & 9 \\
-3 & 0 & 6 \\
\end{matrix} \right] \\
\end{align}\]
\(ix\)
\[3C\]
Solution:
\[\begin{align}
& =3C \\
& =3\left[ \begin{matrix}
1 & -1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\times 3 & -1\times 3 & 2\times 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & -3 & 6 \\
\end{matrix} \right] \\
\end{align}\]
Question#4: Perform the indicated operations and simplify the following.
\(i\)
\[\left( \left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
0 & 2 \\
3 & 0 \\
\end{matrix} \right] \right)+\left[ \begin{matrix}
1 & 1 \\
1 & 0 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left( \left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
0 & 2 \\
3 & 0 \\
\end{matrix} \right] \right)+\left[ \begin{matrix}
1 & 1 \\
1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+0 & 0+2 \\
0+3 & 1+0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 1 \\
1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 2 \\
3 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 1 \\
1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 2+1 \\
3+1 & 1+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 3 \\
4 & 1 \\
\end{matrix} \right] \\
\end{align}\]
\(ii\)
\[\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+\left( \left[ \begin{matrix}
0 & 2 \\
3 & 0 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 1 \\
1 & 0 \\
\end{matrix} \right] \right)\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+\left( \left[ \begin{matrix}
0 & 2 \\
3 & 0 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 1 \\
1 & 0 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
0-1 & 2-1 \\
3-1 & 0-0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1 & 1 \\
2 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-1 & 0+1 \\
0+2 & 1+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & 1 \\
2 & 1 \\
\end{matrix} \right] \\
\end{align}\]
\(iii\)
\[\left[ \begin{matrix}
2 & 3 & 1 \\
\end{matrix} \right]+\left( \left[ \begin{matrix}
1 & 0 & 2 \\
\end{matrix} \right]-\left[ \begin{matrix}
2 & 2 & 2 \\
\end{matrix} \right] \right)\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
2 & 3 & 1 \\
\end{matrix} \right]+\left( \left[ \begin{matrix}
1 & 0 & 2 \\
\end{matrix} \right]-\left[ \begin{matrix}
2 & 2 & 2 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
2 & 3 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1-2 & 0-2 & 2-2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 3 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1 & -2 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2-1 & 3-2 & 1+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 1 & 1 \\
\end{matrix} \right] \\
\end{align}\]
\(iv\)
\[\left[ \begin{matrix}
1 & 2 & 3 \\
-1 & -1 & -1 \\
0 & 1 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 1 & 1 \\
2 & 2 & 2 \\
3 & 3 & 3 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
1 & 2 & 3 \\
-1 & -1 & -1 \\
0 & 1 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 1 & 1 \\
2 & 2 & 2 \\
3 & 3 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 2+1 & 3+1 \\
-1+2 & -1+2 & -1+2 \\
0+3 & 1+3 & 2+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 3 & 4 \\
1 & 1 & 1 \\
3 & 4 & 5 \\
\end{matrix} \right] \\
\end{align}\]
\(v\)
\[\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
3 & 1 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 0 & -2 \\
-2 & -1 & 0 \\
0 & 2 & -1 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
3 & 1 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 0 & -2 \\
-2 & -1 & 0 \\
0 & 2 & -1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 2+0 & 3-2 \\
2-2 & 3-1 & 1+0 \\
3+0 & 1+2 & 2-1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 2 & 1 \\
0 & 2 & 1 \\
3 & 3 & 1 \\
\end{matrix} \right] \\
\end{align}\]
\(vi\)
\[\left( \left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
2 & 1 \\
1 & 0 \\
\end{matrix} \right] \right)+\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left( \left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
2 & 1 \\
1 & 0 \\
\end{matrix} \right] \right)+\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+2 & 2+1 \\
0+1 & 1+0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 3 \\
1 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3+1 & 3+1 \\
1+1 & 1+1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4 & 4 \\
2 & 2 \\
\end{matrix} \right] \\
\end{align}\]
Question#5: For the matrices \(A=\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]\),\(B=\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 2 \\
\end{matrix} \right]\) and \(C=\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right]\), verify the following rules:
\(i\)
\[A+C=C+A\]
Solution:
\[\begin{align}
A+C&=\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-1 & 2+0 & 3+0 \\
2+0 & 3-2 & 1+3 \\
1+1 & -1+1 & 0+2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & 2 & 3 \\
2 & 1 & 4 \\
2 & 0 & 2 \\
\end{matrix} \right] \\
C+A&=\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1+1 & 0+2 & 0+3 \\
0+2 & -2+3 & 3+1 \\
1+1 & 1-1 & 2+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & 2 & 3 \\
2 & 1 & 4 \\
2 & 0 & 2 \\
\end{matrix} \right] \\
So,A+C&=C+A \\
\end{align}\]
\(ii\)
\[A+B=B+A\]
Solution:
\[\begin{align}
A+B&=\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 2-1 & 3+1 \\
2+2 & 3-2 & 1+2 \\
1+3 & -1+1 & 0+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 1 & 4 \\
4 & 1 & 3 \\
4 & 0 & 3 \\
\end{matrix} \right] \\
B+A&=\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & -1+2 & 1+3 \\
2+2 & -2+3 & 2+1 \\
3+1 & 1-1 & 3+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 1 & 4 \\
4 & 1 & 3 \\
4 & 0 & 3 \\
\end{matrix} \right] \\
So,A+B&=B+A \\
\end{align}\]
\(iii\)
\[B+C=C+B\]
Solution:
\[\begin{align}
B+C&=\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-1 & -1+0 & 1+0 \\
2+0 & -2-2 & 2+3 \\
3+1 & 1+1 & 3+2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & -1 & 1 \\
2 & -4 & 5 \\
4 & 2 & 5 \\
\end{matrix} \right] \\
C+B&=\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1+1 & 0-1 & 0+1 \\
0+2 & -2-2 & 3+2 \\
1+3 & 1+1 & 2+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & -1 & 1 \\
2 & -4 & 5 \\
4 & 2 & 5 \\
\end{matrix} \right] \\
So,B+C&=C+B \\
\end{align}\]
\(iv\)
\[A+\left( B+A \right)=2A+B\]
Solution:
\[\begin{align}
A+\left( B+A \right) &=\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left( \left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1+1 & -1+2 & 1+3 \\
2+2 & -2+3 & 2+1 \\
3+1 & 1-1 & 3+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
2 & 1 & 4 \\
4 & 1 & 3 \\
4 & 0 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+2 & 2+1 & 3+4 \\
2+4 & 3+1 & 1+3 \\
1+4 & -1+0 & 0+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 3 & 7 \\
6 & 4 & 4 \\
5 & -1 & 3 \\
\end{matrix} \right] \\
2A+B&=2\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 4 & 6 \\
4 & 6 & 2 \\
2 & -2 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2+1 & 4-1 & 6+1 \\
4+2 & 6-2 & 2+2 \\
2+3 & -2+1 & 0+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 3 & 7 \\
6 & 4 & 4 \\
5 & -1 & 3 \\
\end{matrix} \right] \\
So,A+\left( B+A \right) &=2A+B \\
\end{align}\]
\(v\)
\[\left( C-B \right)+A=C+\left( A-B \right)\]
Solution:
\[\begin{align}
\left( C-B \right)+A &=\left( \left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \right)+\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1-1 & 0+1 & 0-1 \\
0-2 & -2+2 & 3-2 \\
1-3 & 1-1 & 2-3 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2 & 1 & -1 \\
-2 & 0 & 1 \\
-2 & 0 & -1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2+1 & 1+2 & -1+3 \\
-2+2 & 0+3 & 1+1 \\
-2+1 & 0-1 & -1+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 & 3 & 2 \\
0 & 3 & 2 \\
-1 & -1 & -1 \\
\end{matrix} \right] \\
C+\left( A-B \right) &=\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right]+\left( \left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
1-1 & 2+1 & 3-1 \\
2-2 & 3+2 & 1-2 \\
1-3 & -1-1 & 0-3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
0 & 3 & 2 \\
0 & 5 & -1 \\
-2 & -2 & -3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1+0 & 0+3 & 0+2 \\
0+0 & -2+5 & 3-1 \\
1-2 & 1-2 & 2-3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 & 3 & 2 \\
0 & 3 & 2 \\
-1 & -1 & -1 \\
\end{matrix} \right] \\
So,\left( C-B \right)+A&=C+\left( A-B \right) \\
\end{align}\]
\(vi\)
\[2A+B=A+\left( A+B \right)\]
Solution:
\[\begin{align}
2A+B&=2\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 4 & 6 \\
4 & 6 & 2 \\
2 & -2 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2+1 & 4-1 & 6+1 \\
4+2 & 6-2 & 2+2 \\
2+3 & -2+1 & 0+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 3 & 7 \\
6 & 4 & 4 \\
5 & -1 & 3 \\
\end{matrix} \right] \\
A+\left( A+B \right) &=\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left( \left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1+1 & 2-1 & 3+1 \\
2+2 & 3-2 & 1+2 \\
1+3 & -1+1 & 0+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
2 & 1 & 4 \\
4 & 1 & 3 \\
4 & 0 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+2 & 2+1 & 3+4 \\
2+4 & 3+1 & 1+3 \\
1+4 & -1+0 & 0+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 3 & 7 \\
6 & 4 & 4 \\
5 & -1 & 3 \\
\end{matrix} \right] \\
So,2A+B&=A+\left( A+B \right) \\
\end{align}\]
\(vii\)
\[\left( C-B \right)-A=\left( C-A \right)-B\]
Solution:
\[\begin{align}
\left( C-B \right)-A&=\left( \left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \right)-\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1-1 & 0+1 & 0-1 \\
0-2 & -2+2 & 3-2 \\
1-3 & 1-1 & 2-3 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2 & 1 & -1 \\
-2 & 0 & 1 \\
-2 & 0 & -1 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2-1 & 1-2 & -1-3 \\
-2-2 & 0-3 & 1-1 \\
-2-1 & 0+1 & -1-0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-3 & -1 & -4 \\
-4 & -3 & 0 \\
-3 & 1 & -1 \\
\end{matrix} \right] \\
\left( C-A \right)-B&=\left( \left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right] \right)-\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1-1 & 0-2 & 0-3 \\
0-2 & -2-3 & 3-1 \\
1-1 & 1+1 & 2-0 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2 & -2 & -3 \\
-2 & -5 & 2 \\
0 & 2 & 2 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2-1 & -2+1 & -3-1 \\
-2-2 & -5+2 & 2-2 \\
0-3 & 2-1 & 2-3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-3 & -1 & -4 \\
-4 & -3 & 0 \\
-3 & 1 & -1 \\
\end{matrix} \right] \\
So,\left( C-B \right)-A&=\left( C-A \right)-B \\
\end{align}\]
\(viii\)
\[\left( A+B \right)+C=A+\left( B+C \right)\]
Solution:
\[\begin{align}
\left( A+B \right)+C&=\left( \left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \right)+\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 2-1 & 3+1 \\
2+2 & 3-2 & 1+2 \\
1+3 & -1+1 & 0+3 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 1 & 4 \\
4 & 1 & 3 \\
4 & 0 & 3 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2-1 & 1+0 & 4+0 \\
4+0 & 1-2 & 3+3 \\
4+1 & 0+1 & 3+2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 1 & 4 \\
4 & -1 & 6 \\
5 & 1 & 5 \\
\end{matrix} \right] \\
A+\left( B+C \right) &=\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left( \left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1-1 & -1+0 & 1+0 \\
2+0 & -2-2 & 2+3 \\
3+1 & 1+1 & 3+2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
0 & -1 & 1 \\
2 & -4 & 5 \\
4 & 2 & 5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+0 & 2-1 & 3+1 \\
2+2 & 3-4 & 1+5 \\
1+4 & -1+2 & 0+5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 1 & 4 \\
4 & -1 & 6 \\
5 & 1 & 5 \\
\end{matrix} \right] \\
So,\left( A+B \right)+C&=A+\left( B+C \right) \\
\end{align}\]
\(ix\)
\[A+\left( B-C \right)=\left( A-C \right)+B\]
Solution:
\[\begin{align}
A+\left( B-C \right) &=\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left( \left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right]-\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1+1 & -1-0 & 1-0 \\
2-0 & -2+2 & 2-3 \\
3-1 & 1-1 & 3-2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
2 & -1 & 1 \\
2 & 0 & -1 \\
2 & 0 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+2 & 2-1 & 3+1 \\
2+2 & 3+0 & 1-1 \\
1+2 & -1+0 & 0+1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 1 & 4 \\
4 & 3 & 0 \\
3 & -1 & 1 \\
\end{matrix} \right] \\
\left( A-C \right)+B&=\left( \left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]-\left[ \begin{matrix}
-1 & 0 & 0 \\
0 & -2 & 3 \\
1 & 1 & 2 \\
\end{matrix} \right] \right)+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 2-0 & 3-0 \\
2-0 & 3+2 & 1-3 \\
1-1 & -1-1 & 0-2 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 2 & 3 \\
2 & 5 & -2 \\
0 & -2 & -2 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2+1 & 2-1 & 3+1 \\
2+2 & 5-2 & -2+2 \\
0+3 & -2+1 & -2+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 1 & 4 \\
4 & 3 & 0 \\
3 & -1 & 1 \\
\end{matrix} \right] \\
So,A+\left( B-C \right) &=\left( A-C \right)+B \\
\end{align}\]
\(x\)
\[2A+2B=2(A+B)\]
Solution:
\[\begin{align}
2A+2B&=2\left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+2\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 4 & 6 \\
4 & 6 & 2 \\
2 & -2 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
2 & -2 & 2 \\
4 & -4 & 4 \\
6 & 2 & 6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2+2 & 4-2 & 6+2 \\
4+4 & 6-4 & 2+4 \\
2+6 & -2+2 & 0+6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4 & 2 & 8 \\
8 & 2 & 6 \\
8 & 0 & 6 \\
\end{matrix} \right] \\
2(A+B) &=2\left( \left[ \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
1 & -1 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & -1 & 1 \\
2 & -2 & 2 \\
3 & 1 & 3 \\
\end{matrix} \right] \right) \\
& =2\left[ \begin{matrix}
1+1 & 2-1 & 3+1 \\
2+2 & 3-2 & 1+2 \\
1+3 & -1+1 & 0+3 \\
\end{matrix} \right] \\
& =2\left[ \begin{matrix}
2 & 1 & 4 \\
4 & 1 & 3 \\
4 & 0 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4 & 2 & 8 \\
8 & 2 & 6 \\
8 & 0 & 6 \\
\end{matrix} \right] \\
So,2A+2B&=2(A+B) \\
\end{align}\]
Question#6: If \(A=\left[ \begin{matrix}
1 & -2 \\
3 & 4 \\
\end{matrix} \right]\) and \(B=\left[ \begin{matrix}
0 & 7 \\
-3 & 8 \\
\end{matrix} \right]\) then find:
\(i\)
\[3A-2B\]
Solution:
\[\begin{align}
3A-2B&=3\left[ \begin{matrix}
1 & -2 \\
3 & 4 \\
\end{matrix} \right]-2\left[ \begin{matrix}
0 & 7 \\
-3 & 8 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & -6 \\
9 & 12 \\
\end{matrix} \right]-\left[ \begin{matrix}
0 & 14 \\
-6 & 16 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3-0 & -6-14 \\
9+6 & 12-16 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & -20 \\
15 & -4 \\
\end{matrix} \right] \\
\end{align}\]
\(ii\)
\[2{{A}^{t}}-3{{B}^{t}}\]
Solution:
\[\begin{align}
2{{A}^{t}}-3{{B}^{t}}&=2{{\left[ \begin{matrix}
1 & -2 \\
3 & 4 \\
\end{matrix} \right]}^{t}}-3{{\left[ \begin{matrix}
0 & 7 \\
-3 & 8 \\
\end{matrix} \right]}^{t}} \\
& =2\left[ \begin{matrix}
1 & 3 \\
-2 & 4 \\
\end{matrix} \right]-3\left[ \begin{matrix}
0 & -3 \\
7 & 8 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 6 \\
-4 & 8 \\
\end{matrix} \right]-\left[ \begin{matrix}
0 & -9 \\
21 & 24 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2-0 & 6+9 \\
-4-21 & 8-24 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 15 \\
-25 & -16 \\
\end{matrix} \right] \\
\end{align}\]
Question#7: If \(2\left[ \begin{matrix}
2 & 4 \\
-3 & a \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & b \\
8 & -4 \\
\end{matrix} \right]=\left[ \begin{matrix}
7 & 10 \\
18 & 1 \\
\end{matrix} \right]\) then find and \(b\).
Solution:
\[\begin{align}
2\left[ \begin{matrix}
2 & 4 \\
-3 & a \\
\end{matrix} \right]+3\left[ \begin{matrix}
1 & b \\
8 & -4 \\
\end{matrix} \right]&=\left[ \begin{matrix}
7 & 10 \\
18 & 1 \\
\end{matrix} \right] \\
\left[ \begin{matrix}
4 & 8 \\
-6 & 2a \\
\end{matrix} \right]+\left[ \begin{matrix}
3 & 3b \\
24 & -12 \\
\end{matrix} \right] &=\left[ \begin{matrix}
7 & 10 \\
18 & 1 \\
\end{matrix} \right] \\
\left[ \begin{matrix}
7 & 8+3b \\
18 & 2a-12 \\
\end{matrix} \right] &=\left[ \begin{matrix}
7 & 10 \\
18 & 1 \\
\end{matrix} \right] \\
8+3b=10\quad&|\quad2a-12=1 \\
3b=10-8\quad&|\quad2a=12+1 \\
3b=2\quad&|\quad2a=13 \\
b=\frac{2}{3}\quad&|\quad a=\frac{13}{2} \\
\end{align}\]
Question#8: If \(A=\left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]\) and \(B=\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right]\) then verify that.
\(i\)
\({{\left( A+B \right)}^{t}}={{A}^{t}}+{{B}^{t}}\)
Solution:
\[\begin{align}
{{\left( A+B \right)}^{t}}&={{\left( \left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right] \right)}^{t}} \\
& ={{\left[ \begin{matrix}
1+1 & 2+1 \\
0+2 & 1+0 \\
\end{matrix} \right]}^{t}} \\
& ={{\left[ \begin{matrix}
2 & 3 \\
2 & 1 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
2 & 2 \\
3 & 1 \\
\end{matrix} \right] \\
{{A}^{t}}+{{B}^{t}}&={{\left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]}^{t}}+{{\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
1 & 0 \\
2 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 2 \\
1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 0+2 \\
2+1 & 1+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 2 \\
3 & 1 \\
\end{matrix} \right] \\
So,{{\left( A+B \right)}^{t}}&={{A}^{t}}+{{B}^{t}} \\
\end{align}\]
\(ii\)
\[{{\left( A-B \right)}^{t}}={{A}^{t}}-{{B}^{t}}\]
Solution:
\[\begin{align}
{{\left( A-B \right)}^{t}}&={{\left( \left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right] \right)}^{t}} \\
& ={{\left[ \begin{matrix}
1-1 & 2-1 \\
0-2 & 1-0 \\
\end{matrix} \right]}^{t}} \\
& ={{\left[ \begin{matrix}
0 & 1 \\
-2 & 1 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
0 & -2 \\
1 & 1 \\
\end{matrix} \right] \\
{{A}^{t}}-{{B}^{t}}&={{\left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]}^{t}}-{{\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
1 & 0 \\
2 & 1 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 2 \\
1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-1 & 0-2 \\
2-1 & 1-0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & -2 \\
1 & 1 \\
\end{matrix} \right] \\
So,{{\left( A-B \right)}^{t}}&={{A}^{t}}-{{B}^{t}} \\
\end{align}\]
\(iii\)
\[A+{{A}^{t}}\text{ is symmetric }\]
Solution:
\[\begin{align}
A+{{A}^{t}}&=\left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]+{{\left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 0 \\
2 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 2+0 \\
0+2 & 1+1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 2 \\
2 & 2 \\
\end{matrix} \right] \\
{{\left( A+{{A}^{t}} \right)}^{t}}&={{\left[ \begin{matrix}
2 & 2 \\
2 & 2 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
2 & 2 \\
2 & 2 \\
\end{matrix} \right] \\
{{\left( A+{{A}^{t}} \right)}^{t}}&=A+{{A}^{t}} \\
\text{Hence, }&A+{{A}^{t}}\text{ is symmetric}\text{.} \\
\end{align}\]
\(iv\)
\[A-{{A}^{t}}\text{ is skew symmetric}\]
Solution:
\[\begin{align}
A-{{A}^{t}}&=\left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]-{{\left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
1 & 2 \\
0 & 1 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 0 \\
2 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-1 & 2-0 \\
0-2 & 1-1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & 2 \\
-2 & 0 \\
\end{matrix} \right] \\
{{\left( A-{{A}^{t}} \right)}^{t}}&={{\left[ \begin{matrix}
0 & 2 \\
-2 & 0 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
0 & -2 \\
2 & 0 \\
\end{matrix} \right] \\
& =-\left[ \begin{matrix}
0 & 2 \\
-2 & 0 \\
\end{matrix} \right] \\
{{\left( A-{{A}^{t}} \right)}^{t}}&=-\left( A-{{A}^{t}} \right) \\
\text{Hence, }&A-{{A}^{t}}\text{ is skew symmetric} \\
\end{align}\]
\(v\)
\[B+{{B}^{t}}\text{ is symmetric}\]
Solution:
\[\begin{align}
B+{{B}^{t}}&=\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right]+{{\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 2 \\
1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 1+2 \\
2+1 & 0+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 3 \\
3 & 0 \\
\end{matrix} \right] \\
{{\left( B+{{B}^{t}} \right)}^{t}}&={{\left[ \begin{matrix}
2 & 3 \\
3 & 0 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
2 & 3 \\
3 & 0 \\
\end{matrix} \right] \\
{{\left( B+{{B}^{t}} \right)}^{t}}&=B+{{B}^{t}} \\
\text{Hence, }&B+{{B}^{t}}\text{ is symmetric}\text{.} \\
\end{align}\]
\(vi\)
\[B-{{B}^{t}}\text{ is skew symmetric}\]
Solution:
\[\begin{align}
B-{{B}^{t}}&=\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right]-{{\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
1 & 1 \\
2 & 0 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 2 \\
1 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-1 & 1-2 \\
2-1 & 0-0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & -1 \\
1 & 0 \\
\end{matrix} \right] \\
{{\left( B-{{B}^{t}} \right)}^{t}}&={{\left[ \begin{matrix}
0 & -1 \\
1 & 0 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
0 & 1 \\
-1 & 0 \\
\end{matrix} \right] \\
& =-\left[ \begin{matrix}
0 & -1 \\
1 & 0 \\
\end{matrix} \right] \\
{{\left( B-{{B}^{t}} \right)}^{t}}&=-B-{{B}^{t}} \\
\text{Hence, } &B-{{B}^{t}}\text{ is skew symmetric}\text{.} \\
\end{align}\]
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