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If $A=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right], B=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right], C=\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$, $a=4$, and $b=-2$, then show that
(i) $\mathrm{A}+(\mathrm{B}+\mathrm{C})=(\mathrm{A}+\mathrm{B})+\mathrm{C}$
(ii) $\mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C}$
(iii) $\quad(a+b) B=a B+b B$
(iv) $a(C-A)=a C-a A$
(v) $\left(\mathrm{A}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}$
(vi) $(\mathrm{bA})^{\mathrm{T}}=\mathrm{b} \mathrm{A}^{\mathrm{T}}$
(vii) $(\mathrm{AB})^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}$
(viii) $(\mathrm{A}-\mathrm{B}) \mathrm{C}=\mathrm{AC}-\mathrm{BC}$
(ix) $(\mathrm{A}-\mathrm{B})^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}-\mathrm{B}^{\mathrm{T}}$
(i) $\mathrm{A}+(\mathrm{B}+\mathrm{C})=(\mathrm{A}+\mathrm{B})+\mathrm{C}$
(ii) $\mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C}$
(iii) $\quad(a+b) B=a B+b B$
(iv) $a(C-A)=a C-a A$
(v) $\left(\mathrm{A}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}$
(vi) $(\mathrm{bA})^{\mathrm{T}}=\mathrm{b} \mathrm{A}^{\mathrm{T}}$
(vii) $(\mathrm{AB})^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}$
(viii) $(\mathrm{A}-\mathrm{B}) \mathrm{C}=\mathrm{AC}-\mathrm{BC}$
(ix) $(\mathrm{A}-\mathrm{B})^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}-\mathrm{B}^{\mathrm{T}}$
Solution:
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Verified Answer
(i) $A+(B+C)=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]+\left[\begin{array}{ll}6 & 0 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}7 & 2 \\ 1 & 6\end{array}\right]$
and $(\mathrm{A}+\mathrm{B})+\mathrm{C}=\left[\begin{array}{ll}5 & 2 \\ 0 & 8\end{array}\right]+\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$
$=\left[\begin{array}{ll}7 & 2 \\ 1 & 6\end{array}\right]=\mathrm{A}+(\mathrm{B}+\mathrm{C})$ Henceproved.
(ii) $(\mathrm{BC})=\left[\begin{array}{cc}4 & 0 \\ 1 & 5\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]$
and $\quad \mathrm{A}(\mathrm{BC})=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]=\left[\begin{array}{cc}22 & -20 \\ 13 & -30\end{array}\right]$
Also, $\quad(\mathrm{AB})=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right] \cdot\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]$
$$
\text { (AB) } \begin{aligned}
\mathrm{C} &=\left[\begin{array}{cc}
6 & 10 \\
-1 & 15
\end{array}\right]\left[\begin{array}{cc}
2 & 0 \\
1 & -2
\end{array}\right] \\
&=\left[\begin{array}{rr}
22 & -20 \\
13 & -30
\end{array}\right]=\mathrm{A}(\mathrm{BC}) \quad \text { Henceproved }
\end{aligned}
$$
(iii) $(\mathrm{a}+\mathrm{b}) \mathrm{B}=(4-2)\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right] \quad[\because a=4, b=-2]$
$$
=\left[\begin{array}{cc}
8 & 0 \\
2 & 10
\end{array}\right]
$$
and $a B+b B=4 B-2 B$
$$
=\left[\begin{array}{cc}
8 & 0 \\
2 & 10
\end{array}\right]=(a+b) B \quad \text { Henceproved. }
$$
(iv) $(\mathrm{C}-\mathrm{A})=\left[\begin{array}{ll}1 & -2 \\ 2 & -5\end{array}\right]$
and $\mathrm{a}(\mathrm{C}-\mathrm{A})=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right] \quad[\because \mathrm{a}=4]$
Also, $a C-a A=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$
$=\mathrm{a}(\mathrm{C}-\mathrm{A})$
Hence proved
(v) $A^T=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$
Now, $\left(\mathbf{A}^{\mathrm{T}}\right)^{\mathrm{T}}=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]^{\mathrm{T}}=\mathbf{A}$
Hence proved
(vi) $(\mathrm{bA})^{\mathrm{T}}=\left[\begin{array}{cc}-2 & -4 \\ 2 & -6\end{array}\right]^{\mathrm{T}} \quad[\because \mathrm{b}=-2]$
$=\left[\begin{array}{cc}-2 & 2 \\ -4 & -6\end{array}\right]$ and $\mathbf{A}^{\mathrm{T}}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$
$\therefore \quad \mathrm{bA}^{\mathrm{T}}=\left[\begin{array}{cc}-2 & 2 \\ -4 & -6\end{array}\right]=(\mathrm{bA})^{\mathrm{T}}$
Hence proved
(vii) $\mathrm{AB}=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]$
$\therefore \quad(\mathrm{AB})^{\mathrm{T}}=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]$
$\therefore \quad(\mathrm{AB})^{\mathrm{T}}=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]$
Now, $B^{\mathrm{T}} A^{\mathrm{T}}=\left[\begin{array}{cc}4 & 1 \\ 0 & 5\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$ $=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]=(\mathrm{AB})^{\mathrm{T}} \quad$ Hence proved
(viii) $(\mathrm{A}-\mathrm{B})=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right] ;(\mathrm{A}-\mathrm{B}) \mathrm{C}=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$
$$
=\left[\begin{array}{cc}
-4 & -4 \\
-6 & 4
\end{array}\right]
$$
Now, $A C=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$
$$
=\left[\begin{array}{cc}
4 & -4 \\
1 & -6
\end{array}\right]
$$
$$
\text { and } \mathrm{BC}=\left[\begin{array}{cc}
4 & 0 \\
1 & 5
\end{array}\right]\left[\begin{array}{cc}
2 & 0 \\
1 & -2
\end{array}\right]
$$
$\therefore \quad \mathrm{AC}-\mathrm{BC}=\left[\begin{array}{cc}-4 & -4 \\ -6 & 4\end{array}\right] \quad$ using Eqs. (ii) and (iii) $]$ $=(\mathrm{A}-\mathrm{B}) \mathrm{C}[$ using Eq. (i) $]$ Hence proved.
$\begin{aligned}(\mathrm{ix})(\mathrm{A}-\mathrm{B})^{\mathrm{T}} &=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}-3 & -2 \\ 2 & -2\end{array}\right] \\ \mathrm{A}^{\mathrm{T}}-\mathrm{B}^{\mathrm{T}} &=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]-\left[\begin{array}{ll}4 & 1 \\ 0 & 5\end{array}\right] \\ &=\left[\begin{array}{cc}-3 & -2 \\ 2 & -2\end{array}\right]=(\mathrm{A}-\mathrm{B})^{\mathrm{T}} \text { Hence Proved. } \end{aligned}$
and $(\mathrm{A}+\mathrm{B})+\mathrm{C}=\left[\begin{array}{ll}5 & 2 \\ 0 & 8\end{array}\right]+\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$
$=\left[\begin{array}{ll}7 & 2 \\ 1 & 6\end{array}\right]=\mathrm{A}+(\mathrm{B}+\mathrm{C})$ Henceproved.
(ii) $(\mathrm{BC})=\left[\begin{array}{cc}4 & 0 \\ 1 & 5\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]$
and $\quad \mathrm{A}(\mathrm{BC})=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]=\left[\begin{array}{cc}22 & -20 \\ 13 & -30\end{array}\right]$
Also, $\quad(\mathrm{AB})=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right] \cdot\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]$
$$
\text { (AB) } \begin{aligned}
\mathrm{C} &=\left[\begin{array}{cc}
6 & 10 \\
-1 & 15
\end{array}\right]\left[\begin{array}{cc}
2 & 0 \\
1 & -2
\end{array}\right] \\
&=\left[\begin{array}{rr}
22 & -20 \\
13 & -30
\end{array}\right]=\mathrm{A}(\mathrm{BC}) \quad \text { Henceproved }
\end{aligned}
$$
(iii) $(\mathrm{a}+\mathrm{b}) \mathrm{B}=(4-2)\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right] \quad[\because a=4, b=-2]$
$$
=\left[\begin{array}{cc}
8 & 0 \\
2 & 10
\end{array}\right]
$$
and $a B+b B=4 B-2 B$
$$
=\left[\begin{array}{cc}
8 & 0 \\
2 & 10
\end{array}\right]=(a+b) B \quad \text { Henceproved. }
$$
(iv) $(\mathrm{C}-\mathrm{A})=\left[\begin{array}{ll}1 & -2 \\ 2 & -5\end{array}\right]$
and $\mathrm{a}(\mathrm{C}-\mathrm{A})=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right] \quad[\because \mathrm{a}=4]$
Also, $a C-a A=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$
$=\mathrm{a}(\mathrm{C}-\mathrm{A})$
Hence proved
(v) $A^T=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$
Now, $\left(\mathbf{A}^{\mathrm{T}}\right)^{\mathrm{T}}=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]^{\mathrm{T}}=\mathbf{A}$
Hence proved
(vi) $(\mathrm{bA})^{\mathrm{T}}=\left[\begin{array}{cc}-2 & -4 \\ 2 & -6\end{array}\right]^{\mathrm{T}} \quad[\because \mathrm{b}=-2]$
$=\left[\begin{array}{cc}-2 & 2 \\ -4 & -6\end{array}\right]$ and $\mathbf{A}^{\mathrm{T}}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$
$\therefore \quad \mathrm{bA}^{\mathrm{T}}=\left[\begin{array}{cc}-2 & 2 \\ -4 & -6\end{array}\right]=(\mathrm{bA})^{\mathrm{T}}$
Hence proved
(vii) $\mathrm{AB}=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]$
$\therefore \quad(\mathrm{AB})^{\mathrm{T}}=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]$
$\therefore \quad(\mathrm{AB})^{\mathrm{T}}=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]$
Now, $B^{\mathrm{T}} A^{\mathrm{T}}=\left[\begin{array}{cc}4 & 1 \\ 0 & 5\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$ $=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]=(\mathrm{AB})^{\mathrm{T}} \quad$ Hence proved
(viii) $(\mathrm{A}-\mathrm{B})=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right] ;(\mathrm{A}-\mathrm{B}) \mathrm{C}=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$
$$
=\left[\begin{array}{cc}
-4 & -4 \\
-6 & 4
\end{array}\right]
$$
Now, $A C=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$
$$
=\left[\begin{array}{cc}
4 & -4 \\
1 & -6
\end{array}\right]
$$
$$
\text { and } \mathrm{BC}=\left[\begin{array}{cc}
4 & 0 \\
1 & 5
\end{array}\right]\left[\begin{array}{cc}
2 & 0 \\
1 & -2
\end{array}\right]
$$
$\therefore \quad \mathrm{AC}-\mathrm{BC}=\left[\begin{array}{cc}-4 & -4 \\ -6 & 4\end{array}\right] \quad$ using Eqs. (ii) and (iii) $]$ $=(\mathrm{A}-\mathrm{B}) \mathrm{C}[$ using Eq. (i) $]$ Hence proved.
$\begin{aligned}(\mathrm{ix})(\mathrm{A}-\mathrm{B})^{\mathrm{T}} &=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}-3 & -2 \\ 2 & -2\end{array}\right] \\ \mathrm{A}^{\mathrm{T}}-\mathrm{B}^{\mathrm{T}} &=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]-\left[\begin{array}{ll}4 & 1 \\ 0 & 5\end{array}\right] \\ &=\left[\begin{array}{cc}-3 & -2 \\ 2 & -2\end{array}\right]=(\mathrm{A}-\mathrm{B})^{\mathrm{T}} \text { Hence Proved. } \end{aligned}$
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