Question

If $A=\left[\begin{array}{c}12\\ -13\end{array}\right]B=\left[\begin{array}{c}40\\ 15\end{array}\right]C=\left[\begin{array}{c}20\\ 1-2\end{array}\right]$,
a = 4, and b = - 2, then show that:

(i) A + (B + C) = (A + B) + C

(ii) A (BC) = (AB) C

(iii) (a + b)B = aB + bB

(iv) a (C - A) = aC - aA

(v) $(A^T{)}^T$ = A

(vi) $(bA{)}^T$ = b $A^T$

(vii) $(AB{)}^T=B^T{A}^T$

(viii) (A - B)C = AC - BC

(ix) $(A-B{)}^T=A^T-B^T$

Answer 1

We have, $A=\left[\begin{array}{c}12\\ -13\end{array}\right]B=\left[\begin{array}{c}40\\ 15\end{array}\right]$

$C=\left[\begin{array}{c}20\\ 1-2\end{array}\right]$ and a = 4, b = - 2.

(i) A + (B + C) = $\left[\begin{array}{c}12\\ -13\end{array}\right]+\left[\begin{array}{c}60\\ 23\end{array}\right]=\left[\begin{array}{c}72\\ 16\end{array}\right]$

and (A + B) + C = $\left[\begin{array}{c}52\\ 08\end{array}\right]+\left[\begin{array}{c}20\\ 1-2\end{array}\right]$

$=\left[\begin{array}{c}72\\ 16\end{array}\right]$ = A + (B + C)
Hence proved.

(ii) (BC) = $\left[\begin{array}{c}40\\ 15\end{array}\right]\left[\begin{array}{c}20\\ 1-2\end{array}\right]=\left[\begin{array}{c}80\\ 7-10\end{array}\right]$

and A (BC) $\left[\begin{array}{c}12\\ -13\end{array}\right]\left[\begin{array}{c}80\\ 7-10\end{array}\right]$

$\left[\begin{array}{c}8+140-20\\ -8+210-30\end{array}\right]=\left[\begin{array}{c}22-20\\ 13-30\end{array}\right]$

Also, (AB) =$\left[\begin{array}{c}12\\ -13\end{array}\right]\left[\begin{array}{c}40\\ 15\end{array}\right]=\left[\begin{array}{c}610\\ -115\end{array}\right]$

(AB) C = $\left[\begin{array}{c}610\\ -115\end{array}\right]\left[\begin{array}{c}20\\ 1-2\end{array}\right]$

$=\left[\begin{array}{c}22-20\\ 13-30\end{array}\right]=A\left(BC\right)$

(iii) (a + b) B = (4 -2)$\left[\begin{array}{c}40\\ 15\end{array}\right]$

=$\left[\begin{array}{c}80\\ 210\end{array}\right]$

and aB + bB = 4B -2B

$\left[\begin{array}{c}160\\ 420\end{array}\right]-\left[\begin{array}{c}80\\ 210\end{array}\right]$

=$\left[\begin{array}{c}80\\ 210\end{array}\right]$

= (a + b) B

(iv) (C - A) $=\left[\begin{array}{c}2-10-2\\ 1+1-2-3\end{array}\right]=\left[\begin{array}{c}1-2\\ 2-5\end{array}\right]$

and a(C - A)=$\left[\begin{array}{c}4-8\\ 8-20\end{array}\right]$

Also, aC - aA $=\left[\begin{array}{c}80\\ 4-8\end{array}\right]-\left[\begin{array}{c}48\\ -412\end{array}\right]=\left[\begin{array}{c}4-8\\ 8-20\end{array}\right]$

= a (C - A)

(v) $A^T$ =${\left[\begin{array}{c}12\\ -13\end{array}\right]}^T=\left[\begin{array}{c}1-1\\ 23\end{array}\right]$

Now, $(A^T{)}^T={\left[\begin{array}{c}12\\ -13\end{array}\right]}^T$

= A

(vi) $(bA{)}^T={\left[\begin{array}{c}-2-4\\ 2-6\end{array}\right]}^T$ [$\because$ b = - 2]

$=\left[\begin{array}{c}-2-4\\ 2-6\end{array}\right]$

and $A^T=\left[\begin{array}{c}1-1\\ 23\end{array}\right]$

$\therefore b{A}^T=\left[\begin{array}{c}-22\\ -4-6\end{array}\right]=(bA{)}^T$

(vii) AB = $\left[\begin{array}{c}12\\ -13\end{array}\right]\left[\begin{array}{c}40\\ 15\end{array}\right]=\left[\begin{array}{c}4+20+10\\ -4+30+15\end{array}\right]=\left[\begin{array}{c}610\\ -115\end{array}\right]$

$\therefore (AB{)}^T=\left[\begin{array}{c}6-1\\ 1015\end{array}\right]$

Now, $B^T{A}^T=\left[\begin{array}{c}40\\ 15\end{array}\right]\left[\begin{array}{c}1-1\\ 23\end{array}\right]=\left[\begin{array}{c}6-1\\ 1015\end{array}\right]$

=$(AB{)}^T$

(viii) (A - B) = $\left[\begin{array}{c}1-42-0\\ -1-13-5\end{array}\right]=\left[\begin{array}{c}-32\\ -2-2\end{array}\right]$

(A - B) C = $\left[\begin{array}{c}-32\\ -2-2\end{array}\right]\left[\begin{array}{c}20\\ 1-2\end{array}\right]=\left[\begin{array}{c}-4-4\\ -64\end{array}\right]$ ...(i)

Now, AC = $\left[\begin{array}{c}12\\ -13\end{array}\right]\left[\begin{array}{c}20\\ 1-2\end{array}\right]=\left[\begin{array}{c}4-4\\ 1-6\end{array}\right]$ ...(ii)

and BC = $\left[\begin{array}{c}40\\ 15\end{array}\right]\left[\begin{array}{c}20\\ 1-2\end{array}\right]=\left[\begin{array}{c}80\\ 7-10\end{array}\right]$ ...(iii)

$\therefore AC-BC=\left[\begin{array}{c}4-8-4-0\\ 1-7-6+10\end{array}\right]$ [using Eqs. (ii) & (iii)]

$=\left[\begin{array}{c}-4-4\\ -64\end{array}\right]$

= (A - B)C [using Eq. (i)]