Question:

Which of the following product of matrices are conformable for multiplication?

 (i) $\left[ \begin{matrix} 1 & -1 \\ 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} -2 \\ 3 \\ \end{matrix} \right]$ (ii) $\left[ \begin{matrix} 1 & -1 \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -1 \\ 1 & 3 \\ \end{matrix} \right]$ (iii) $\left[ \begin{matrix} 1 \\ -1 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 \\ -1 & 2 \\ \end{matrix} \right]$ (iv) $\left[ \begin{matrix} 1 & 2 \\ 0 & -1 \\ -1 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ \end{matrix} \right]$ (v) $\left[ \begin{matrix} 3 & 2 & 1 \\ 0 & 1 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -1 \\ 0 & 2 \\ -2 & 3 \\ \end{matrix} \right]$
Difficulty: Easy

Solution:
Two matrices are conformable for multiplication if the numbers of columns of first matrix are equal to number of rows of second matrix. So, according to this definition:

(i) is conformable for multiplication (because the first matrix has two columns and second matrix has same number of rows).

(ii) is conformable for multiplication (because the first matrix has two columns and second matrix has same number of rows).

(iii) is not conformable for multiplication (because the first matrix has just one column and second matrix has two rows).

(iv) is conformable for multiplication (because the first matrix has just two columns and second matrix has the same number of rows).

(v) is conformable for multiplication (because the first matrix has three columns and second matrix has same number of rows).

Question:

If A = $\left[ \begin{matrix} 3 & 0 \\ -1 & 2 \\ \end{matrix} \right]$, B = $\left[ \begin{matrix} 6 \\ 5 \\ \end{matrix} \right]$, find

 (i) AB (ii) BA (if possible)

Difficulty: Easy

Solution:

(i) AB

= $\left[ \begin{matrix} 3 & 0 \\ -1 & 2 \\ \end{matrix} \right]\times \left[ \begin{matrix} 6 \\ 5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times 6+0\times 5 \\ \left( -1 \right)\times 6+2\times 5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 18+0 \\ -6+10 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 18 \\ 4 \\ \end{matrix} \right]$

So, AB = $\left[ \begin{matrix} 18 \\ 4 \\ \end{matrix} \right]$

(ii) BA

BA is not possible as it is not conformable for multiplication (number of columns of B is not equal to number of rows of A).

Question:

Find the following products

 (i) $\left[ \begin{matrix} 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 0 \\ \end{matrix} \right]$ (ii) $\left[ \begin{matrix} 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 5 \\ -4 \\ \end{matrix} \right]$ (iii) $\left[ \begin{matrix} -3 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 0 \\ \end{matrix} \right]$ (iv) $\left[ \begin{matrix} 6 & -0 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 0 \\ \end{matrix} \right]$ (v) $\left[ \begin{matrix} 1 & 2 \\ -3 & 0 \\ 6 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 5 \\ 0 & -4 \\ \end{matrix} \right]$
Difficulty: Easy

Solution:

(i) $\left[ \begin{matrix} 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 0 \\ \end{matrix} \right]$

= $\left[ 1\times 4+2\times 0 \right]$

= $\left[ 4+0 \right]$

= $\left[ 4 \right]$

So, $\left[ \begin{matrix} 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 0 \\ \end{matrix} \right]~$= $\left[ 4 \right]$

(ii) $\left[ \begin{matrix} 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 5 \\ -4 \\ \end{matrix} \right]$

= $\left[ 1\times 5+2\times \left( -4 \right) \right]$

= $\left[ 5-8 \right]$

= $\left[ -3 \right]$

So, $\left[ \begin{matrix} 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 5 \\ -4 \\ \end{matrix} \right]$ = $\left[ -3 \right]$

(iii) $\left[ \begin{matrix} -3 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 0 \\ \end{matrix} \right]$

= $\left[ \left( -3 \right)\times 4+0\times 0 \right]$

= $\left[ -12+0 \right]$

= $\left[ -12 \right]$

So, $\left[ \begin{matrix} -3 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 0 \\ \end{matrix} \right]$=$\left[ -12 \right]$

(iv) $\left[ \begin{matrix} 6 & -0 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 0 \\ \end{matrix} \right]$

= $\left[ 6\times 4+0\times 0 \right]$

= $\left[ 24+0 \right]$

= $\left[ 24 \right]$

So, $\left[ \begin{matrix} 6 & -0 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 0 \\ \end{matrix} \right]$ = $\left[ 24 \right]$

(v) $\left[ \begin{matrix} 1 & 2 \\ -3 & 0 \\ 6 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 5 \\ 0 & -4 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1\times 4+2\times 0 & 1\times 5+2\times \left( -4 \right) \\ -3\times 4+0\times 0 & -3\times 5+0\times \left( -4 \right) \\ 6\times 4+\left( -1 \right)\times 0 & 6\times 5+\left( -1 \right)\times \left( -4 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4+0 & 5+\left( -8 \right) \\ -12+0 & -15+0 \\ 24+0 & 30+4 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & -3 \\ -12 & -15 \\ 24 & 34 \\ \end{matrix} \right]$

So, $\left[ \begin{matrix} 1 & 2 \\ -3 & 0 \\ 6 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 5 \\ 0 & -4 \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 & -3 \\ -12 & -15 \\ 24 & 34 \\ \end{matrix} \right]$

Question:

Multiply the following matrices.

 (a) $\left[ \begin{matrix} 2 & 3 \\ 1 & 1 \\ 0 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -1 \\ 3 & 0 \\ \end{matrix} \right]$ (b)$~\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ -1 & 1 \\ \end{matrix} \right]$ (c) $\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]$ (d) $\left[ \begin{matrix} 8 & 5 \\ 6 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -\frac{5}{2} \\ -4 & 4 \\ \end{matrix} \right]$ (e) $\left[ \begin{matrix} -1 & 2 \\ 1 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$
Difficulty: Easy

Solution:

(a) $\left[ \begin{matrix} 2 & 3 \\ 1 & 1 \\ 0 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -1 \\ 3 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2\times 2+3\times 3 & 2\times \left( -1 \right)+3\times 0 \\ 1\times 2+1\times 3 & 1\times \left( -1 \right)+1\times 0 \\ 0\times 2+\left( -2 \right)\times 3 & 0\times \left( -1 \right)+\left( -2 \right)\times 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4+9 & -2+0 \\ 2+3 & -1+0 \\ 0-6 & 0+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 13 & -2 \\ 5 & -1 \\ -6 & 0 \\ \end{matrix} \right]$

So, $\left[ \begin{matrix} 2 & 3 \\ 1 & 1 \\ 0 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -1 \\ 3 & 0 \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 13 & -2 \\ 5 & -1 \\ -6 & 0 \\ \end{matrix} \right]$

(b) $\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ -1 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1\times 1+2\times 3+3\times \left( -1 \right) & 1\times 2+2\times 4+3\times 1 \\ 4\times 1+5\times 3+6\times \left( -1 \right) & 4\times 2+5\times 4+6\times 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+6-3 & 2+8+3 \\ 4+15-6 & 8+20+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & 13 \\ 13 & 34 \\ \end{matrix} \right]$

So, $\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ -1 & 1 \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 & 13 \\ 13 & 34 \\ \end{matrix} \right]$

(c) $\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1\times 1+2\times 4 & 1\times 2+2\times 5 & 1\times 3+2\times 6 \\ 3\times 1+4\times 4 & 3\times 2+4\times 5 & 3\times 3+4\times 6 \\ -1\times 1+1\times 4 & -1\times 2+1\times 5 & -1\times 3+1\times 6 \\ \end{matrix} \right]$

. = $\left[ \begin{matrix} 1+8 & 2+10 & 3+12 \\ 3+16 & 6+20 & 9+24 \\ -1+4 & -2+5 & -3+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 9 & 12 & 15 \\ 19 & 26 & 33 \\ 3 & 3 & 3 \\ \end{matrix} \right]$

So, $\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 9 & 12 & 15 \\ 19 & 26 & 33 \\ 3 & 3 & 3 \\ \end{matrix} \right]$

(d) $\left[ \begin{matrix} 8 & 5 \\ 6 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -\frac{5}{2} \\ -4 & 4 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 8\times 2+5\times \left( -4 \right) & 8\times \left( -\frac{5}{2} \right)+5\times 4 \\ 6\times 2+4\times \left( -4 \right) & 6\times \left( -\frac{5}{2} \right)+4\times 4 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 16-20 & -20+20 \\ 12-16 & -15+16 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -4 & 0 \\ -4 & 1 \\ \end{matrix} \right]$

So, $\left[ \begin{matrix} 8 & 5 \\ 6 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & \frac{-5}{2} \\ -4 & 4 \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -4 & 0 \\ -4 & 1 \\ \end{matrix} \right]$

(e) $\left[ \begin{matrix} -1 & 2 \\ 1 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 0+2\times 0 & -1\times 0+2\times 0 \\ 1\times 0+3\times 0 & 1\times 0+3\times 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0+0 & 0+0 \\ 0+0 & 0+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$

So, $\left[ \begin{matrix} -1 & 2 \\ 1 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]=~\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$

Question:

Let A = $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]$, B= $\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]$ and C = $\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

Verify whether

 (i) AB = BA (ii) A(BC) = (AB)C (iii) A(B $+$ C) = AB $+$ AC (iv) A(B $-$ C) = AB $-$ AC

Difficulty: Easy

Solution:

(i) AB = BA

Taking Left Hand Side: AB

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: BA

= $\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]\times \left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1\times \left( -1 \right)+2\times 2 & 1\times 3+2\times 0 \\ -3\times \left( -1 \right)+\left( -5 \right)\times 2 & -3\times 3+\left( -5 \right)\times 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1+4 & 3+0 \\ 3-10 & -9+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 3 \\ -7 & -9 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(1)$

From $(1)$ and $(2)$, it is proved that: AB $\neq$ BA

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

Taking Left Hand Side: A(BC)

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]~\times ~\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1\times 2+2\times 1 & 1\times 1+2\times 3 \\ -3\times 2+\left( -5 \right)\times 1 & -3\times 1+\left( -5 \right)\times 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 2+2 & 1+6 \\ -6-5 & -3-15 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 4 & 7 \\ -11 & -18 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 4+3\times \left( -11 \right) & -1\times 7+3\times \left( -18 \right) \\ 2\times 4+0\times \left( -11 \right) & 2\times 7+0\times \left( -18 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -4-33 & -7-54 \\ 8+0 & 14+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -37 & -61 \\ 8 & 14 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: (AB)C

= $\left( \left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]~ \right)\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left( \left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right] \right)\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10\times 2+\left( -17 \right)\times 1 & -10\times 1+\left( -17 \right)\times 3 \\ 2\times 2+4\times 1 & 2\times 1+4\times 3 \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -20-17 & -10-51 \\ 4+4 & 2+12 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -37 & -61 \\ 8 & 14 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that: A(BC) = (AB)C

(iii) A(B $+$ C) = AB $+$ AC

Taking Left Hand Side: A(B $+$ C)

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]+~\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left( \left[ \begin{matrix} 1+2 & 2+1 \\ -3+1 & -5+3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 3 & 3 \\ -2 & -2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 3+3\times \left( -2 \right) & -1\times 3+3\times \left( -2 \right) \\ 2\times 3+0\times \left( -2 \right) & 2\times 3+0\times \left( -2 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -3-6 & -3-6 \\ 6+0 & 6+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -9 & -9 \\ 6 & 6 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: AB $+$ AC

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right]+\left[ \begin{matrix} -1\times 2+3\times 1 & -1\times 1+3\times 3 \\ 2\times 2+0\times 1 & 2\times 1+0\times 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right]+\left[ \begin{matrix} -2+3 & -1+9 \\ 4+0 & 2+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right]+~\left[ \begin{matrix} 1 & 8 \\ 4 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10+1 & -17+8 \\ 2+4 & 4+2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -9 & -9 \\ -6 & 6 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that: A(B $+$ C) = AB $+$ AC

(iv) A(B $-$ C) = AB $-$ AC

Taking Left Hand Side: A(B $-$ C)

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]-~\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left( \left[ \begin{matrix} 1-2 & 2-1 \\ -3-1 & -5-3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} -1 & 1 \\ -4 & -8 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times \left( -1 \right)+3\times \left( -4 \right) & -1\times 1+3\times \left( -8 \right) \\ 2\times \left( -1 \right)+0\times \left( -4 \right) & 2\times 1+0\times \left( -8 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1-12 & -1-24 \\ -2+0 & 2+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -11 & -25 \\ -2 & 2 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: AB $-$ AC

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]-\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right]-\left[ \begin{matrix} -1\times 2+3\times 1 & -1\times 1+3\times 3 \\ 2\times 2+0\times 1 & 2\times 1+0\times 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right]-\left[ \begin{matrix} -2+3 & -1+9 \\ 4+0 & 2+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 8 \\ 4 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10-1 & -17-8 \\ 2-4 & 4-2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -11 & -25 \\ -2 & 2 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that: A(B $-$ C) = AB $-$ AC

Question:

For the matrices A = $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]$, B= $\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]$, C = $\left[ \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right]$

Verify that

 (i) (AB)$^{t}$ = B$^{t}$ A$^{t}$ (ii) (BC)$^{t}$ = C$^{t}$B$^{t}$
Difficulty: Easy

Solution:

(i) (AB)$^{t}$ = B$^{t}$ A$^{t}$

Taking Left Hand Side: (AB)$^{t}$

= ${{\left( \left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right] \right)}^{t}}$

= $\left[ \begin{matrix} -10 & 2 \\ -17 & 4 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: B$^{t}$ A$^{t}$

= ${{\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]}^{t}}\times {{\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 1 & -3 \\ 2 & -5 \\ \end{matrix} \right]\times \left[ \begin{matrix} -1 & 2 \\ 3 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1\times \left( -1 \right)+\left( -3 \right)\times 3 & 1\times 2+\left( -3 \right)\times 0 \\ 2\times \left( -1 \right)+\left( -5 \right)\times 3 & 2\times 2+\left( -5 \right)\times 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & 2+0 \\ -2-15 & 4+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & 2 \\ -17 & 4 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that: (AB)$^{t}$ = B$^{t}$ A$^{t}$

(ii) (BC)$^{t}$ = C$^{t}$B$^{t}$

Taking Left Hand Side: (BC)$^{t}$

= ${{\left( \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]~\times ~\left[ \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right]~ \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} 1\times \left( -2 \right)+2\times 3 & 1\times 6+2\times \left( -9 \right) \\ -3\times \left( -2 \right)+\left( -5 \right)\times 3 & -3\times 6+\left( -5 \right)\times \left( -9 \right) \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} -2+6 & 6-18 \\ 6-15 & -18+45 \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} 4 & -12 \\ -9 & 27 \\ \end{matrix} \right] \right)}^{t}}$

= $\left[ \begin{matrix} 4 & -9 \\ -12 & 27 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: C$^{t}$B$^{t}$

= ${{\left[ \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right]}^{t}}\times {{\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} -2 & 3 \\ 6 & -9 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & -3 \\ 2 & -5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2\times 1+3\times 2 & -2\times \left( -3 \right)+3\times \left( -5 \right) \\ 6\times 1+\left( -9 \right)\times 2 & 6\times \left( -3 \right)+\left( -9 \right)\times \left( -5 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2+6 & 6-15 \\ 6-18 & -18+45 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & -9 \\ -12 & 27 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that: (BC)$^{t}$ = C$^{t}$B$^{t}$