Question#1: Which of the following product of matrices is conformable for multiplications?
\(i\)
\[\left[ \begin{matrix}
1 & -1 \\
0 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
-2 \\
3 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left[ \begin{matrix}
1 & -1 \\
0 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
-2 \\
3 \\
\end{matrix} \right]\\
\text{Order of first Matrix}&=2-by-\bbox[5px, border: 1px solid red]{2} \\
\text{Order of second Matrix}&=\bbox[5px, border: 1px solid red]{2}-by-1 \\
\end{align}\]
Multiplication is possible since the columns of the first matrix are equal to the rows of the second matrix.
\(ii\)
\[\left[ \begin{matrix}
1 & -1 \\
1 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
-2 & -1 \\
1 & 3 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left[ \begin{matrix}
1 & -1 \\
1 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
-2 & -1 \\
1 & 3 \\
\end{matrix} \right]\\
\text{Order of first Matrix}&=2-by-\bbox[5px, border: 1px solid red]{2} \\
\text{Order of second Matrix}&=\bbox[5px, border: 1px solid red]{2}-by-2 \\
\end{align}\]
Multiplication is possible since the columns of the first matrix are equal to the rows of the second matrix.
\(iii\)
\[\left[ \begin{matrix}
1 \\
-1 \\
\end{matrix} \right]\left[ \begin{matrix}
0 & 1 \\
-1 & 2 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left[ \begin{matrix}
1 \\
-1 \\
\end{matrix} \right]\left[ \begin{matrix}
0 & 1 \\
-1 & 2 \\
\end{matrix} \right]\\
\text{Order of first Matrix}&=2-by-\bbox[5px, border: 1px solid red]{1} \\
\text{Order of second Matrix}&=\bbox[5px, border: 1px solid red]{2}-by-2 \\
\end{align}\]
Multiplication is not possible since the columns of the first matrix are not equal to the rows of the second matrix.
\(iv\)
\[\left[ \begin{matrix}
1 & 2 \\
0 & -1 \\
-1 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 0 & -1 \\
0 & 1 & 2 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left[ \begin{matrix}
1 & 2 \\
0 & -1 \\
-1 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 0 & -1 \\
0 & 1 & 2 \\
\end{matrix} \right]\\
\text{Order of first Matrix}&=3-by-\bbox[5px, border: 1px solid red]{2} \\
\text{Order of second Matrix}&=\bbox[5px, border: 1px solid red]{2}-by-3 \\
\end{align}\]
Multiplication is possible since the columns of the first matrix are equal to the rows of the second matrix.
\(v\)
\[\left[ \begin{matrix}
3 & 2 & 1 \\
0 & 1 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & -1 \\
0 & 2 \\
-2 & 3 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
\left[ \begin{matrix}
3 & 2 & 1 \\
0 & 1 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & -1 \\
0 & 2 \\
-2 & 3 \\
\end{matrix} \right]\\
\text{Order of first Matrix}&=2-by-\bbox[5px, border: 1px solid red]{3} \\
\text{Order of second Matrix}&=\bbox[5px, border: 1px solid red]{3}-by-2 \\
\end{align}\]
Multiplication is possible since the columns of the first matrix are equal to the rows of the second matrix.
Question#2: If \(A=\left[ \begin{matrix}
3 & 0 \\
-1 & 2 \\
\end{matrix} \right]\) , \(B=\left[ \begin{matrix}
6 \\
5 \\
\end{matrix} \right]\) find \(AB\) and \(BA\) if possible.
\[AB\]
Solution:
\[\begin{align}
AB&=\left[ \begin{matrix}
3 & 0 \\
-1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
6 \\
5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
18+0 \\
-6+10 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
18 \\
4 \\
\end{matrix} \right] \\
\end{align}\]
\(BA\)
\(BA\) is not possible since the columns of the first matrix aren’t equal to the rows of the second matrix.
Question#3: Find the following products.
\(i\)
\[\left[ \begin{matrix}
1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
4 \\
0 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
4 \\
0 \\
\end{matrix} \right] \\
& =\left[ 1\times 4+2\times 0 \right] \\
& =\left[ 4+0 \right] \\
& =\left[ 4 \right] \\
\end{align}\]
\(ii\)
\[\left[ \begin{matrix}
1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
5 \\
-4 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
5 \\
-4 \\
\end{matrix} \right] \\
& =\left[ 1\times 5+2\times -4 \right] \\
& =\left[ 5-8 \right] \\
& =\left[ -3 \right] \\
\end{align}\]
\(iii\)
\[\left[ \begin{matrix}
-3 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
4 \\
0 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
-3 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
4 \\
0 \\
\end{matrix} \right] \\
& =\left[ -3\times 4+0\times 0 \right] \\
& =\left[ -12+0 \right] \\
& =\left[ -12 \right] \\
\end{align}\]
\(iv\)
\[\left[ \begin{matrix}
6 & -0 \\
\end{matrix} \right]\left[ \begin{matrix}
4 \\
0 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
6 & -0 \\
\end{matrix} \right]\left[ \begin{matrix}
4 \\
0 \\
\end{matrix} \right] \\
& =\left[ 6\times 4-0\times 0 \right] \\
& =\left[ 24-0 \right] \\
& =\left[ 24 \right] \\
\end{align}\]
\(v\)
\[\left[ \begin{matrix}
1 & 2 \\
-3 & 0 \\
6 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
4 & 5 \\
0 & -4 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
1 & 2 \\
-3 & 0 \\
6 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
4 & 5 \\
0 & -4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( 1 \right)\left( 4 \right)+\left( 2 \right)\left( 0 \right) & \left( 1 \right)\left( 5 \right)+\left(2
\right)\left( -4 \right) \\
\left( -3 \right)\left( 4 \right)+\left( 0 \right)\left( 0 \right) & \left( -3 \right)\left( 5 \right)+\left(0
\right)\left( -4 \right) \\
\left( 6 \right)\left( 4 \right)+\left( -1 \right)\left( 0 \right) & \left( 6 \right)\left( 5 \right)+\left(-1
\right)\left( -4 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4+0 & 5-8 \\
-12+0 & -15-0 \\
24-0 & 30+4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4 & -3 \\
-12 & -15 \\
24 & 34 \\
\end{matrix} \right] \\
\end{align}\]
Question#4: Multiply the following matrices.
\(a\)
\[\left[ \begin{matrix}
2 & 3 \\
1 & 1 \\
0 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & -1 \\
3 & 0 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
2 & 3 \\
1 & 1 \\
0 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & -1 \\
3 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( 2 \right)\left( 2 \right)+\left( 3 \right)\left( 3 \right) & \left( 2 \right)\left( -1 \right)+\left( 3 \right)\left( 0 \right) \\
\left( 1 \right)\left( 2 \right)+\left( 1 \right)\left( 3 \right) & \left( 1 \right)\left( -1 \right)+\left( 1 \right)\left( 0 \right) \\
\left( 0 \right)\left( 2 \right)+\left( -2 \right)\left( 3 \right) & \left( 0 \right)\left( -1 \right)+\left( -2 \right)\left( 0 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4+9 & -2+0 \\
2+3 & -1+0 \\
0-6 & 0+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
13 & -2 \\
5 & -1 \\
-6 & 0 \\
\end{matrix} \right] \\
\end{align}\]
\(b\)
\[\left[ \begin{matrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 \\
3 & 4 \\
-1 & 1 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 \\
3 & 4 \\
-1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\times 1+2\times 3+3\times -1 & 1\times 2+2\times 4+3\times 1 \\
4\times 1+5\times 3+6\times -1 & 4\times 2+5\times 4+6\times 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+6-3 & 2+8+3 \\
4+15-6 & 8+20+6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4 & 13 \\
13 & 34 \\
\end{matrix} \right] \\
\end{align}\]
\(c\)
\[\left[ \begin{matrix}
1 & 2 \\
3 & 4 \\
-1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
1 & 2 \\
3 & 4 \\
-1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( 1 \right)\left( 1 \right)+\left( 2 \right)\left( 4 \right) & \left( 1 \right)\left( 2 \right)+\left( 2 \right)\left( 5 \right) & \left( 1 \right)\left( 3 \right)+\left( 2 \right)\left( 6 \right) \\
\left( 3 \right)\left( 1 \right)+\left( 4 \right)\left( 4 \right) & \left( 3 \right)\left( 2 \right)+\left( 4 \right)\left( 5 \right) & \left( 3 \right)\left( 3 \right)+\left( 4 \right)\left( 6 \right) \\
\left( -1 \right)\left( 1 \right)+\left( 1 \right)\left( 4 \right) & \left( -1 \right)\left( 2 \right)+\left( 1 \right)\left( 5 \right) & \left( -1 \right)\left( 3 \right)+\left( 1 \right)\left( 6 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+8 & 2+10 & 3+12 \\
3+16 & 6+20 & 9+24 \\
-1+4 & -2+5 & -3+6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
9 & 12 & 15 \\
19 & 26 & 33 \\
3 & 3 & 3 \\
\end{matrix} \right] \\
\end{align}\]
\(d\)
\[\left[ \begin{matrix}
8 & 5 \\
6 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & -\frac{5}{2} \\
-4 & 4 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
8 & 5 \\
6 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & -\frac{5}{2} \\
-4 & 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( 8 \right)\left( 2 \right)+\left( 5 \right)\left( -4 \right) & \left( \cancelto{4}{8} \right)\left( -\frac{5}{\cancel{2}} \right)+\left( 5 \right)\left( 4 \right) \\
\left( 6 \right)\left( 2 \right)+\left( 4 \right)\left( -4 \right) & \left( \cancelto{3}{6} \right)\left( -\frac{5}{\cancel{2}} \right)+\left( 4 \right)\left( 4 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
16-20 & -20+20 \\
12-16 & -15+16 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-4 & 0 \\
-4 & 1 \\
\end{matrix} \right] \\
\end{align}\]
\(e\)
\[\left[ \begin{matrix}
-1 & 2 \\
1 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right]\]
Solution:
\[\begin{align}
& =\left[ \begin{matrix}
-1 & 2 \\
1 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -1 \right)\left( 0 \right)+\left( 2 \right)\left( 0 \right) & \left( -1 \right)\left( 0 \right)+\left( 2 \right)\left( 0 \right) \\
\left( 1 \right)\left( 0 \right)+\left( 3 \right)\left( 0 \right) & \left( 1 \right)\left( 0 \right)+\left( 3 \right)\left( 0 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0+0 & 0+0 \\
0+0 & 0+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right] \\
\end{align}\]
Question#5: 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.
\(i\)
\[AB=BA\]
Solution:
\[\begin{align}
AB&=\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -1 \right)\left( 1 \right)+\left( 3 \right)\left( -3 \right) & \left( -1 \right)\left( 2 \right)+\left( 3 \right)\left( -5 \right) \\
\left( 2 \right)\left( 1 \right)+\left( 0 \right)\left( -3 \right) & \left( 2 \right)\left( 2 \right)+\left( 0 \right)\left( -5 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1-9 & -2-15 \\
2+0 & 4+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10 & -17 \\
2 & 4 \\
\end{matrix} \right] \\
BA &=\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( 1 \right)\left( -1 \right)+\left( 2 \right)\left( 2 \right) & \left( 1 \right)\left( 3 \right)+\left( 2 \right)\left( 0 \right) \\
\left( -3 \right)\left( -1 \right)+\left( -5 \right)\left( 2 \right) & \left( -3 \right)\left( 3 \right)+\left( -5 \right)\left( 0 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1+4 & 3+0 \\
3-10 & -9+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 3 \\
-7 & -9 \\
\end{matrix} \right] \\
& So,AB\ne BA \\
\end{align}\]
\(ii\)
\[A\left( BC \right)=\left( AB \right)C\]
Solution:
\[\begin{align}
A\left( BC \right) &=\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]\left( \left[ \begin{matrix}
\left( 1 \right)\left( 2 \right)+\left( 2 \right)\left( 1 \right) & \left( 1 \right)\left( 1 \right)+\left( 2 \right)\left( 3 \right) \\
\left( -3 \right)\left( 2 \right)+\left( -5 \right)\left( 1 \right) & \left( -3 \right)\left( 1 \right)+\left( -5 \right)\left( 3 \right) \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
2+2 & 1+6 \\
-6-5 & -3-15 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
4 & 7 \\
-11 & -18 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -1 \right)\left( 4 \right)+\left( 3 \right)\left( -11 \right) & \left( -1 \right)\left( 7 \right)+\left( 3 \right)\left( -18 \right) \\
\left( 2 \right)\left( 4 \right)+\left( 0 \right)\left( -11 \right) & \left( 2 \right)\left( 7 \right)+\left( 0 \right)\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] \\
\left( AB \right)C&=\left( \left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right] \right)\left[ \begin{matrix}
2 & 1 \\
1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -1 \right)\left( 1 \right)+\left( 3 \right)\left( -3 \right) & \left( -1 \right)\left( 2 \right)+\left( 3 \right)\left( -5 \right) \\
\left( 2 \right)\left( 1 \right)+\left( 0 \right)\left( -3 \right) & \left( 2 \right)\left( 2 \right)+\left( 0 \right)\left( -5 \right) \\
\end{matrix} \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]\left[ \begin{matrix}
2 & 1 \\
1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10 & -17 \\
2 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & 1 \\
1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -10 \right)\left( 2 \right)+\left( -17 \right)\left( 1 \right) & \left( -10 \right)\left( 1 \right)+\left( -17 \right)\left( 3 \right) \\
\left( 2 \right)\left( 2 \right)+\left( 4 \right)\left( 1 \right) & \left( 2 \right)\left( 1 \right)+\left( 4 \right)\left( 3 \right) \\
\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] \\
So,A\left( BC \right) &=\left( AB \right)C \\
\end{align}\]
\(iii\)
\[A\left( B+C \right)=AB+AC\]
Solution:
\[\begin{align}
A\left( B+C \right) &=\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left( \left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]+\left[ \begin{matrix}
2 & 1 \\
1 & 3 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
1+2 & 2+1 \\
-3+1 & -5+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
3 & 3 \\
-2 & -2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1\times 3+3\times -2 & -1\times 3+3\times -2 \\
2\times 3+0\times -2 & 2\times 3+0\times -2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-3-6 & -3-6 \\
6+0 & 6+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-9 & -9 \\
6 & 6 \\
\end{matrix} \right] \\
AB+AC&=\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & 1 \\
1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1\times 1+3\times -3 & -1\times 2+3\times -5 \\
2\times 1+0\times -3 & 2\times 2+0\times -5 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1\times 2+3\times 1 & -1\times 1+3\times 3 \\
2\times 2+0\times 1 & 2\times 1+0\times 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1-9 & -2-15 \\
2+0 & 4+0 \\
\end{matrix} \right]+\left[ \begin{matrix}
-2+3 & -1+9 \\
4+0 & 2+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10 & -17 \\
2 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 8 \\
4 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10+1 & -17+8 \\
2+4 & 4+2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-9 & -9 \\
6 & 6 \\
\end{matrix} \right] \\
So,A\left( B+C \right) &=AB+AC \\
\end{align}\]
\(iv\)
\[A\left( B-C \right)=AB-AC\]
Solution:
\[\begin{align}
A\left( B-C \right) &=\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left( \left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]-\left[ \begin{matrix}
2 & 1 \\
1 & 3 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
1-2 & 2-1 \\
-3-1 & -5-3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
-1 & 1 \\
-4 & -8 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1\times -1+3\times -4 & -1\times 1+3\times -8 \\
2\times -1+0\times -4 & 2\times 1+0\times -8 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-12 & -1-24 \\
-2+0 & 2+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-11 & -25 \\
-2 & 2 \\
\end{matrix} \right] \\
AB-AC&=\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]-\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & 1 \\
1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1\times 1+3\times -3 & -1\times 2+3\times -5 \\
2\times 1+0\times -3 & 2\times 2+0\times -5 \\
\end{matrix} \right]-\left[ \begin{matrix}
-1\times 2+3\times 1 & -1\times 1+3\times 3 \\
2\times 2+0\times 1 & 2\times 1+0\times 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1-9 & -2-15 \\
2+0 & 4+0 \\
\end{matrix} \right]-\left[ \begin{matrix}
-2+3 & -1+9 \\
4+0 & 2+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10 & -17 \\
2 & 4 \\
\end{matrix} \right]-\left[ \begin{matrix}
1 & 8 \\
4 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10-1 & -17-8 \\
2-4 & 4-2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-11 & -25 \\
-2 & 2 \\
\end{matrix} \right] \\
So,A\left( B-C \right) &=AB-AC \\
\end{align}\]
Question#6: For the matrices \(A=\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\) , \(B=\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]\), and \(C=\left[ \begin{matrix}
-2 & 6 \\
3 & -9 \\
\end{matrix} \right]\) verify that.
\(i\)
\[{{\left( AB \right)}^{t}}={{B}^{t}}{{A}^{t}}\]
Solution:
\[\begin{align}
{{\left( AB \right)}^{t}}&={{\left( \left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right] \right)}^{t}} \\
& ={{\left( \left[ \begin{matrix}
-1\times 1+3\times -3 & -1\times 2+3\times -5 \\
2\times 1+0\times -3 & 2\times 2+0\times -5 \\
\end{matrix} \right] \right)}^{t}} \\
& ={{\left( \left[ \begin{matrix}
-1-9 & -2-15 \\
2+0 & 4+0 \\
\end{matrix} \right] \right)}^{t}} \\
& ={{\left[ \begin{matrix}
-10 & -17 \\
2 & 4 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
-10 & 2 \\
-17 & 4 \\
\end{matrix} \right] \\
{{B}^{t}}{{A}^{t}}&={{\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]}^{t}}{{\left[ \begin{matrix}
-1 & 3 \\
2 & 0 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
1 & -3 \\
2 & -5 \\
\end{matrix} \right]\left[ \begin{matrix}
-1 & 2 \\
3 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\times -1+-3\times 3 & 1\times 2+-3\times 0 \\
2\times -1+-5\times 3 & 2\times 2+-5\times 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1-9 & 2-0 \\
-2-15 & 4-0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10 & 2 \\
-17 & 4 \\
\end{matrix} \right] \\
So,{{\left( AB \right)}^{t}}&={{B}^{t}}{{A}^{t}} \\
\end{align}\]
\(ii\)
\[{{\left( BC \right)}^{t}}={{C}^{t}}{{B}^{t}}\]
Solution:
\[\begin{align}
{{\left( BC \right)}^{t}}&={{\left( \left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]\left[ \begin{matrix}
-2 & 6 \\
3 & -9 \\
\end{matrix} \right] \right)}^{t}} \\
& ={{\left( \left[ \begin{matrix}
1\times -2+2\times 3 & 1\times 6+2\times -9 \\
-3\times -2+-5\times 3 & -3\times 6+-5\times -9 \\
\end{matrix} \right] \right)}^{t}} \\
& ={{\left( \left[ \begin{matrix}
-2+6 & 6-18 \\
6-15 & -18+45 \\
\end{matrix} \right] \right)}^{t}} \\
& ={{\left[ \begin{matrix}
4 & -12 \\
-9 & 27 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
4 & -9 \\
-12 & 27 \\
\end{matrix} \right] \\
{{C}^{t}}{{B}^{t}}&={{\left[ \begin{matrix}
-2 & 6 \\
3 & -9 \\
\end{matrix} \right]}^{t}}{{\left[ \begin{matrix}
1 & 2 \\
-3 & -5 \\
\end{matrix} \right]}^{t}} \\
& =\left[ \begin{matrix}
-2 & 3 \\
6 & -9 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & -3 \\
2 & -5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2\times 1+3\times 2 & -2\times -3+3\times -5 \\
6\times 1+-9\times 2 & 6\times -3+-9\times -5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2+6 & 6-15 \\
6-18 & -18+45 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4 & -9 \\
-12 & 27 \\
\end{matrix} \right] \\
So,{{\left( BC \right)}^{t}}&={{C}^{t}}{{B}^{t}} \\
\end{align}\]
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