Question:
Which of the following matrices are conformable for addition?
Solution:
Solving D
D = $\left[ \begin{matrix} 2+1 \\ 3 \\ \end{matrix} \right]$
D = $\left[ \begin{matrix} 3 \\ 3 \\ \end{matrix} \right]$
Solving F
F = $\left[ \begin{matrix} 3 & 2 \\ 1+1 & -4 \\ 3+2 & 2+1 \\ \end{matrix} \right]$
F = $\left[ \begin{matrix} 3 & 2 \\ 2 & -4 \\ 5 & 3 \\ \end{matrix} \right]$
Matrices that are of same order are conformable for addition. So, according to this definition:
(i) Matrices A and E are conformable for addition (because both have order 2-by-2).
(ii) Matrices B and D are conformable for addition (because both have order 1-by-1).
(iii) Matrices C and F are conformable for addition (because both have order 3-by-2).
Question:
Find the additive inverse of following matrices.
Solution:
The additive inverse of a matrix is obtained by changing the sign of each entry. So, according to the definition:
(i) Additive inverse of A = $-$A = $\left[ \begin{matrix} -2 & -4 \\ 2 & -1 \\ \end{matrix} \right]$
(ii) Additive inverse of B = $-$B = $\left[ \begin{matrix} -1 & 0 & 1 \\ -2 & 1 & -3 \\ -3 & 2 & -1 \\ \end{matrix} \right]$
(iii) Additive inverse of C = $-$C = $\left[ \begin{matrix} -4 \\ 2 \\ \end{matrix} \right]$
(iv) Additive inverse of D = $-$D = $\left[ \begin{matrix} -1 & 0 \\ 3 & 2 \\ -2 & -1 \\ \end{matrix} \right]$
(v) Additive inverse of E = $-$E = $\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right]$
(vi) Additive inverse of F = $-$F = $\left[ \begin{matrix} -\sqrt{3} & -1 \\ 1 & -\sqrt{2} \\ \end{matrix} \right]$
Question:
If
then find:
Solution:
(i) A + $\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$
= A + $\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} -1 & 2 \\ 2 & 1 \\ \end{matrix} \right]+~\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} -1+1 & 2+1 \\ 2+1 & 1+1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 0 & 3 \\ 3 & 2 \\ \end{matrix} \right]$
So
A + $~\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]=~\left[ \begin{matrix} 0 & 3 \\ 3 & 2 \\ \end{matrix} \right]$
(ii) B + $\left[ \begin{matrix} -2 \\ 3 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 1 \\ -1 \\ \end{matrix} \right]$ + $\left[ \begin{matrix} -2 \\ 3 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 1+\left( -2 \right) \\ \left( -1 \right)+3 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} -1 \\ 2 \\ \end{matrix} \right]$
So
B + $\left[ \begin{matrix} 1 \\ -1 \\ \end{matrix} \right]=~\left[ \begin{matrix} -1 \\ 2 \\ \end{matrix} \right]$
(iii) C + $\left[ \begin{matrix} -2 & 1 & 3 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 1 & -1 & 2 \\ \end{matrix} \right]$ + $\left[ \begin{matrix} -2 & 1 & 3 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 1+\left( -2 \right) & -1+1 & 2+3 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} -1 & 0 & 5 \\ \end{matrix} \right]$
So
C + $\left[ \begin{matrix} -2 & 1 & 3 \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -1 & 0 & 5 \\ \end{matrix} \right]$
(iv) D + $\left[ \begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \\ \end{matrix} \right]~+~\left[ \begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 1+0 & 2+1 & 3+0 \\ -1+2 & 0+0 & 2+1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 1 & 3 & 3 \\ 1 & 0 & 3 \\ \end{matrix} \right]$
So
D + $\left[ \begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ \end{matrix} \right]=~$ $\left[ \begin{matrix} 1 & 3 & 3 \\ 1 & 0 & 3 \\ \end{matrix} \right]$
(v) $2$A
= $2 \times \left[ \begin{matrix} -1 & 2 \\ 2 & 1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 2~\times \left( -1 \right) & 2~\times 2 \\ 2\times 2 & 2\times 1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} -2 & 4 \\ 4 & 2 \\ \end{matrix} \right]$
So
$2$A = $\left[ \begin{matrix} -2 & 4 \\ 4 & 2 \\ \end{matrix} \right]$
(vi) $(-1)$B
= $(-1) \times \left[ \begin{matrix} 1 \\ -1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} \left( -1 \right)\times 1 \\ \left( -1 \right)\times -1 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} -1 \\ 1 \\ \end{matrix} \right]$
So
$(-1)$B = $\left[ \begin{matrix} -1 \\ 1 \\ \end{matrix} \right]$
(vii) $(-2)$C
= $(-2) \times \left[ \begin{matrix} 1 & -1 & 2 \\ \end{matrix} \right]$
= $\left[ \left( -2 \right)\times \begin{matrix} 1 & \left( -2 \right)\times -1 & \left( -2 \right)\times 2 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} -2 & 2 & -4 \\ \end{matrix} \right]$
So
$(-2)$C = $\left[ \begin{matrix} -2 & 2 & -4 \\ \end{matrix} \right]$
(viii) $3$D
= $3 \times \left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 3\times 1 & 3\times 2 & 3\times 3 \\ 3\times -1 & 3\times 0 & 3\times 2 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 3 & 6 & 9 \\ -3 & 0 & 6 \\ \end{matrix} \right]$
So
$3$D = $\left[ \begin{matrix} 3 & 6 & 9 \\ -3 & 0 & 6 \\ \end{matrix} \right]$
(ix) $3$C
= $3 \times \left[ \begin{matrix} 1 & -1 & 2 \\ \end{matrix} \right]$
= $\left[ 3\times \begin{matrix} 1 & 3\times -1 & 3\times 2 \\ \end{matrix} \right]$
= $\left[ \begin{matrix} 3 & -3 & 6 \\ \end{matrix} \right]$
So
$3$C = $\left[ \begin{matrix} 3 & -3 & 6 \\ \end{matrix} \right]$