Let A= \(\begin {bmatrix} 2&4\\3&2\end {bmatrix}\),B=\(\begin {bmatrix} 1&3\\-2&5\end {bmatrix}\),C=\(\begin {bmatrix} -2&5\\3&4\end {bmatrix}\).Find each of the following
I. A+B
II. A-B
III. 3A-C
IV. AB
V. BA
Let A= \(\begin {bmatrix} 2&4\\3&2\end {bmatrix}\),B=\(\begin {bmatrix} 1&3\\-2&5\end {bmatrix}\),C=\(\begin {bmatrix} -2&5\\3&4\end {bmatrix}\).Find each of the following
I. A+B
II. A-B
III. 3A-C
IV. AB
V. BA
Solution and Explanation
(i) A+B=\(\begin {bmatrix} 2&4\\3&2\end {bmatrix}\)+\(\begin {bmatrix} 1&3\\-2&5\end {bmatrix}\)=\(\begin{bmatrix}2+1&4+3\\3-2&2+5\end{bmatrix}\)=\(\begin {bmatrix} 3&7\\1&7\end {bmatrix}\)
(ii) A-B= \(\begin {bmatrix} 2&4\\3&2\end {bmatrix}\)-\(\begin {bmatrix} 1&3\\-2&5\end {bmatrix}\)=\(\begin{bmatrix}2-1&4-3\\3-(-2)&2-5\end{bmatrix}\)=\(\begin {bmatrix} 1&1\\5&-3\end {bmatrix}\)
(iii) 3A-C=3\(\begin {bmatrix} 2&4\\3&2\end {bmatrix}\)-\(\begin {bmatrix} -2&5\\3&4\end {bmatrix}\)
=\(\begin{bmatrix}3*2&3*4\\3*3&3*2\end{bmatrix}\)-\(\begin {bmatrix} -2&5\\3&4\end {bmatrix}\)
=\(\begin{bmatrix}6&12\\9&6\end{bmatrix}\)-\(\begin {bmatrix} -2&5\\3&4\end {bmatrix}\)=\(\begin{bmatrix}6+2&12-5\\9-3&6-4\end{bmatrix}\)=\(\begin {bmatrix} 8&7\\6&2\end {bmatrix}\)
(iv)Matrix A has 2 columns. This number is equal to the number of rows in matrix B.
Therefore, AB is defined as:
AB=\(\begin {bmatrix} 2&4\\3&2\end {bmatrix}\)\(\begin {bmatrix} 1&3\\-2&5\end {bmatrix}\)
=\(\begin{bmatrix}2(1)+4(-2)&2(3)+4(5)\\3(1)+2(-2)&3(3)+2(5)\end{bmatrix}\)
=\(\begin{bmatrix}2-8&6+20\\3-4&9+10\end{bmatrix}\)=\(\begin {bmatrix} -6&26\\-1&19\end {bmatrix}\)
(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A.
Therefore, BA is defined as:
BA=\(\begin {bmatrix} 1&3\\-2&5\end {bmatrix}\)\(\begin {bmatrix} 2&4\\3&2\end {bmatrix}\)
=\(\begin{bmatrix}1(2)+3(3)&1(4)+3(2)\\-2(2)+5(3)&-2(4)+5(2)\end{bmatrix}\)
=\(\begin{bmatrix}2+9&4+6\\-4+15&-8+10\end{bmatrix}=\begin{bmatrix}11&10\\11&2\end{bmatrix}\)
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