Question
Given $A=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right], B=\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]$ and that $\ce{AB = A + B}.$ Find the values of $a, b$ and $c$
✓
Answer
$AB = [(3, 0),(0, 4)][(a, b),(0, c)] $
$= [(3a + 0, 3b + 0),(0 + 0, 0 = 4c)] $
$= [(3a, 3b),(0, 4c)]$
$A+B=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]=\left[\begin{array}{cc}3+a & b \\ 0 & 4+c\end{array}\right]$
Given $AB = A + B$
$\therefore\left[\begin{array}{cc}3 a & 3 b \\ 0 & 4 c\end{array}\right]=\left[\begin{array}{cc}3+a & b \\ 0 & 4+c\end{array}\right]$
Comparing the correspomding elememts we get
$3a = 3 + a$
$2a = 3$
$\Rightarrow a=\frac{3}{2}$
$3 b=b $
$\Rightarrow b=0$
$4 c=4+c $
$\Rightarrow 3 c=4 $
$\Rightarrow c=\frac{4}{3}$
$= [(3a + 0, 3b + 0),(0 + 0, 0 = 4c)] $
$= [(3a, 3b),(0, 4c)]$
$A+B=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]=\left[\begin{array}{cc}3+a & b \\ 0 & 4+c\end{array}\right]$
Given $AB = A + B$
$\therefore\left[\begin{array}{cc}3 a & 3 b \\ 0 & 4 c\end{array}\right]=\left[\begin{array}{cc}3+a & b \\ 0 & 4+c\end{array}\right]$
Comparing the correspomding elememts we get
$3a = 3 + a$
$2a = 3$
$\Rightarrow a=\frac{3}{2}$
$3 b=b $
$\Rightarrow b=0$
$4 c=4+c $
$\Rightarrow 3 c=4 $
$\Rightarrow c=\frac{4}{3}$
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