Question
If $2\left[\begin{array}{ll}3 & 4 \\ 5 & x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}z & 0 \\ 10 & 5\end{array}\right]$ Find the values of $x$ and $y$
✓
Answer
$
\begin{aligned}
& 2\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right] \\
& {\left[\begin{array}{cc}
6 & 8 \\
10 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
6+1 & 8+y \\
10+0 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
7 & 8+y \\
10 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right]
\end{aligned}
$
Comparing,
$
\begin{aligned}
& 2 x+1=5 \\
& \Rightarrow 2 x=5-1=4 \\
& \therefore x=\frac{4}{2}=2 \\
& 8+y=0 \\
& \Rightarrow y=-8 \\
& z=7
\end{aligned}
$
Hence $x=2, y=-8, z=7$.
\begin{aligned}
& 2\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right] \\
& {\left[\begin{array}{cc}
6 & 8 \\
10 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
6+1 & 8+y \\
10+0 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
7 & 8+y \\
10 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right]
\end{aligned}
$
Comparing,
$
\begin{aligned}
& 2 x+1=5 \\
& \Rightarrow 2 x=5-1=4 \\
& \therefore x=\frac{4}{2}=2 \\
& 8+y=0 \\
& \Rightarrow y=-8 \\
& z=7
\end{aligned}
$
Hence $x=2, y=-8, z=7$.
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