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Matrices (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Matrices (p-1)3 Marks
Question
Find $x$ and $y$, if
$
\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]
$
✓

Answer

$
\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{rrr}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]
$
$
\begin{aligned}
& \therefore\left[\begin{array}{lrr}
2 x+y-1 & -1+6 & 1+4 \\
3+3 & 4 y+0 & 4+3
\end{array}\right]=\left[\begin{array}{rrr}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right] \\
& \therefore\left[\begin{array}{lrr}
2 x+y-1 & 5 & 5 \\
6 & 4 y & 7
\end{array}\right]=\left[\begin{array}{rrr}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]
\end{aligned}
$
By equality of matrices, we get
$
2 x+y-1=3
$
and $4 y =18$
From (2), $y=\frac{9}{2}$
Substituting $y=\frac{9}{2}$ in (1), we get
$
\begin{aligned}
& 2 x+\frac{9}{2}-1=3 \\
& \therefore 2 x=3-\frac{7}{2}=\frac{-1}{2} \\
& \therefore x=\frac{-1}{4}
\end{aligned}
$
Hence, $x=\frac{-1}{4}$ and $y=\frac{9}{2}$.

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