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

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Matrices (p-1)4 Marks
Question
If $A=\left[\begin{array}{cc}-3 & 2 \\ 2 & 4\end{array}\right], B=\left[\begin{array}{ll}1 & a \\ b & 0\end{array}\right]$ and $(A+B)(A-B)=A^2-B^2$, find $a$ and $b$.
✓

Answer

$
\begin{aligned}
& (A+B)(A-B)=A^2-B^2 \\
& \therefore A^2-A B+B A-B^2=A^2-B^2 \\
& \therefore-A B+B A=0 \\
& \therefore A B=B A \\
& \therefore\left[\begin{array}{rr}
-3 & 2 \\
2 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & a \\
b & 0
\end{array}\right]=\left[\begin{array}{ll}
1 & a \\
b & 0
\end{array}\right]\left[\begin{array}{rr}
-3 & 2 \\
2 & 4
\end{array}\right] \\
& \therefore\left[\begin{array}{rr}
-3+2 b & -3 a+0 \\
2+4 b & 2 a+0
\end{array}\right]=\left[\begin{array}{ll}
-3+2 a & 2+4 a \\
-3 b+0 & 2 b+0
\end{array}\right]
\end{aligned}
$
$
\therefore\left[\begin{array}{rr}
-3+2 b & -3 a \\
2+4 b & 2 a
\end{array}\right]=\left[\begin{array}{rr}
-3+2 a & 2+4 a \\
-3 b & 2 b
\end{array}\right]
$
By equality of matrices,
$
\begin{aligned}
& -3+2 b=-3+2 a \\
& -3 a=2+4 a \ldots \ldots . . \\
& 2+4 b=-3 b . . . \ldots . . \\
& 2 a=2 b . . . . . .(4)
\end{aligned}
$
From (2), 7a $=-2$
$
\therefore a=\frac{-2}{7}
$
From (3), $7 b =-2$
$
\therefore b =\frac{-2}{7}
$
These values of $a$ and $b$ also satisfy equations (1) and (4). Hence, $a=\frac{-2}{7}$ and $b=\frac{-2}{7}$

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