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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
$
\begin{aligned}
& \text { Show that } A B=B A \text {, where } A=\left[\begin{array}{lll}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right], B= \\
& {\left[\begin{array}{lll}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]}
\end{aligned}
$
✓

Answer

$
\begin{aligned}
AB & =\left[\begin{array}{lll}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right]\left[\begin{array}{lll}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right] \\
& =\left[\begin{array}{rrr}
-2+6-3 & -6+6-0 & 2-3+1 \\
-1+4-3 & -3+4-0 & 1-2+1 \\
-6+18-12 & -18+18-0 & 6-9+4
\end{array}\right] \\
& =\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right)\\
B & =\left[\begin{array}{lll}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]\left[\begin{array}{lll}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right] \\
& =\left[\begin{array}{lll}
-2-3+6 & 3+6-9 & -1-3+4 \\
-4-2+6 & 6+4-9 & -2-2+4 \\
-6-0+6 & 9+0-9 & -3-0+4
\end{array}\right] \\
& =\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$
From (1) and (2), $AB = BA$.

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