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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 $A=\left[\begin{array}{cc}2 & -3 \\ 3 & -2 \\ -1 & 4\end{array}\right], B=\left[\begin{array}{ccc}-3 & 4 & 1 \\ 2 & -1 & -3\end{array}\right]$ verify
(i) $\left( A +2 B ^{\top}\right)^{\top}= A ^{\top}+2 B$
(ii) $\left(3 A -5 B ^{\top}\right)^{\top}=3 A ^{\top}-5 B$
✓

Answer

$
\begin{aligned}
& A =\left[\begin{array}{rr}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right], B =\left[\begin{array}{rrr}
-3 & 4 & 1 \\
2 & -1 & -3
\end{array}\right] \\
& \therefore A ^{ T }=\left[\begin{array}{rrr}
2 & 3 & -1 \\
-3 & -2 & 4
\end{array}\right], B ^{ T }=\left[\begin{array}{rr}
-3 & 2 \\
4 & -1 \\
1 & -3
\end{array}\right]
\end{aligned}
$
(i)
$
\begin{aligned}
& A+2 B^T=\left[\begin{array}{rr}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right]+2\left[\begin{array}{rr}
-3 & 2 \\
4 & -1 \\
1 & -3
\end{array}\right] \\
& =\left[\begin{array}{rr}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right]+\left[\begin{array}{rr}
-6 & 4 \\
8 & -2 \\
2 & -6
\end{array}\right] \\
& =\left(\begin{array}{rr}
2-6 & -3+4 \\
3+8 & -2-2 \\
-1+2 & 4-6
\end{array}\right]=\left[\begin{array}{rr}
-4 & 1 \\
11 & -4 \\
1 & -2
\end{array}\right) \\
& \therefore\left( A +2 B ^{ T }\right)^{ T }=\left(\begin{array}{rrr}
-4 & 11 & 1 \\
1 & -4 & -2
\end{array}\right) \\
&
\end{aligned}
$
Also, $A ^{ T }+2 B$
$
\begin{aligned}
& =\left[\begin{array}{rrr}
2 & 3 & -1 \\
-3 & -2 & 4
\end{array}\right]+2\left(\begin{array}{rrr}
-3 & 4 & 1 \\
2 & -1 & -3
\end{array}\right) \\
& =\left[\begin{array}{rrr}
2 & 3 & -1 \\
-3 & -2 & 4
\end{array}\right]+\left[\begin{array}{rrr}
-6 & 8 & 2 \\
4 & -2 & -6
\end{array}\right]
\end{aligned}
$
$
\begin{aligned}
& =\left(\begin{array}{rrr}
2-6 & 3+8 & -1+2 \\
-3+4 & -2-2 & 4-6
\end{array}\right) \\
& =\left(\begin{array}{rrr}
-4 & 11 & 1 \\
1 & -4 & -2
\end{array}\right)
\end{aligned}
$
From (1) and (2), $\left(A+2 B^T\right)^T=A^T+2 B$.
$
\begin{aligned}
& 3 A-5 B^T=3\left(\begin{array}{rr}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right]-5\left[\begin{array}{rr}
-3 & 2 \\
4 & -1 \\
1 & -3
\end{array}\right) \\
& =\left[\begin{array}{rr}
6 & -9 \\
9 & -6 \\
-3 & 12
\end{array}\right]-\left[\begin{array}{rr}
-15 & 10 \\
20 & -5 \\
5 & -15
\end{array}\right) \\
& =\left(\begin{array}{rr}
6-(-15) & -9-10 \\
9-20 & -6-(-5) \\
-3-5 & 12-(-15)
\end{array}\right) \\
& =\left(\begin{array}{rr}
21 & -19 \\
-11 & -1 \\
-8 & 27
\end{array}\right) \\
& \therefore\left(3 A -5 B ^{ T }\right)^{ T }=\left[\begin{array}{rrr}
21 & -11 & -8 \\
-19 & -1 & 27
\end{array}\right] \text {. } \\
&
\end{aligned}
$
Also, $3 A ^{ T }-5 B$
$
\begin{aligned}
& =3\left[\begin{array}{rrr}
2 & 3 & -1 \\
-3 & -2 & 4
\end{array}\right]-5\left[\begin{array}{rrr}
-3 & 4 & 1 \\
2 & -1 & -3
\end{array}\right] \\
= & {\left[\begin{array}{rrr}
6 & 9 & -3 \\
-9 & -6 & 12
\end{array}\right]-\left[\begin{array}{rrr}
-15 & 20 & 5 \\
10 & -5 & -15
\end{array}\right] } \\
= & \left(\begin{array}{rrr}
6-(-15) & 9-20 & -3-5 \\
-9-10 & -6-(-5) & 12-(-15)
\end{array}\right] \\
= & \left(\begin{array}{rrr}
21 & -11 & -8 \\
-19 & -1 & 27
\end{array}\right]
\end{aligned}
$
From (1) and (2),
$
\left(3 A -5 B ^{ T }\right)^{ T }=3 A ^{ T }=3 A ^{ T }-5 B .
$

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