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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 $A=\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 1 \\ 4 & -1 \\ -3 & 3\end{array}\right], C=\left[\begin{array}{cc}1 & 2 \\ -1 & 4 \\ -2 & 3\end{array}\right]$, then show that
(i) $(A+B)^{\top}=A^{\top}+B^{\top}$
(ii) $(A-C)^{\top}=A^{\top}-C^{\top}$
✓

Answer

(i)
$
\begin{aligned}
& A+B=\left(\begin{array}{rr}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right)+\left[\begin{array}{rr}
2 & 1 \\
4 & -1 \\
-3 & 3
\end{array}\right) \\
& {\left[\begin{array}{rr}
2+2 & -3+1 \\
5+4 & -4-1 \\
-6-3 & 1+3
\end{array}\right]=\left(\begin{array}{rr}
4 & -2 \\
9 & -5 \\
-9 & 4
\end{array}\right)} \\
& \therefore( A + B )^{ T }=\left(\begin{array}{rrr}
& 9 \\
-2 & -5 & -9 \\
-2 & 4
\end{array}\right] \\
& A ^{ T }=\left[\begin{array}{rrr}
2 & 5 & -6 \\
-3 & -4 & 1
\end{array}\right], B ^{ T }=\left(\begin{array}{rrr}
2 & 4 & -3 \\
1 & -1 & 3
\end{array}\right] \\
& \therefore A ^{ T }+ B ^{ T }=\left[\begin{array}{rrr}
2 & 5 & -6 \\
-3 & -4 & 1
\end{array}\right]+\left(\begin{array}{rrr}
2 & 4 & -3 \\
1 & -1 & 3
\end{array}\right) \\
& =\left(\begin{array}{rrr}
2+2 & 5+4 & -6-3 \\
-3+1 & -4-1 & 1+3
\end{array}\right) \\
& =\left(\begin{array}{rrr}
4 & 9 & -9 \\
-2 & -5 & 4
\end{array}\right) \\
&
\end{aligned}
$
From (1) and (2),
$
( A + B )^{ T }= A ^{ T }+ B ^{ T } \text {. }
$
(ii) $A - C =\left(\begin{array}{rr}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right]-\left[\begin{array}{rr}1 & 2 \\ -1 & 4 \\ -2 & 3\end{array}\right]$
$
\begin{aligned}
&=\left[\begin{array}{rr}
2-1 & -3-2 \\
5-(-1) & -4-4 \\
-6-(-2) & 1-3
\end{array}\right]=\left[\begin{array}{rr}
1 & -5 \\
6 & -8 \\
-4 & -2
\end{array}\right] \\
& \therefore(A-C)^{ T }=\left[\begin{array}{rrr}
1 & 6 & -4 \\
-5 & -8 & -2
\end{array}\right] \\
& A ^{ T }=\left[\begin{array}{rrr}
2 & 5 & -6 \\
-3 & -4 & 1
\end{array}\right], C ^{ T }=\left[\begin{array}{rrr}
1 & -1 & -2 \\
2 & 4 & 3
\end{array}\right] \\
& \therefore A ^{ T }- C ^{ T }=\left[\begin{array}{rrr}
2 & 5 & -6 \\
-3 & -4 & 1
\end{array}\right]-\left(\begin{array}{rrr}
1 & -1 & -2 \\
2 & 4 & 3
\end{array}\right] \\
&=\left[\begin{array}{rrr}
2-1 & 5-(-1) & -6-(-2) \\
-3-2 & -4-4 & 1-3
\end{array}\right] \\
&=\left[\begin{array}{rrr}
1 & 6 & -4 \\
-5 & -8 & -2
\end{array}\right]
\end{aligned}
$
From (1) and (2),
$
(A-C)^T=A^T-C^T .
$

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