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if a bmatrix 1 3 3 2 2 5 bmatrix and b bmatrix 1 2
Question:
If
$A = \begin{bmatrix}1&3\\ 3&2\\ 2&5\end{bmatrix}$
and
$ B = \begin{bmatrix}-1&-2\\ 0&5\\ 3&1\end{bmatrix} $
and
$A + B - D = 0$
(zero matrix), then
$D$
matrix will be -
BITSAT - 2010
BITSAT
Updated On:
Jan 30, 2025
$ \begin{bmatrix} 0 & 2 \\ 3&7 \\ 6 &5\end{bmatrix}$
$ \begin{bmatrix} 0 & 2 \\ 3&7 \\ 5 & 6 \end{bmatrix}$
$ \begin{bmatrix} 0 &1 \\ 3&7 \\ 5 & 6 \end{bmatrix}$
$ \begin{bmatrix} 0 & -2 \\ -3 & -7 \\ -5 & -6 \end{bmatrix}$
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The Correct Option is
C
Approach Solution - 1
Let
$D = \begin{bmatrix}a&b\\ c&d\\ e&f\end{bmatrix}$
$\therefore A + B - C = \begin{bmatrix}1&3\\ 3&2\\ 2&5\end{bmatrix}+ \begin{bmatrix}-1&-2\\ 0&5\\ 3&1\end{bmatrix} - \begin{bmatrix}a&b\\ c&d\\ e&f\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1-1-a&3-2-b\\ 3+0-c&2+5-d\\ 2+3-e&5+1-f\end{bmatrix} =\begin{bmatrix}0&0\\ 0&0\\ 0&0\end{bmatrix}$
$-a=0 \Rightarrow a=0,1-b=0$
$\Rightarrow b=1$
$3-c=0 \Rightarrow c=3,7-d=0$
$\Rightarrow d=7$
,
$5-e=0 \Rightarrow e=5,6-f=0$
$\Rightarrow f=6$
,
$\therefore D = \begin{bmatrix}0&1\\ 3&7\\ 5&6\end{bmatrix}$
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Approach Solution -2
The correct answer is:D
Given that;
\(A=\begin{vmatrix}1&3\\3&2\\2&5\end{vmatrix},B=\begin{vmatrix}-1&-2&\\0&5\\3&1\end{vmatrix}\)
\( A+B-D=0\)
⇒
\(D=A+B\)
\(\therefore D=\begin{vmatrix}1&3\\3&2\\2&5\end{vmatrix}+\begin{vmatrix}-1&-2\\0&5\\3&1\end{vmatrix}\)
\(=\begin{vmatrix}0&1\\3&7\\5&6\end{vmatrix}\)
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