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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Determinants

BITSAT 2010 · Mathematics · Determinants

If (A= [ array ll 1 & 3 3 & 2 2 & 5 array ]) and (B=[ array cc -1 &...

If (A=\left[133225\begin{array}{ll}1 & 3 \\ 3 & 2 \\ 2 & 5\end{array}\right]) and (B=\left[−1−20531\begin{array}{cc}-1 & -2 \\ 0 & 5 \\ 3 & 1\end{array}\right]) and (\mathrm{A}+\mathrm{B}-\mathrm{D}=0)(zeromatrix),then(0) (zero matrix), then (\mathrm{D}$) matrix will be -

Options

  1. A(\left[023765\begin{array}{ll}0 & 2 \\ 3 & 7 \\ 6 & 5\end{array}$\right])
  2. B(\left[023756\begin{array}{ll}0 & 2 \\ 3 & 7 \\ 5 & 6\end{array}$\right])
  3. C(\left[013756\begin{array}{ll}0 & 1 \\ 3 & 7 \\ 5 & 6\end{array}$\right])
  4. D(\left[0−2−3−7−5−6\begin{array}{cc}0 & -2 \\ -3 & -7 \\ -5 & -6\end{array}$\right])

Correct answer

C. (\left[013756\begin{array}{ll}0 & 1 \\ 3 & 7 \\ 5 & 6\end{array}$\right])

Step-by-step solution

Let (D=\left[abcdef\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right])∴A+B−C=[133225]+[−1−20531]−[abcdef]⇒[1−1−a3−2−b3+0−c2+5−d2+3−e5+1−f]=[000000]\begin{aligned}& \therefore A+B-C=\left[\begin{array}{ll}1 & 3 \\ 3 & 2 \\ 2 & 5\end{array}\right]+\left[\begin{array}{cc}-1 & -2 \\ 0 & 5 \\ 3 & 1\end{array}\right]-\left[\begin{array}{ll}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\\ \mathrm{e}& \mathrm{f}\end{array}\right]\\ & \Rightarrow \left[\begin{array}{ll}1-1-\mathrm{a}& 3-2-\mathrm{b}\\ 3+0-\mathrm{c}& 2+5-\mathrm{d}\\ 2+3-\mathrm{e}& 5+1-\mathrm{f}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0 \\ 0 & 0\end{array}\right]\end{aligned}))−a=0⇒a=0,1−b=0⇒b=1,3−c=0⇒c=3,7−d=0⇒d=7,5−e=0⇒e=5,6−f=0⇒f=6,\begin{array}{ll}-\mathrm{a}=0 \Rightarrow \mathrm{a}=0,1-\mathrm{b}=0 \Rightarrow \mathrm{b}=1, \\ 3-\mathrm{c}=0 \Rightarrow \mathrm{c}=3, & 7-\mathrm{d}=0 \Rightarrow \mathrm{d}=7, \\ 5-\mathrm{e}=0 \Rightarrow \mathrm{e}=5, & 6-\mathrm{f}=0 \Rightarrow \mathrm{f}=6,\end{array}(∴D=(\therefore D=\left[013756\begin{array}{ll}0 & 1 \\ 3 & 7 \\ 5 & 6\end{array}\right])
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