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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Matrices

NDA 2024 · Mathematics · Matrices

Paragraph: Let A=[ array lll 3 & -3 & 4 2 & -3 & 4 0 & -1 & 1 array...

Paragraph: Let \mathrm{A}=\left[3−342−340−11\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]
Question: What is A−1A^{-1} equal to?

Options

  1. A\left[1−10−23−4−23−3\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]
  2. B\left[12−120−132−2−132−32\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & 0 \\ -1 & \frac{3}{2} & -2 \\ -1 & \frac{3}{2} & \frac{-3}{2}\end{array}\right]
  3. C\left[2−20−46−8−46−6\begin{array}{ccc}2 & -2 & 0 \\ -4 & 6 & -8 \\ -4 & 6 & -6\end{array}\right]
  4. D\left[15−150−2535−45−2535−35\begin{array}{ccc}\frac{1}{5} & \frac{-1}{5} & 0 \\ \frac{-2}{5} & \frac{3}{5} & \frac{-4}{5} \\ \frac{-2}{5} & \frac{3}{5} & \frac{-3}{5}\end{array}\right]

Correct answer

A. \left[1−10−23−4−23−3\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]

Step-by-step solution

We have : A=\left[3−342−340−11\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right] So, Determinant ∣A∣=3(−3+4)−2(−3+4)+0=3−2=1∵A(adj⁡A)=∣A∣I=1.I=I.\begin{aligned} & |A|=3(-3+4)-2(-3+4)+0=3-2=1 \\ & \because \quad A(\operatorname{adj} A)=|A| I=1 . I=I . \end{aligned} Now, \operatorname{adj}(\mathrm{A})=\left[1−10−23−4−23−3\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] So, A−1=1∣A∣(Adj⁡(A))=11[1−10−23−4−23−3]⇒A−1=[1−10−23−4−23−3]\begin{aligned} & \mathrm{A}^{-1}=\frac{1}{|\mathrm{ A}|}(\operatorname{Adj}(\mathrm{A}))=\frac{1}{1}\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] \\ & \Rightarrow \mathrm{A}^{-1}=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] \end{aligned}
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