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NDA 2 2024 Maths, question 50

Topic: Matrices and determinants

Passage · Q49 to Q50

Let A=[3−342−340−11]A=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}

Q50

What is A−1A^{-1} equal to?

  1. (a) [1−10−23−4−23−3]\begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}Correct answer
  2. (b) [1/2−1/20−13/2−2−13/2−3/2]\begin{bmatrix} 1/2 & -1/2 & 0 \\ -1 & 3/2 & -2 \\ -1 & 3/2 & -3/2 \end{bmatrix}
  3. (c) [2−20−46−8−46−6]\begin{bmatrix} 2 & -2 & 0 \\ -4 & 6 & -8 \\ -4 & 6 & -6 \end{bmatrix}
  4. (d) [1/5−1/50−2/53/5−4/5−2/53/5−3/5]\begin{bmatrix} 1/5 & -1/5 & 0 \\ -2/5 & 3/5 & -4/5 \\ -2/5 & 3/5 & -3/5 \end{bmatrix}

Solution.

Answer: (a) [1−10−23−4−23−3]\begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}

  1. The determinant is 1.
  2. Computing cofactors and transposing gives the matrix in (a): rows (1,−1,0), (−2,3,−4), (−2,3,−3).
  3. Verification by multiplication with A produces rows (1,0,0), (0,1,0), (0,0,1), so it is the inverse.