ICSE Class 10 Mathematics Question 18 of 49

Multiple-Choice Questions — Question 18

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Question 18

Matrix A = [694k] such that A2=[0000]\begin{bmatrix*}[r] 6 & 9 \\ -4 & k \end{bmatrix*} \text{ such that } A^2 = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}. Then k is :

  1. 6

  2. -6

  3. 36

  4. ±6

Answer

Given, A2=[0000]A^2 = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}.

[694k][694k]=[0000][6×6+9×(4)6×9+9×k4×6+k×(4)(4)×9+k×k]=[0000][36+(36)54+9k244k36+k2]=[0000][054+9k244k36+k2]=[0000]54+9k=09k=54k=549=6.\therefore \begin{bmatrix*}[r] 6 & 9 \\ -4 & k \end{bmatrix*}\begin{bmatrix*}[r] 6 & 9 \\ -4 & k \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 \times 6 + 9 \times (-4) & 6 \times 9 + 9 \times k \\ -4 \times 6 + k \times (-4) & (-4) \times 9 + k \times k \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 36 + (-36) & 54 + 9k \\ -24 - 4k & -36 + k^2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & 54 + 9k \\ -24 - 4k & -36 + k^2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow 54 + 9k = 0 \\[1em] \Rightarrow 9k = -54 \\[1em] \Rightarrow k = -\dfrac{54}{9} = -6.

Hence, Option 2 is the correct option.