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NCERT Exemplar · Class 12 Mathematics Matrices

101 questions · 101 still being checked

EXERCISE 3.3 51–60 (part 6 of 10)

  1. Exercise 51

    If possible, using elementary row transformations, find the inverse of the following matrices
    (i)
    [2−13−531−323]\displaystyle \left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]
    (ii)
    [23−3−1−2211−1]\displaystyle \left[\begin{array}{ccc}2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1\end{array}\right]
    (iii)
    [20−1510013]\displaystyle \left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]

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    NCERT’s answer
    (i)
    \(\displaystyle \left[\begin{array}{ccc}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1\end{array}\right]\)
    (ii)
    inverse does not exist
    (iii)
    \(\displaystyle \left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]\)
    (i) \[[A\,|\,I]=\left[\begin{array}{ccc|ccc}2&-1&3&1&0&0\\-5&3&1&0&1&0\\-3&2&3&0&0&1\end{array}\right] \] \[R_1\to-(R_1+R_3):\ \left[\begin{array}{ccc|ccc}1&-1&-6&-1&0&-1\\-5&3&1&0&1&0\\-3&2&3&0&0&1\end{array}\right] \] \[R_2\to R_2+5R_1,\ R_3\to R_3+3R_1:\ \left[\begin{array}{ccc|ccc}1&-1&-6&-1&0&-1\\0&-2&-29&-5&1&-5\\0&-1&-15&-3&0&-2\end{array}\right] \] \[R_2\to R_2-2R_3,\ R_2\leftrightarrow R_3,\ R_2\to-R_2:\ \left[\begin{array}{ccc|ccc}1&-1&-6&-1&0&-1\\0&1&15&3&0&2\\0&0&1&1&1&-1\end{array}\right] \] \[R_1\to R_1+R_2:\ \left[\begin{array}{ccc|ccc}1&0&9&2&0&1\\0&1&15&3&0&2\\0&0&1&1&1&-1\end{array}\right] \] \[R_1\to R_1-9R_3,\ R_2\to R_2-15R_3:\ \left[\begin{array}{ccc|ccc}1&0&0&-7&-9&10\\0&1&0&-12&-15&17\\0&0&1&1&1&-1\end{array}\right] \] \[A^{-1}=\begin{bmatrix}-7&-9&10\\-12&-15&17\\1&1&-1\end{bmatrix} \](ii) \[\left[\begin{array}{ccc|ccc}2&3&-3&1&0&0\\-1&-2&2&0&1&0\\1&1&-1&0&0&1\end{array}\right] \] \[R_1\to R_1+R_2:\ \left[\begin{array}{ccc|ccc}1&1&-1&1&1&0\\-1&-2&2&0&1&0\\1&1&-1&0&0&1\end{array}\right] \] \[R_3\to R_3-R_1:\ \left[\begin{array}{ccc|ccc}1&1&-1&1&1&0\\-1&-2&2&0&1&0\\0&0&0&-1&-1&1\end{array}\right] \] The left half has a zero row, so the inverse does not exist.(iii) \[\left[\begin{array}{ccc|ccc}2&0&-1&1&0&0\\5&1&0&0&1&0\\0&1&3&0&0&1\end{array}\right] \] \[R_2\to R_2-R_3,\ R_2\to R_2-2R_1:\ \left[\begin{array}{ccc|ccc}2&0&-1&1&0&0\\1&0&-1&-2&1&-1\\0&1&3&0&0&1\end{array}\right] \] \[R_1\to R_1-2R_2:\ \left[\begin{array}{ccc|ccc}0&0&1&5&-2&2\\1&0&-1&-2&1&-1\\0&1&3&0&0&1\end{array}\right] \] \[R_2\to R_2+R_1,\ R_3\to R_3-3R_1:\ \left[\begin{array}{ccc|ccc}0&0&1&5&-2&2\\1&0&0&3&-1&1\\0&1&0&-15&6&-5\end{array}\right] \] Reorder rows as \(\displaystyle R_2,R_3,R_1\): \[A^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix} \]Answer: (i) \(\displaystyle \begin{bmatrix}-7&-9&10\\-12&-15&17\\1&1&-1\end{bmatrix}\); (ii) no inverse; (iii) \(\displaystyle \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}\)

  2. Exercise 52

    Express the matrix [2311−12412]\displaystyle \left[\begin{array}{ccc}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2\end{array}\right] as the sum of a symmetric and a skew symmetric matrix.

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    NCERT’s answer
    \(\displaystyle \left[\begin{array}{ccc}2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2\end{array}\right]+\left[\begin{array}{ccc}0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0\end{array}\right]\)
    \[A=P+Q,\quad P=\tfrac12(A+A'),\quad Q=\tfrac12(A-A') \] \[A'=\begin{bmatrix}2&1&4\\3&-1&1\\1&2&2\end{bmatrix} \] \[P=\tfrac12\begin{bmatrix}4&4&5\\4&-2&3\\5&3&4\end{bmatrix}=\begin{bmatrix}2&2&\frac52\\2&-1&\frac32\\\frac52&\frac32&2\end{bmatrix} \] \[Q=\tfrac12\begin{bmatrix}0&2&-3\\-2&0&1\\3&-1&0\end{bmatrix}=\begin{bmatrix}0&1&-\frac32\\-1&0&\frac12\\\frac32&-\frac12&0\end{bmatrix} \] \[P'=P,\quad Q'=-Q \]Answer: \(\displaystyle A=P+Q\) with \(\displaystyle P\) symmetric and \(\displaystyle Q\) skew symmetric as above.
  3. Choose the correct answer from the given four options in each of the Exercises $\displaystyle 53$ to 67.

    Exercise 53

    The matrix P=[004040400]\displaystyle \mathrm{P}=\left[\begin{array}{lll}0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0\end{array}\right] is a
    (A)
    square matrix
    (B)
    diagonal matrix
    (C)
    unit matrix
    (D)
    none

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    NCERT’s answer
    A
    (A) square matrix. The order is \(\displaystyle 3\times3\), so \(\displaystyle P\) is square. Two entries off the main diagonal are non-zero, so it is not diagonal, and hence not a unit matrix. \[p_{13}=p_{31}=4\neq0 \]

  4. Exercise 54

    Total number of possible matrices of order 3×3\displaystyle 3 \times 3 with each entry 2\displaystyle 2 or 0\displaystyle 0 is
    (A)
    9\displaystyle 9 (B) 27\displaystyle 27
    (C)
    81\displaystyle 81
    (D)
    512\displaystyle 512

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    NCERT’s answer
    D
    (D) \(\displaystyle 512\). Each of the $\displaystyle 9$ entries has $\displaystyle 2$ choices. \[2^9=512 \]

  5. Exercise 55

    If [2x+y4x5x−74x]=[77y−13yx+6]\displaystyle \left[\begin{array}{cc}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right], then the value of x+y\displaystyle x+y is
    (A)
    x=3,y=1\displaystyle x=3, y=1
    (B)
    x=2,y=3\displaystyle x=2, y=3
    (C)
    x=2,y=4\displaystyle x=2, y=4
    (D)
    x=3,y=3\displaystyle x=3, y=3

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    NCERT’s answer
    B
    (B) \(\displaystyle x=2,\ y=3\). Equal matrices have equal entries; the \(\displaystyle (2,2)\) entries give \(\displaystyle x\), the \(\displaystyle (2,1)\) entries then give \(\displaystyle y\). \[4x=x+6\ \Rightarrow\ x=2 \] \[5x-7=y\ \Rightarrow\ y=3 \] \[2x+y=7,\quad 4x=7y-13 \ \Rightarrow\ 7=7,\quad 8=8 \] \[x+y=5 \]

  6. Exercise 56

    If A=1π[sin⁡−1(xπ)tan⁡−1(xπ)sin⁡−1(xπ)cot⁡−1(πx)],B=1π[−cos⁡−1(xπ)tan⁡−1(xπ)sin⁡−1(xπ)−tan⁡−1(πx)]\displaystyle \mathrm{A}=\frac{1}{\pi}\left[\begin{array}{cc}\sin ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)\end{array}\right], \mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right], then A - B is equal to
    (A)
    I (B) O (C) 2I
    (D)
    12I\displaystyle \frac{1}{2} \mathrm{I}

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    NCERT’s answer
    D
    (D) \(\displaystyle \tfrac12 I\). The off-diagonal entries cancel; the diagonal entries are complementary inverse-trigonometric pairs. \[A-B=\frac1\pi\begin{bmatrix}\sin^{-1}(x\pi)+\cos^{-1}(x\pi)&0\\0&\cot^{-1}(\pi x)+\tan^{-1}(\pi x)\end{bmatrix} \] \[\sin^{-1}t+\cos^{-1}t=\frac\pi2,\quad \tan^{-1}t+\cot^{-1}t=\frac\pi2 \] \[A-B=\frac1\pi\begin{bmatrix}\frac\pi2&0\\0&\frac\pi2\end{bmatrix}=\frac12I \]

  7. Exercise 57

    If A and B are two matrices of the order 3×m\displaystyle 3 \times m and 3×n\displaystyle 3 \times n, respectively, and m=n\displaystyle m=n, then the order of matrix (5A - 2B) is
    (A)
    m×3\displaystyle m \times 3
    (B)
    3×3\displaystyle 3 \times 3
    (C)
    m×n\displaystyle m \times n
    (D)
    3×n\displaystyle 3 \times n

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    NCERT’s answer
    D
    (D) \(\displaystyle 3 \times n\)Since \(\displaystyle m=n\), both matrices have the same order, so the difference exists and keeps that order.\[5A : 3 \times m, \qquad 2B : 3 \times n \] \[m = n \;\Rightarrow\; 3 \times m = 3 \times n \] \[5A - 2B : 3 \times n \]

  8. Exercise 58

    If A=[0110]\displaystyle \mathrm{A}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right], then A2\displaystyle \mathrm{A}^2 is equal to
    (A)
    [0110]\displaystyle \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]
    (B)
    [1010]\displaystyle \left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right]
    (C)
    [0101]\displaystyle \left[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right]
    (D)
    [1001]\displaystyle \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]

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    NCERT’s answer
    D
    (D) \(\displaystyle \begin{bmatrix}1&0\\0&1\end{bmatrix}\)\[A^2 = \begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix} \] \[= \begin{bmatrix}0\cdot 0+1\cdot 1 & 0\cdot 1+1\cdot 0\\ 1\cdot 0+0\cdot 1 & 1\cdot 1+0\cdot 0\end{bmatrix} \] \[= \begin{bmatrix}1&0\\0&1\end{bmatrix} \]

  9. Exercise 59

    If matrix A=[aij]2×2\displaystyle \mathrm{A}=\left[a_{i j}\right]_{2 \times 2}, where aij=1\displaystyle a_{i j}=1 if i≠j\displaystyle i \neq j =0 if i=j=0 \text { if } i=j then A2\displaystyle \mathrm{A}^2 is equal to
    (A)
    I (B) A (C) 0\displaystyle 0 (D) None of these

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    NCERT’s answer
    A
    (A) \(\displaystyle I\)Off-diagonal entries are \(\displaystyle 1\), diagonal entries are \(\displaystyle 0\).\[A = \begin{bmatrix}0&1\\1&0\end{bmatrix} \] \[A^2 = \begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = I \]

  10. Exercise 60

    The matrix [100020004]\displaystyle \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4\end{array}\right] is a
    (A)
    identity matrix
    (B)
    symmetric matrix
    (C)
    skew symmetric matrix
    (D)
    none of these

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    NCERT’s answer
    B
    (B) symmetric matrixThe matrix is diagonal, so every off-diagonal entry is \(\displaystyle 0\).\[a_{ij} = a_{ji} = 0 \quad (i \neq j) \;\Rightarrow\; A' = A \]It is not the identity since \(\displaystyle a_{22}=2\), and not skew symmetric since \(\displaystyle a_{11}=1 \neq 0\).