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NCERT Exemplar · Class 12 Mathematics Matrices

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EXERCISE 3.3 61–70 (part 7 of 10)

  1. Choose the correct answer from the given four options in each of the Exercises $\displaystyle 53$ to 67.

    Exercise 61

    The matrix [0−585012−8−120]\displaystyle \left[\begin{array}{ccc}0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0\end{array}\right] is a
    (A)
    diagonal matrix
    (B)
    symmetric matrix
    (C)
    skew symmetric matrix
    (D)
    scalar matrix

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    NCERT’s answer
    C
    (C) skew symmetric matrix\[A = \begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix} \] \[A' = \begin{bmatrix}0&5&-8\\-5&0&-12\\8&12&0\end{bmatrix} \] \[-A = \begin{bmatrix}0&5&-8\\-5&0&-12\\8&12&0\end{bmatrix} = A' \]Hence \(\displaystyle A' = -A\).

  2. Exercise 62

    If A is matrix of order m×n\displaystyle m \times n and B is a matrix such that AB' and B'A are both defined, then order of matrix B is
    (A)
    m×m\displaystyle m \times m
    (B)
    n×n\displaystyle n \times n
    (C)
    n×m\displaystyle n \times m
    (D)
    m×n\displaystyle m \times n

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    NCERT’s answer
    D
    (D) \(\displaystyle m \times n\)\[A : m \times n, \quad \text{let } B' : p \times q \] \[AB' \text{ defined} \;\Rightarrow\; p = n \] \[B'A \text{ defined} \;\Rightarrow\; q = m \] \[B' : n \times m \;\Rightarrow\; B : m \times n \]

  3. Exercise 63

    If A and B are matrices of same order, then (AB′−BA′)\displaystyle \left(\mathrm{AB}^{\prime}-\mathrm{BA}^{\prime}\right) is a
    (A)
    skew symmetric matrix
    (B)
    null matrix
    (C)
    symmetric matrix
    (D)
    unit matrix

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    NCERT’s answer
    A
    (A) skew symmetric matrixTake the transpose of \(\displaystyle AB' - BA'\).\[(AB' - BA')' = (AB')' - (BA')' \] \[(AB')' = (B')'A' = BA', \qquad (BA')' = (A')'B' = AB' \] \[(AB' - BA')' = BA' - AB' = -(AB' - BA') \]

  4. Exercise 64

    If A is a square matrix such that A2=I\displaystyle \mathrm{A}^2=\mathrm{I}, then (A−I)3+(A+I)3−7A\displaystyle (\mathrm{A}-\mathrm{I})^3+(\mathrm{A}+\mathrm{I})^3-7 \mathrm{A} is equal to
    (A)
    A (B) I−A\displaystyle \mathrm{I}-\mathrm{A}
    (C)
    I+A\displaystyle \mathrm{I}+\mathrm{A}
    (D)
    3A

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    NCERT’s answer
    A
    (A) \(\displaystyle A\)Use \(\displaystyle A^2 = I\), so \(\displaystyle A^3 = A\cdot A^2 = A\).\[(A-I)^3 = A^3 - 3A^2 + 3A - I = A - 3I + 3A - I = 4A - 4I \] \[(A+I)^3 = A^3 + 3A^2 + 3A + I = A + 3I + 3A + I = 4A + 4I \] \[(A-I)^3 + (A+I)^3 - 7A = 8A - 7A = A \]

  5. Exercise 65

    For any two matrices A and B, we have
    (A)
    AB=BA\displaystyle \mathrm{AB}=\mathrm{BA}
    (B)
    AB≠BA\displaystyle \mathrm{AB} \neq \mathrm{BA}
    (C)
    AB=O\displaystyle \mathrm{AB}=\mathrm{O}
    (D)
    None of the above

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    NCERT’s answer
    D
    (D) None of the above.Each of (A), (B), (C) fails for some pair of matrices.\[\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&1\\0&0\end{bmatrix} \] \[\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}=O \] \[AB\neq BA \quad \Rightarrow \quad \text{(A) false} \] \[A=B=I:\quad AB=I=BA,\quad AB\neq O \quad \Rightarrow \quad \text{(B), (C) false} \]

  6. Exercise 66

    On using elementary column operations C2→C2−2C1\displaystyle \mathrm{C}_2 \rightarrow \mathrm{C}_2-2 \mathrm{C}_1 in the following matrix equation [1−324]=[1−101][3124]\displaystyle \left[\begin{array}{cc}1 & -3 \\ 2 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}3 & 1 \\ 2 & 4\end{array}\right], we have :
    (A)
    [1−504]=[1−1−22][3−520]\displaystyle \left[\begin{array}{cc}1 & -5 \\ 0 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]
    (B)
    [1−504]=[1−101][3−5−02]\displaystyle \left[\begin{array}{cc}1 & -5 \\ 0 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -0 & 2\end{array}\right]
    (C)
    [1−520]=[1−301][31−24]\displaystyle \left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -2 & 4\end{array}\right]
    (D)
    [1−520]=[1−101][3−520]\displaystyle \left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]

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    NCERT’s answer
    D
    (D)Apply \(\displaystyle C_2\to C_2-2C_1\) to the left side and to the second matrix on the right; the first factor stays unchanged.\[\begin{bmatrix}1&-3\\2&4\end{bmatrix}\to\begin{bmatrix}1&-3-2(1)\\2&4-2(2)\end{bmatrix}=\begin{bmatrix}1&-5\\2&0\end{bmatrix} \] \[\begin{bmatrix}3&1\\2&4\end{bmatrix}\to\begin{bmatrix}3&1-2(3)\\2&4-2(2)\end{bmatrix}=\begin{bmatrix}3&-5\\2&0\end{bmatrix} \] \[\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix} \]

  7. Exercise 67

    On using elementary row operation R1→R1−3R2\displaystyle \mathrm{R}_1 \rightarrow \mathrm{R}_1-3 \mathrm{R}_2 in the following matrix equation: [4233]=[1203][2011]\displaystyle \left[\begin{array}{ll}4 & 2 \\ 3 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right], we have :
    (A)
    [−5−733]=[1−703][2011]\displaystyle \left[\begin{array}{cc}-5 & -7 \\ 3 & 3\end{array}\right]=\left[\begin{array}{cc}1 & -7 \\ 0 & 3\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & 1\end{array}\right]
    (B)
    [−5−733]=[1203][−1−311]\displaystyle \left[\begin{array}{cc}-5 & -7 \\ 3 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{cc}-1 & -3 \\ 1 & 1\end{array}\right]
    (C)
    [−5−733]=[121−7][2011]\displaystyle \left[\begin{array}{cc}-5 & -7 \\ 3 & 3\end{array}\right]=\left[\begin{array}{cc}1 & 2 \\ 1 & -7\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]
    (D)
    [42−5−7]=[12−3−3][2011]\displaystyle \left[\begin{array}{cc}4 & 2 \\ -5 & -7\end{array}\right]=\left[\begin{array}{cc}1 & 2 \\ -3 & -3\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & 1\end{array}\right]

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    NCERT’s answer
    A
    (A)Apply \(\displaystyle R_1\to R_1-3R_2\) to the left side and to the first matrix on the right; the second factor stays unchanged.\[\begin{bmatrix}4&2\\3&3\end{bmatrix}\to\begin{bmatrix}4-3(3)&2-3(3)\\3&3\end{bmatrix}=\begin{bmatrix}-5&-7\\3&3\end{bmatrix} \] \[\begin{bmatrix}1&2\\0&3\end{bmatrix}\to\begin{bmatrix}1-3(0)&2-3(3)\\0&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix} \] \[\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix} \]
  8. Fill in the blanks in each of the Exercises $\displaystyle 68$–81.

    Exercise 68

    ____ matrix is both symmetric and skew symmetric matrix.

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    NCERT’s answer
    Null matrix
    Zero (null)For a matrix \(\displaystyle A\) that is both symmetric and skew symmetric:\[A'=A \quad \text{and} \quad A'=-A \] \[A=-A \;\Rightarrow\; 2A=O \;\Rightarrow\; A=O \]

  9. Exercise 69

    Sum of two skew symmetric matrices is always ____ matrix.

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    NCERT’s answer
    Skew symmetric matrix
    Skew symmetric\[A'=-A,\qquad B'=-B \] \[(A+B)'=A'+B'=-A-B=-(A+B) \]

  10. Exercise 70

    The negative of a matrix is obtained by multiplying it by ____.

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    NCERT’s answer
    \(\displaystyle -1\)
    \(\displaystyle -1\)\[(-1)\,[a_{ij}]=[-a_{ij}]=-A \]