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NCERT Exemplar · Class 12 Mathematics Matrices

101 questions · 101 still being checked

EXERCISE 3.3 91–101 (part 10 of 10)

  1. State Exercises $\displaystyle 82$ to $\displaystyle 101$ which of the following statements are True or False

    Exercise 91

    If matrix AB=O\displaystyle \mathrm{AB}=\mathrm{O}, then A=O\displaystyle \mathrm{A}=\mathrm{O} or B=O\displaystyle \mathrm{B}=\mathrm{O} or both A and B are null matrices.

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    NCERT’s answer
    False
    False. The product of two non-null matrices can be the null matrix. \[A=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad B=\begin{bmatrix}0&0\\0&1\end{bmatrix} \] \[AB=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O \] \[A \neq O,\quad B \neq O \]

  2. Exercise 92

    Transpose of a column matrix is a column matrix.

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    NCERT’s answer
    False
    False. A column matrix has order \(\displaystyle m\times 1\); its transpose has order \(\displaystyle 1\times m\), a row matrix. \[A=\begin{bmatrix}1\\2\\3\end{bmatrix}_{3\times 1} \;\Rightarrow\; A'=\begin{bmatrix}1&2&3\end{bmatrix}_{1\times 3} \]

  3. Exercise 93

    If A and B are two square matrices of the same order, then AB=BA\displaystyle \mathrm{AB}=\mathrm{BA}.

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    NCERT’s answer
    False
    False. Matrix multiplication is not commutative in general. \[A=\begin{bmatrix}1&2\\0&1\end{bmatrix},\quad B=\begin{bmatrix}1&0\\3&1\end{bmatrix} \] \[AB=\begin{bmatrix}7&2\\3&1\end{bmatrix} \] \[BA=\begin{bmatrix}1&2\\3&7\end{bmatrix} \] \[AB \neq BA \]

  4. Exercise 94

    If each of the three matrices of the same order are symmetric, then their sum is a symmetric matrix.

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    NCERT’s answer
    True
    True. The transpose of a sum is the sum of the transposes. \[A'=A,\quad B'=B,\quad C'=C \quad \text{(symmetric)} \] \[(A+B+C)' = A'+B'+C' \] \[A'+B'+C' = A+B+C \] \[\therefore\; (A+B+C)' = A+B+C \]

  5. Exercise 95

    If A and B are any two matrices of the same order, then (AB)′=A′B′\displaystyle (\mathrm{AB})^{\prime}=\mathrm{A}^{\prime} \mathrm{B}^{\prime}.

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    NCERT’s answer
    False
    False. The correct rule is \(\displaystyle (AB)'=B'A'\), and \(\displaystyle B'A' \neq A'B'\) in general. \[A=\begin{bmatrix}1&2\\0&1\end{bmatrix},\quad B=\begin{bmatrix}1&0\\3&1\end{bmatrix} \] \[(AB)'=\begin{bmatrix}7&2\\3&1\end{bmatrix}'=\begin{bmatrix}7&3\\2&1\end{bmatrix} \] \[A'B'=\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}1&3\\0&1\end{bmatrix}=\begin{bmatrix}1&3\\2&7\end{bmatrix} \] \[(AB)' \neq A'B' \]

  6. Exercise 96

    If (AB)′=B′A′\displaystyle (\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}, where A and B are not square matrices, then number of rows in A is equal to number of columns in B and number of columns in A is equal to number of rows in B.

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    False. \(\displaystyle (AB)'=B'A'\) holds whenever \(\displaystyle AB\) is defined, so it imposes no extra condition. \[A_{m\times n},\; B_{n\times p} \;\Rightarrow\; AB_{\,m\times p},\quad B'A'_{\,p\times m} \] Columns of \(\displaystyle A\) equal rows of \(\displaystyle B\) (\(\displaystyle =n\)), but \(\displaystyle m\) and \(\displaystyle p\) need not be equal. \[A_{2\times 3},\; B_{3\times 4}:\quad m = 2 \neq 4 = p \]

  7. Exercise 97

    If A,B\displaystyle \mathrm{A}, \mathrm{B} and C are square matrices of same order, then AB=AC\displaystyle \mathrm{AB}=\mathrm{AC} always implies that B=C\displaystyle \mathrm{B}=\mathrm{C}.

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    NCERT’s answer
    False
    False — a matrix \(\displaystyle A\) that is not invertible cannot be cancelled. Counterexample:\[A=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad B=\begin{bmatrix}1&2\\3&4\end{bmatrix},\quad C=\begin{bmatrix}1&2\\5&6\end{bmatrix} \]\[AB=\begin{bmatrix}1&2\\0&0\end{bmatrix} \]\[AC=\begin{bmatrix}1&2\\0&0\end{bmatrix}=AB \]\[B\neq C \]

  8. Exercise 98

    AA′\displaystyle \mathrm{AA}^{\prime} is always a symmetric matrix for any matrix A.

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    NCERT’s answer
    True
    True — if \(\displaystyle A\) is \(\displaystyle m\times n\), then \(\displaystyle AA'\) is \(\displaystyle m\times m\), a square matrix.\[(AA')' = (A')'\,A' \quad \text{(reversal law)} \]\[(AA')' = AA' \quad \text{(since } (A')'=A\text{)} \]So \(\displaystyle AA'\) is symmetric.

  9. Exercise 99

    If A=[23−1142]\displaystyle \mathrm{A}=\left[\begin{array}{ccc}2 & 3 & -1 \\ 1 & 4 & 2\end{array}\right] and B=[234521]\displaystyle \mathrm{B}=\left[\begin{array}{cc}2 & 3 \\ 4 & 5 \\ 2 & 1\end{array}\right], then AB and BA are defined and equal.

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    NCERT’s answer
    False
    False — both products exist, but they have different orders.\[A_{2\times 3}\,B_{3\times 2}=(AB)_{2\times 2} \]\[B_{3\times 2}\,A_{2\times 3}=(BA)_{3\times 3} \]\[AB=\begin{bmatrix}4+12-2 & 6+15-1\\ 2+16+4 & 3+20+2\end{bmatrix}=\begin{bmatrix}14&20\\22&25\end{bmatrix} \]A \(\displaystyle 2\times 2\) matrix cannot equal a \(\displaystyle 3\times 3\) matrix.

  10. Exercise 100

    If A is skew symmetric matrix, then A2\displaystyle \mathrm{A}^2 is a symmetric matrix.

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    NCERT’s answer
    True
    True — use the transpose of a product.\[A'=-A \quad \text{(skew symmetric)} \]\[(A^2)'=(AA)'=A'A' \quad \text{(reversal law)} \]\[(A^2)'=(-A)(-A)=A^2 \]So \(\displaystyle A^2\) is symmetric.

  11. Exercise 101

    (AB)−1=A−1.B−1\displaystyle (\mathrm{AB})^{-1}=\mathrm{A}^{-1} . \mathrm{B}^{-1}, where A and B are invertible matrices satisfying commutative property with respect to multiplication.

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    NCERT’s answer
    True
    True — the reversal law, then \(\displaystyle AB=BA\) swaps the order back.\[(AB)^{-1}=B^{-1}A^{-1} \quad \text{(reversal law)} \]\[AB=BA \quad \text{(commutative)} \]\[B^{-1}A^{-1}=(AB)^{-1}=(BA)^{-1}=A^{-1}B^{-1} \]\[(AB)^{-1}=A^{-1}B^{-1} \]