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NCERT Exemplar · Class 12 Mathematics Matrices

101 questions · 101 still being checked

EXERCISE 3.3 21–30 (part 3 of 10)

  1. Exercise 21

    Give an example of matrices A,B\displaystyle \mathrm{A}, \mathrm{B} and C such that AB=AC\displaystyle \mathrm{AB}=\mathrm{AC}, where A is non-zero matrix, but B≠C\displaystyle \mathrm{B} \neq \mathrm{C}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Take a non-zero \(\displaystyle A\) that wipes out the second row of whatever it multiplies: \[A=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad B=\begin{bmatrix}1&2\\3&4\end{bmatrix},\quad C=\begin{bmatrix}1&2\\5&6\end{bmatrix} \] \[AB=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}=\begin{bmatrix}1&2\\0&0\end{bmatrix} \] \[AC=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}1&2\\5&6\end{bmatrix}=\begin{bmatrix}1&2\\0&0\end{bmatrix} \] \[AB=AC,\qquad A\neq O,\qquad B\neq C \] Answer: \(\displaystyle A=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ B=\begin{bmatrix}1&2\\3&4\end{bmatrix},\ C=\begin{bmatrix}1&2\\5&6\end{bmatrix}\).

  2. Exercise 22

    If A=[12−21],B=[233−4]\displaystyle \mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right] and C=[10−10]\displaystyle \mathrm{C}=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right], verify :
    (i)
    (AB)C=A(BC)\displaystyle (\mathrm{AB}) \mathrm{C}=\mathrm{A}(\mathrm{BC})
    (ii)
    A(B+C)=AB+AC\displaystyle \mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    (i) \[AB=\begin{bmatrix}1&2\\-2&1\end{bmatrix}\begin{bmatrix}2&3\\3&-4\end{bmatrix}=\begin{bmatrix}2+6&3-8\\-4+3&-6-4\end{bmatrix}=\begin{bmatrix}8&-5\\-1&-10\end{bmatrix} \] \[(AB)C=\begin{bmatrix}8&-5\\-1&-10\end{bmatrix}\begin{bmatrix}1&0\\-1&0\end{bmatrix}=\begin{bmatrix}8+5&0\\-1+10&0\end{bmatrix}=\begin{bmatrix}13&0\\9&0\end{bmatrix} \] \[BC=\begin{bmatrix}2&3\\3&-4\end{bmatrix}\begin{bmatrix}1&0\\-1&0\end{bmatrix}=\begin{bmatrix}2-3&0\\3+4&0\end{bmatrix}=\begin{bmatrix}-1&0\\7&0\end{bmatrix} \] \[A(BC)=\begin{bmatrix}1&2\\-2&1\end{bmatrix}\begin{bmatrix}-1&0\\7&0\end{bmatrix}=\begin{bmatrix}-1+14&0\\2+7&0\end{bmatrix}=\begin{bmatrix}13&0\\9&0\end{bmatrix} \] \[(AB)C=A(BC) \] (ii) \[B+C=\begin{bmatrix}3&3\\2&-4\end{bmatrix} \] \[A(B+C)=\begin{bmatrix}1&2\\-2&1\end{bmatrix}\begin{bmatrix}3&3\\2&-4\end{bmatrix}=\begin{bmatrix}3+4&3-8\\-6+2&-6-4\end{bmatrix}=\begin{bmatrix}7&-5\\-4&-10\end{bmatrix} \] \[AC=\begin{bmatrix}1&2\\-2&1\end{bmatrix}\begin{bmatrix}1&0\\-1&0\end{bmatrix}=\begin{bmatrix}1-2&0\\-2-1&0\end{bmatrix}=\begin{bmatrix}-1&0\\-3&0\end{bmatrix} \] \[AB+AC=\begin{bmatrix}8-1&-5+0\\-1-3&-10+0\end{bmatrix}=\begin{bmatrix}7&-5\\-4&-10\end{bmatrix} \] \[A(B+C)=AB+AC \] Answer: both results are verified.

  3. Exercise 23

    If P=[x000y000z]\displaystyle \mathrm{P}=\left[\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right] and Q=[a000b000c]\displaystyle \mathrm{Q}=\left[\begin{array}{ccc}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right], prove that PQ=[xa000yb000zc]=QP.\mathrm{PQ}=\left[\begin{array}{ccc} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{array}\right]=\mathrm{QP} .

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Each entry of the product is a row of \(\displaystyle P\) times a column of \(\displaystyle Q\); the zeros kill every cross term. \[PQ=\begin{bmatrix}xa+0+0&0+0\cdot b+0&0+0+0\cdot c\\0\cdot a+0+0&0+yb+0&0+0+0\cdot c\\0\cdot a+0+0&0+0\cdot b+0&0+0+zc\end{bmatrix}=\begin{bmatrix}xa&0&0\\0&yb&0\\0&0&zc\end{bmatrix} \] \[QP=\begin{bmatrix}ax+0+0&0+0\cdot y+0&0+0+0\cdot z\\0\cdot x+0+0&0+by+0&0+0+0\cdot z\\0\cdot x+0+0&0+0\cdot y+0&0+0+cz\end{bmatrix}=\begin{bmatrix}ax&0&0\\0&by&0\\0&0&cz\end{bmatrix} \] \[xa=ax,\quad yb=by,\quad zc=cz \quad \text{(real numbers commute)} \] \[PQ=\begin{bmatrix}xa&0&0\\0&yb&0\\0&0&zc\end{bmatrix}=QP \]

  4. Exercise 24

    If : [213][−10−1−110011][10−1]=A\displaystyle \left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=\mathrm{A}, find A.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \mathrm{A}=[-4]\)
    \[\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}=\begin{bmatrix}-2-1+0&0+1+3&-2+0+3\end{bmatrix}=\begin{bmatrix}-3&4&1\end{bmatrix} \] \[A=\begin{bmatrix}-3&4&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=\begin{bmatrix}-3+0-1\end{bmatrix}=\begin{bmatrix}-4\end{bmatrix} \] Answer: \(\displaystyle A=[-4]\).

  5. Exercise 25

    If A=[21],B=[534876]\displaystyle \mathrm{A}=\left[\begin{array}{ll}2 & 1\end{array}\right], \mathrm{B}=\left[\begin{array}{ccc}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right] and C=[−121102]\displaystyle \mathrm{C}=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 2\end{array}\right], verify that A(B+C)=(AB+AC)\displaystyle \mathrm{A}(\mathrm{B}+\mathrm{C})=(\mathrm{AB}+\mathrm{AC}).

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[B + C = \begin{bmatrix} 5-1 & 3+2 & 4+1 \\ 8+1 & 7+0 & 6+2 \end{bmatrix} = \begin{bmatrix} 4 & 5 & 5 \\ 9 & 7 & 8 \end{bmatrix} \] \[A(B+C) = \begin{bmatrix} 2\cdot 4 + 1\cdot 9 & 2\cdot 5 + 1\cdot 7 & 2\cdot 5 + 1\cdot 8 \end{bmatrix} = \begin{bmatrix} 17 & 17 & 18 \end{bmatrix} \] \[AB = \begin{bmatrix} 2\cdot 5 + 1\cdot 8 & 2\cdot 3 + 1\cdot 7 & 2\cdot 4 + 1\cdot 6 \end{bmatrix} = \begin{bmatrix} 18 & 13 & 14 \end{bmatrix} \] \[AC = \begin{bmatrix} 2(-1) + 1\cdot 1 & 2\cdot 2 + 1\cdot 0 & 2\cdot 1 + 1\cdot 2 \end{bmatrix} = \begin{bmatrix} -1 & 4 & 4 \end{bmatrix} \] \[AB + AC = \begin{bmatrix} 18-1 & 13+4 & 14+4 \end{bmatrix} = \begin{bmatrix} 17 & 17 & 18 \end{bmatrix} \] Answer: \(\displaystyle A(B+C) = AB + AC = \begin{bmatrix} 17 & 17 & 18 \end{bmatrix} \)

  6. Exercise 26

    If A=[10−1213011]\displaystyle \mathrm{A}=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right], then verify that A2+A=A(A+I)\displaystyle \mathrm{A}^2+\mathrm{A}=\mathrm{A}(\mathrm{A}+\mathrm{I}), where I is 3×3\displaystyle 3 \times 3 unit matrix.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[A^2 = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{bmatrix} \] \[A^2 + A = \begin{bmatrix} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix} \] \[A + I = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} \] \[A(A+I) = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix} \] Answer: \(\displaystyle A^2 + A = A(A+I) = \begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix} \)

  7. Exercise 27

    If A=[0−1243−4]\displaystyle \mathrm{A}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right] and B=[401326]\displaystyle \mathrm{B}=\left[\begin{array}{cc}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right], then verify that :
    (i)
    (A′)′=A\displaystyle \left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}
    (ii)
    (AB)′=B′A′\displaystyle (\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}
    (iii)
    (k A)′=(k A′)\displaystyle (k \mathrm{~A})^{\prime}=\left(k \mathrm{~A}^{\prime}\right).

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    (i) \[A' = \begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{bmatrix} \] \[(A')' = \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4 \end{bmatrix} = A \] (ii) \[AB = \begin{bmatrix} 0\cdot 4 - 1\cdot 1 + 2\cdot 2 & 0\cdot 0 - 1\cdot 3 + 2\cdot 6 \\ 4\cdot 4 + 3\cdot 1 - 4\cdot 2 & 4\cdot 0 + 3\cdot 3 - 4\cdot 6 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ 11 & -15 \end{bmatrix} \] \[(AB)' = \begin{bmatrix} 3 & 11 \\ 9 & -15 \end{bmatrix} \] \[B'A' = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 6 \end{bmatrix}\begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} 3 & 11 \\ 9 & -15 \end{bmatrix} \] \[(AB)' = B'A' \] (iii) \[kA = \begin{bmatrix} 0 & -k & 2k \\ 4k & 3k & -4k \end{bmatrix} \] \[(kA)' = \begin{bmatrix} 0 & 4k \\ -k & 3k \\ 2k & -4k \end{bmatrix} \] \[kA' = k\begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} 0 & 4k \\ -k & 3k \\ 2k & -4k \end{bmatrix} \] \[(kA)' = kA' \] Answer: all three results hold.

  8. Exercise 28

    If A=[124156],B=[126473]\displaystyle \mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right], then verify that :
    (i)
    (2 A+B)′=2 A′+B′\displaystyle (2 \mathrm{~A}+\mathrm{B})^{\prime}=2 \mathrm{~A}^{\prime}+\mathrm{B}^{\prime}
    (ii)
    (A−B)′=A′−B′\displaystyle (\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    (i) \[2A + B = \begin{bmatrix} 2 & 4 \\ 8 & 2 \\ 10 & 12 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 14 & 6 \\ 17 & 15 \end{bmatrix} \] \[(2A+B)' = \begin{bmatrix} 3 & 14 & 17 \\ 6 & 6 & 15 \end{bmatrix} \] \[2A' + B' = \begin{bmatrix} 2 & 8 & 10 \\ 4 & 2 & 12 \end{bmatrix} + \begin{bmatrix} 1 & 6 & 7 \\ 2 & 4 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 14 & 17 \\ 6 & 6 & 15 \end{bmatrix} \] \[(2A+B)' = 2A' + B' \] (ii) \[A - B = \begin{bmatrix} 0 & 0 \\ -2 & -3 \\ -2 & 3 \end{bmatrix} \] \[(A-B)' = \begin{bmatrix} 0 & -2 & -2 \\ 0 & -3 & 3 \end{bmatrix} \] \[A' - B' = \begin{bmatrix} 1 & 4 & 5 \\ 2 & 1 & 6 \end{bmatrix} - \begin{bmatrix} 1 & 6 & 7 \\ 2 & 4 & 3 \end{bmatrix} = \begin{bmatrix} 0 & -2 & -2 \\ 0 & -3 & 3 \end{bmatrix} \] \[(A-B)' = A' - B' \] Answer: both results hold.

  9. Exercise 29

    Show that A′A\displaystyle \mathrm{A}^{\prime} \mathrm{A} and AA′\displaystyle \mathrm{AA}^{\prime} are both symmetric matrices for any matrix A.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[A_{m\times n} \;\Rightarrow\; A'A \text{ is } n\times n, \quad AA' \text{ is } m\times m \] \[(A'A)' = A'(A')' \quad \text{(reversal law } (XY)' = Y'X') \] \[(A'A)' = A'A \] \[(AA')' = (A')'A' \quad \text{(reversal law)} \] \[(AA')' = AA' \] Answer: each equals its own transpose, so \(\displaystyle A'A \) and \(\displaystyle AA' \) are symmetric.

  10. Exercise 30

    Let A and B be square matrices of the order 3×3\displaystyle 3 \times 3. Is (AB)2=A2 B2\displaystyle (\mathrm{AB})^2=\mathrm{A}^2 \mathrm{~B}^2 ? Give reasons.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    True when \(\displaystyle \mathrm{AB}=\mathrm{BA}\)
    \[(AB)^2 = ABAB \] \[A^2B^2 = AABB \] The two agree when \(\displaystyle AB = BA \), but matrix multiplication is not commutative, so in general they differ. Counter-example: \[A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] \[AB = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \quad (AB)^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] \[A^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \;\Rightarrow\; A^2B^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] \[(AB)^2 \neq A^2B^2 \] Answer: No, not in general; \(\displaystyle (AB)^2 = A^2B^2 \) is guaranteed only when \(\displaystyle AB = BA \).