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NCERT Solutions · Class 12 Mathematics Matrices

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EXERCISE 3.3 1–12 (part 4 of 6)

  1. Exercise 1

    Find the transpose of each of the following matrices:
    (i)
    [512−1]\displaystyle \left[\begin{array}{c}5 \\ \frac{1}{2} \\ -1\end{array}\right]
    (ii)
    [1−123]\displaystyle \left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]
    (iii)
    [−15635623−1]\displaystyle \left[\begin{array}{ccc}-1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1\end{array}\right]

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    By definition, if \(\displaystyle A=[a_{ij}]_{m\times n}\) then its transpose is \(\displaystyle A'=[a_{ji}]_{n\times m}\): the \(\displaystyle i\)th row of \(\displaystyle A\) becomes the \(\displaystyle i\)th column of \(\displaystyle A'\), and the order flips from \(\displaystyle m\times n\) to \(\displaystyle n\times m\).
    (i)
    The matrix is of order \(\displaystyle 3\times 1\), so its transpose is of order \(\displaystyle 1\times 3\):
    \[\left[\begin{array}{r}5 \\ \frac{1}{2} \\ -1\end{array}\right]'=\left[\begin{array}{rrr}5 & \frac{1}{2} & -1\end{array}\right]\]
    (ii)
    Interchanging rows and columns (the diagonal entries \(\displaystyle 1\) and \(\displaystyle 3\) stay where they are, the off-diagonal entries swap):
    \[\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]'=\left[\begin{array}{rr}1 & 2 \\ -1 & 3\end{array}\right]\]
    (iii)
    Row \(\displaystyle 1\) \(\displaystyle (-1,\,5,\,6)\) becomes column \(\displaystyle 1\), row \(\displaystyle 2\) \(\displaystyle (\sqrt{3},\,5,\,6)\) becomes column \(\displaystyle 2\), row \(\displaystyle 3\) \(\displaystyle (2,\,3,\,-1)\) becomes column \(\displaystyle 3\):
    \[\left[\begin{array}{rrr}-1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1\end{array}\right]'=\left[\begin{array}{rrr}-1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1\end{array}\right]\]

  2. Exercise 2

    If A=[−123579−211]\displaystyle \mathrm{A}=\left[\begin{array}{rrr}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right] and B=[−41−5120131]\displaystyle \mathrm{B}=\left[\begin{array}{rrr}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right], then verify that
    (i)
    (A+B)′=A′+B′\displaystyle (\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime},
    (ii)
    (A−B)′=A′−B′\displaystyle (\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}

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    Both parts test the property \(\displaystyle (A\pm B)'=A'\pm B'\). Compute each side separately and compare.
    First write down the two transposes:
    \[A'=\left[\begin{array}{rrr}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right],\qquad B'=\left[\begin{array}{rrr}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right]\]
    (i)
    Adding \(\displaystyle A\) and \(\displaystyle B\) entrywise (they are both \(\displaystyle 3\times 3\), so the sum is defined):
    \[A+B=\left[\begin{array}{rrr}-5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2\end{array}\right]\quad\Rightarrow\quad (A+B)'=\left[\begin{array}{rrr}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{array}\right]\]
    On the other side,
    \[A'+B'=\left[\begin{array}{rrr}-1-4 & 5+1 & -2+1 \\ 2+1 & 7+2 & 1+3 \\ 3-5 & 9+0 & 1+1\end{array}\right]=\left[\begin{array}{rrr}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{array}\right]\]
    The two matrices agree entry by entry, so \(\displaystyle (A+B)'=A'+B'\).
    (ii)
    Subtracting entrywise (watch the double negatives: \(\displaystyle -1-(-4)=3\) and \(\displaystyle 3-(-5)=8\)):
    \[A-B=\left[\begin{array}{rrr}3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0\end{array}\right]\quad\Rightarrow\quad (A-B)'=\left[\begin{array}{rrr}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]\]
    And
    \[A'-B'=\left[\begin{array}{rrr}-1+4 & 5-1 & -2-1 \\ 2-1 & 7-2 & 1-3 \\ 3+5 & 9-0 & 1-1\end{array}\right]=\left[\begin{array}{rrr}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]\]
    Hence \(\displaystyle (A-B)'=A'-B'\).

  3. Exercise 3

    If A′=[34−1201]\displaystyle \mathrm{A}^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right] and B=[−121123]\displaystyle \mathrm{B}=\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right], then verify that
    (i)
    (A+B)′=A′+B′\displaystyle (\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}
    (ii)
    (A−B)′=A′−B′\displaystyle (\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}

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    Here \(\displaystyle A'\) is given, not \(\displaystyle A\). Use \(\displaystyle (A')'=A\) first, otherwise the orders will not match: \(\displaystyle A'\) is \(\displaystyle 3\times 2\), so \(\displaystyle A\) is \(\displaystyle 2\times 3\), which is the order of \(\displaystyle B\).
    \[A=(A')'=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right],\qquad B'=\left[\begin{array}{rr}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]\]
    (i)
    \[A+B=\left[\begin{array}{rrr}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right]\quad\Rightarrow\quad (A+B)'=\left[\begin{array}{rr}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]\]
    \[A'+B'=\left[\begin{array}{rr}3-1 & 4+1 \\ -1+2 & 2+2 \\ 0+1 & 1+3\end{array}\right]=\left[\begin{array}{rr}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]\]
    The two sides are equal, so \(\displaystyle (A+B)'=A'+B'\).
    (ii)
    \[A-B=\left[\begin{array}{rrr}3+1 & -1-2 & 0-1 \\ 4-1 & 2-2 & 1-3\end{array}\right]=\left[\begin{array}{rrr}4 & -3 & -1 \\ 3 & 0 & -2\end{array}\right]\quad\Rightarrow\quad (A-B)'=\left[\begin{array}{rr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\]
    \[A'-B'=\left[\begin{array}{rr}3+1 & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3\end{array}\right]=\left[\begin{array}{rr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\]
    Hence \(\displaystyle (A-B)'=A'-B'\).

  4. Exercise 4

    If A′=[−2312]\displaystyle \mathrm{A}^{\prime}=\left[\begin{array}{cc}-2 & 3 \\ 1 & 2\end{array}\right] and B=[−1012]\displaystyle \mathrm{B}=\left[\begin{array}{rr}-1 & 0 \\ 1 & 2\end{array}\right], then find (A+2 B)′\displaystyle (\mathrm{A}+2 \mathrm{~B})^{\prime}

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    NCERT’s answer
    \(\displaystyle \left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]\)
    Since \(\displaystyle A'\) is given, recover \(\displaystyle A\) using \(\displaystyle (A')'=A\): \[A=\left[\begin{array}{rr}-2 & 1 \\ 3 & 2\end{array}\right]\] Scalar multiplication multiplies every entry: \[2B=2\left[\begin{array}{rr}-1 & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{rr}-2 & 0 \\ 2 & 4\end{array}\right]\] Add entrywise: \[A+2B=\left[\begin{array}{rr}-2-2 & 1+0 \\ 3+2 & 2+4\end{array}\right]=\left[\begin{array}{rr}-4 & 1 \\ 5 & 6\end{array}\right]\] Finally transpose: \[(A+2B)'=\left[\begin{array}{rr}-4 & 5 \\ 1 & 6\end{array}\right]\]

  5. Exercise 5

    For the matrices A and B, verify that (AB)′=B′A′\displaystyle (\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}, where
    (i)
    A=[1−43],B=[−121]\displaystyle \mathrm{A}=\left[\begin{array}{r}1 \\ -4 \\ 3\end{array}\right], \mathrm{B}=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]
    (ii)
    A=[012],B=[157]\displaystyle \mathrm{A}=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right], \mathrm{B}=\left[\begin{array}{lll}1 & 5 & 7\end{array}\right]

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    The property to verify is the reversal law for the transpose of a product: \(\displaystyle (AB)'=B'A'\). Note the order reverses, which is also what makes the orders compatible: \(\displaystyle A\) is \(\displaystyle 3\times 1\) and \(\displaystyle B\) is \(\displaystyle 1\times 3\), so \(\displaystyle AB\) is \(\displaystyle 3\times 3\), while \(\displaystyle B'\) is \(\displaystyle 3\times 1\) and \(\displaystyle A'\) is \(\displaystyle 1\times 3\), so \(\displaystyle B'A'\) is also \(\displaystyle 3\times 3\).
    (i)
    Multiplying the column by the row:
    \[AB=\left[\begin{array}{r}1 \\ -4 \\ 3\end{array}\right]\left[\begin{array}{rrr}-1 & 2 & 1\end{array}\right]=\left[\begin{array}{rrr}-1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3\end{array}\right]\]
    \[\Rightarrow\quad (AB)'=\left[\begin{array}{rrr}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{array}\right]\]
    Now \(\displaystyle B'=\left[\begin{array}{r}-1 \\ 2 \\ 1\end{array}\right]\) and \(\displaystyle A'=\left[\begin{array}{rrr}1 & -4 & 3\end{array}\right]\), so
    \[B'A'=\left[\begin{array}{r}-1 \\ 2 \\ 1\end{array}\right]\left[\begin{array}{rrr}1 & -4 & 3\end{array}\right]=\left[\begin{array}{rrr}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{array}\right]\]
    The two agree, so \(\displaystyle (AB)'=B'A'\).
    (ii)
    \[AB=\left[\begin{array}{r}0 \\ 1 \\ 2\end{array}\right]\left[\begin{array}{rrr}1 & 5 & 7\end{array}\right]=\left[\begin{array}{rrr}0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14\end{array}\right]\quad\Rightarrow\quad (AB)'=\left[\begin{array}{rrr}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{array}\right]\]
    \[B'A'=\left[\begin{array}{r}1 \\ 5 \\ 7\end{array}\right]\left[\begin{array}{rrr}0 & 1 & 2\end{array}\right]=\left[\begin{array}{rrr}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{array}\right]\]
    Hence \(\displaystyle (AB)'=B'A'\).

  6. Exercise 6

    If (i) A=[cos⁡αsin⁡α−sin⁡αcos⁡α]\displaystyle \mathrm{A}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right], then verify that A′A=I\displaystyle \mathrm{A}^{\prime} \mathrm{A}=\mathrm{I}
    (ii)
    If A=[sin⁡αcos⁡α−cos⁡αsin⁡α]\displaystyle \mathrm{A}=\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right], then verify that A′A=I\displaystyle \mathrm{A}^{\prime} \mathrm{A}=\mathrm{I}

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    In each part write \(\displaystyle A'\) by interchanging rows and columns, multiply \(\displaystyle A'A\) in that order, and simplify using the identity \(\displaystyle \sin^{2}\alpha+\cos^{2}\alpha=1\).
    (i)
    With \(\displaystyle A=\left[\begin{array}{rr}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{array}\right]\) we get \(\displaystyle A'=\left[\begin{array}{rr}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{array}\right]\). Then
    \[A'A=\left[\begin{array}{rr}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{array}\right]\left[\begin{array}{rr}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{array}\right]=\left[\begin{array}{rr}\cos^{2}\alpha+\sin^{2}\alpha & \cos\alpha\sin\alpha-\sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha-\cos\alpha\sin\alpha & \sin^{2}\alpha+\cos^{2}\alpha\end{array}\right]\]
    The diagonal entries are \(\displaystyle 1\) and the off-diagonal entries cancel to \(\displaystyle 0\), so
    \[A'A=\left[\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right]=\mathrm{I}\]
    (ii)
    With \(\displaystyle A=\left[\begin{array}{rr}\sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha\end{array}\right]\) we get \(\displaystyle A'=\left[\begin{array}{rr}\sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha\end{array}\right]\). Then
    \[A'A=\left[\begin{array}{rr}\sin^{2}\alpha+\cos^{2}\alpha & \sin\alpha\cos\alpha-\cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha-\sin\alpha\cos\alpha & \cos^{2}\alpha+\sin^{2}\alpha\end{array}\right]=\left[\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right]=\mathrm{I}\]
    In both cases \(\displaystyle A'A=\mathrm{I}\), so each matrix is orthogonal.

  7. Exercise 7

    (i)
    Show that the matrix A=[1−15−121513]\displaystyle \mathrm{A}=\left[\begin{array}{rrr}1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3\end{array}\right] is a symmetric matrix.
    (ii)
    Show that the matrix A=[01−1−1011−10]\displaystyle \mathrm{A}=\left[\begin{array}{rrr}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{array}\right] is a skew symmetric matrix.

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    By definition a square matrix \(\displaystyle A\) is symmetric if \(\displaystyle A'=A\), and skew symmetric if \(\displaystyle A'=-A\). So in each part the test is: transpose the matrix and compare with \(\displaystyle A\) (or with \(\displaystyle -A\)).
    (i)
    For \(\displaystyle A=\left[\begin{array}{rrr}1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3\end{array}\right]\), writing the rows as columns gives
    \[A'=\left[\begin{array}{rrr}1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3\end{array}\right]=A\]
    (equivalently \(\displaystyle a_{12}=a_{21}=-1\), \(\displaystyle a_{13}=a_{31}=5\), \(\displaystyle a_{23}=a_{32}=1\)). Since \(\displaystyle A'=A\), the matrix is symmetric.
    (ii)
    For \(\displaystyle A=\left[\begin{array}{rrr}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{array}\right]\),
    \[A'=\left[\begin{array}{rrr}0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0\end{array}\right]=-\left[\begin{array}{rrr}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{array}\right]=-A\]
    Since \(\displaystyle A'=-A\), the matrix is skew symmetric. (Consistently, every diagonal entry is \(\displaystyle 0\), as \(\displaystyle a_{ii}=-a_{ii}\) forces.)

  8. Exercise 8

    For the matrix A=[1567]\displaystyle \mathrm{A}=\left[\begin{array}{cc}1 & 5 \\ 6 & 7\end{array}\right], verify that
    (i)
    (A+A′)\displaystyle \left(\mathrm{A}+\mathrm{A}^{\prime}\right) is a symmetric matrix
    (ii)
    (A−A′)\displaystyle \left(\mathrm{A}-\mathrm{A}^{\prime}\right) is a skew symmetric matrix

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    For \(\displaystyle A=\left[\begin{array}{rr}1 & 5 \\ 6 & 7\end{array}\right]\), \(\displaystyle A'=\left[\begin{array}{rr}1 & 6 \\ 5 & 7\end{array}\right]\).
    (i)
    Adding,
    \[A+A'=\left[\begin{array}{rr}1+1 & 5+6 \\ 6+5 & 7+7\end{array}\right]=\left[\begin{array}{rr}2 & 11 \\ 11 & 14\end{array}\right]\]
    Transposing this result interchanges the two \(\displaystyle 11\)s, which leaves it unchanged:
    \[(A+A')'=\left[\begin{array}{rr}2 & 11 \\ 11 & 14\end{array}\right]=A+A'\]
    Hence \(\displaystyle A+A'\) is symmetric. (In general \(\displaystyle (A+A')'=A'+(A')'=A'+A=A+A'\).)
    (ii)
    Subtracting,
    \[A-A'=\left[\begin{array}{rr}1-1 & 5-6 \\ 6-5 & 7-7\end{array}\right]=\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]\]
    and
    \[(A-A')'=\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]=-\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]=-(A-A')\]
    Hence \(\displaystyle A-A'\) is skew symmetric. (In general \(\displaystyle (A-A')'=A'-A=-(A-A')\).)

  9. Exercise 9

    Find 12( A+A′)\displaystyle \frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right) and 12( A−A′)\displaystyle \frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right), when A=[0ab−a0c−b−c0]\displaystyle \mathrm{A}=\left[\begin{array}{rrr}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]

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    First transpose \(\displaystyle A\): \[A'=\left[\begin{array}{rrr}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right]=-A\] so the given \(\displaystyle A\) is skew symmetric. That single observation settles both parts.Since \(\displaystyle A'=-A\), \[\frac{1}{2}(A+A')=\frac{1}{2}(A-A)=\frac{1}{2}\,O=\left[\begin{array}{rrr}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\] and \[\frac{1}{2}(A-A')=\frac{1}{2}\bigl(A-(-A)\bigr)=\frac{1}{2}(2A)=A=\left[\begin{array}{rrr}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]\] Thus the symmetric part of \(\displaystyle A\) is the null matrix of order \(\displaystyle 3\) and the skew symmetric part is \(\displaystyle A\) itself.

  10. Exercise 10

    Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
    (i)
    [351−1]\displaystyle \left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]
    (ii)
    [6−22−23−12−13]\displaystyle \left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]
    (iii)
    [33−1−2−21−4−52]\displaystyle \left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]
    (iv)
    [15−12]\displaystyle \left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]

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    NCERT’s answer
    (i)
    \(\displaystyle \mathrm{A}=\left[\begin{array}{cc}3 & 3 \\ 3 & -1\end{array}\right]+\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]\) (ii) \(\displaystyle \mathrm{A}=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]+\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\) (iii) \[A=\begin{array}{cccccc} 3 & \frac{1}{2} & \frac{-5}{2} & 0 & \frac{5}{2} & \frac{3}{2} \\ \frac{1}{2} & -2 & -2 & +\frac{-5}{2} & 0 & 3 \\ \frac{-5}{2} & -2 & 2 & \frac{-3}{2} & -3 & 0 \end{array} \] (iv) \(\displaystyle \mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]+\left[\begin{array}{cc}0 & 3 \\ -3 & 0\end{array}\right]\)
    Theorem used: any square matrix \(\displaystyle A\) can be written as
    \[A=\underbrace{\tfrac{1}{2}(A+A')}_{P,\ \text{symmetric}}+\underbrace{\tfrac{1}{2}(A-A')}_{Q,\ \text{skew symmetric}}\]
    because \(\displaystyle P'=\tfrac{1}{2}(A'+A)=P\) and \(\displaystyle Q'=\tfrac{1}{2}(A'-A)=-Q\). So in each part compute \(\displaystyle A'\), then \(\displaystyle P\) and \(\displaystyle Q\). The step most often botched is halving: every entry of \(\displaystyle A+A'\) and of \(\displaystyle A-A'\) gets divided by \(\displaystyle 2\), fractions included.
    (i)
    \(\displaystyle A=\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]\), \(\displaystyle A'=\left[\begin{array}{rr}3 & 1 \\ 5 & -1\end{array}\right]\).
    \[P=\frac{1}{2}\left[\begin{array}{rr}6 & 6 \\ 6 & -2\end{array}\right]=\left[\begin{array}{rr}3 & 3 \\ 3 & -1\end{array}\right],\qquad Q=\frac{1}{2}\left[\begin{array}{rr}0 & 4 \\ -4 & 0\end{array}\right]=\left[\begin{array}{rr}0 & 2 \\ -2 & 0\end{array}\right]\]
    \[\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]=\left[\begin{array}{rr}3 & 3 \\ 3 & -1\end{array}\right]+\left[\begin{array}{rr}0 & 2 \\ -2 & 0\end{array}\right]\]
    (ii)
    \(\displaystyle A=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]\). Here \(\displaystyle A'=A\), i.e. \(\displaystyle A\) is already symmetric, so
    \[P=\frac{1}{2}(A+A)=A,\qquad Q=\frac{1}{2}(A-A)=O\]
    \[\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]+\left[\begin{array}{rrr}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\]
    (iii)
    \(\displaystyle A=\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]\), \(\displaystyle A'=\left[\begin{array}{rrr}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]\).
    \[A+A'=\left[\begin{array}{rrr}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]\quad\Rightarrow\quad P=\left[\begin{array}{rrr}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]\]
    \[A-A'=\left[\begin{array}{rrr}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]\quad\Rightarrow\quad Q=\left[\begin{array}{rrr}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]\]
    and \(\displaystyle P+Q=A\), e.g. \(\displaystyle \frac{1}{2}+\frac{5}{2}=3\) recovers \(\displaystyle a_{12}\).
    (iv)
    \(\displaystyle A=\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]\), \(\displaystyle A'=\left[\begin{array}{rr}1 & -1 \\ 5 & 2\end{array}\right]\).
    \[P=\frac{1}{2}\left[\begin{array}{rr}2 & 4 \\ 4 & 4\end{array}\right]=\left[\begin{array}{rr}1 & 2 \\ 2 & 2\end{array}\right],\qquad Q=\frac{1}{2}\left[\begin{array}{rr}0 & 6 \\ -6 & 0\end{array}\right]=\left[\begin{array}{rr}0 & 3 \\ -3 & 0\end{array}\right]\]
    \[\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]=\left[\begin{array}{rr}1 & 2 \\ 2 & 2\end{array}\right]+\left[\begin{array}{rr}0 & 3 \\ -3 & 0\end{array}\right]\]
  11. Choose the correct answer in the Exercises $\displaystyle 11$ and $\displaystyle 12$ .

    Exercise 11

    If A,B\displaystyle \mathrm{A}, \mathrm{B} are symmetric matrices of same order, then AB−BA\displaystyle \mathrm{AB}-\mathrm{BA} is a (A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix

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    NCERT’s answer
    A
    Given \(\displaystyle A'=A\) and \(\displaystyle B'=B\), test \(\displaystyle AB-BA\) against the definitions by transposing it.Using \(\displaystyle (X-Y)'=X'-Y'\) and the reversal law \(\displaystyle (XY)'=Y'X'\), \[(AB-BA)'=(AB)'-(BA)'=B'A'-A'B'\] Now substitute \(\displaystyle A'=A\), \(\displaystyle B'=B\): \[(AB-BA)'=BA-AB=-(AB-BA)\] Since the transpose equals the negative of the matrix, \(\displaystyle AB-BA\) is skew symmetric. (It need not be the zero matrix: matrix multiplication is not commutative, so \(\displaystyle AB\neq BA\) in general even for symmetric \(\displaystyle A,B\).)Hence the correct option is (A) Skew symmetric matrix.

  12. Exercise 12

    If A=[cos⁡α−sin⁡αsin⁡αcos⁡α]\displaystyle \mathrm{A}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right], and A+A′=I\displaystyle \mathrm{A}+\mathrm{A}^{\prime}=\mathrm{I}, then the value of α\displaystyle \alpha is (A) π6\displaystyle \frac{\pi}{6} (B) π3\displaystyle \frac{\pi}{3} (C) π\displaystyle \pi (D) 3π2\displaystyle \frac{3 \pi}{2}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    B
    With \(\displaystyle A=\left[\begin{array}{rr}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{array}\right]\), interchanging rows and columns gives \[A'=\left[\begin{array}{rr}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{array}\right]\] Adding, the sine terms cancel: \[A+A'=\left[\begin{array}{rr}2\cos\alpha & 0 \\ 0 & 2\cos\alpha\end{array}\right]\] Equating this to \(\displaystyle \mathrm{I}=\left[\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right]\) and comparing corresponding entries: \[2\cos\alpha=1\quad\Rightarrow\quad \cos\alpha=\frac{1}{2}\quad\Rightarrow\quad \alpha=\frac{\pi}{3}\] (Checking the given options: \(\displaystyle \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\), \(\displaystyle \cos\pi=-1\), \(\displaystyle \cos\frac{3\pi}{2}=0\), none of which is \(\displaystyle \frac{1}{2}\).)Hence the correct option is (B) \(\displaystyle \frac{\pi}{3}\).