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NCERT Exemplar · Class 12 Mathematics Matrices

101 questions · 101 still being checked

EXERCISE 3.3 11–20 (part 2 of 10)

  1. Exercise 11

    Show that A=[53−1−2]\displaystyle \mathrm{A}=\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right] satisfies the equation A2−3 A−7I=O\displaystyle \mathrm{A}^2-3 \mathrm{~A}-7 \mathrm{I}=\mathrm{O} and hence find A−1\displaystyle \mathrm{A}^{-1}.

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    NCERT’s answer
    \(\displaystyle \mathrm{A}^{-1}=\frac{-1}{7}\left[\begin{array}{cc}-2 & -3 \\ 1 & 5\end{array}\right]\)
    \[A^2=\begin{bmatrix}5&3\\-1&-2\end{bmatrix}\begin{bmatrix}5&3\\-1&-2\end{bmatrix}=\begin{bmatrix}22&9\\-3&1\end{bmatrix} \] \[A^2-3A-7I=\begin{bmatrix}22&9\\-3&1\end{bmatrix}-\begin{bmatrix}15&9\\-3&-6\end{bmatrix}-\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O \] So \(\displaystyle A\) satisfies the equation. Since \(\displaystyle |A|\neq0\), \(\displaystyle A^{-1}\) exists: \[|A|=-10+3=-7 \] Multiply the equation by \(\displaystyle A^{-1}\): \[A^{-1}A^2-3A^{-1}A-7A^{-1}I=O \] \[A-3I-7A^{-1}=O \] \[A^{-1}=\tfrac17(A-3I)=\tfrac17\begin{bmatrix}2&3\\-1&-5\end{bmatrix} \] Answer: \(\displaystyle A^{-1}=\dfrac17\begin{bmatrix}2&3\\-1&-5\end{bmatrix}\)

  2. Exercise 12

    Find the matrix A satisfying the matrix equation: [2132]A[−325−3]=[1001]\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]

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    NCERT’s answer
    \(\displaystyle \mathrm{A}=\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\)
    \[P=\begin{bmatrix}2&1\\3&2\end{bmatrix},\qquad Q=\begin{bmatrix}-3&2\\5&-3\end{bmatrix},\qquad PAQ=I \] \[|P|=4-3=1,\qquad P^{-1}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix} \] \[|Q|=9-10=-1,\qquad Q^{-1}=\begin{bmatrix}3&2\\5&3\end{bmatrix} \] Multiply by \(\displaystyle P^{-1}\) on the left and \(\displaystyle Q^{-1}\) on the right: \[A=P^{-1}IQ^{-1}=P^{-1}Q^{-1} \] \[A=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}\begin{bmatrix}3&2\\5&3\end{bmatrix}=\begin{bmatrix}6-5&4-3\\-9+10&-6+6\end{bmatrix} \] Answer: \(\displaystyle A=\begin{bmatrix}1&1\\1&0\end{bmatrix}\)

  3. Exercise 13

    Find A\displaystyle \mathrm{A}, if [413]A=[−484−121−363]\displaystyle \left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right] \mathrm{A}=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]

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    NCERT’s answer
    \(\displaystyle \mathrm{A}=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]\)
    A \(\displaystyle 3\times1\) matrix times \(\displaystyle A\) gives a \(\displaystyle 3\times3\) matrix, so \(\displaystyle A\) is \(\displaystyle 1\times3\): \[A=\begin{bmatrix}a&b&c\end{bmatrix} \] \[\begin{bmatrix}4\\1\\3\end{bmatrix}\begin{bmatrix}a&b&c\end{bmatrix}=\begin{bmatrix}4a&4b&4c\\a&b&c\\3a&3b&3c\end{bmatrix}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix} \] \[4a=-4,\quad 4b=8,\quad 4c=4 \] \[a=-1,\quad b=2,\quad c=1 \] The other two rows, \(\displaystyle \begin{bmatrix}-1&2&1\end{bmatrix}\) and \(\displaystyle 3\begin{bmatrix}-1&2&1\end{bmatrix}=\begin{bmatrix}-3&6&3\end{bmatrix}\), agree. Answer: \(\displaystyle A=\begin{bmatrix}-1&2&1\end{bmatrix}\)

  4. Exercise 14

    If A=[3−41120]\displaystyle \mathrm{A}=\left[\begin{array}{cc}3 & -4 \\ 1 & 1 \\ 2 & 0\end{array}\right] and B=[212124]\displaystyle \mathrm{B}=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right], then verify (BA)2≠B2A2\displaystyle (\mathrm{BA})^2 \neq \mathrm{B}^2 \mathrm{A}^2

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

    \[BA=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}\begin{bmatrix}3&-4\\1&1\\2&0\end{bmatrix}=\begin{bmatrix}6+1+4&-8+1+0\\3+2+8&-4+2+0\end{bmatrix}=\begin{bmatrix}11&-7\\13&-2\end{bmatrix} \] \[(BA)^2=\begin{bmatrix}11&-7\\13&-2\end{bmatrix}\begin{bmatrix}11&-7\\13&-2\end{bmatrix}=\begin{bmatrix}121-91&-77+14\\143-26&-91+4\end{bmatrix}=\begin{bmatrix}30&-63\\117&-87\end{bmatrix} \] \(\displaystyle B\) is \(\displaystyle 2\times3\): its $\displaystyle 3$ columns do not match its $\displaystyle 2$ rows, so \(\displaystyle B^2\) is not defined. Likewise \(\displaystyle A\) is \(\displaystyle 3\times2\), so \(\displaystyle A^2\) is not defined. Hence \(\displaystyle B^2A^2\) does not exist, while \(\displaystyle (BA)^2\) does. Answer: \(\displaystyle (BA)^2=\begin{bmatrix}30&-63\\117&-87\end{bmatrix}\), but \(\displaystyle B^2A^2\) is not defined, so \(\displaystyle (BA)^2\neq B^2A^2\).

  5. Exercise 15

    If possible, find BA and AB, where A=[212124],B=[412312].\mathrm{A}=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right], \mathrm{B}=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right] .

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    NCERT’s answer
    \(\displaystyle \mathrm{AB}=\left[\begin{array}{cc}12 & 9 \\ 12 & 15\end{array}\right]\) \(\displaystyle \mathrm{BA}=\left[\begin{array}{lll}9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10\end{array}\right]\)
    \(\displaystyle A\) is \(\displaystyle 2\times3\) and \(\displaystyle B\) is \(\displaystyle 3\times2\), so both products exist: \(\displaystyle AB\) is \(\displaystyle 2\times2\), \(\displaystyle BA\) is \(\displaystyle 3\times3\). \[AB=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}=\begin{bmatrix}8+2+2&2+3+4\\4+4+4&1+6+8\end{bmatrix}=\begin{bmatrix}12&9\\12&15\end{bmatrix} \] \[BA=\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}=\begin{bmatrix}8+1&4+2&8+4\\4+3&2+6&4+12\\2+2&1+4&2+8\end{bmatrix}=\begin{bmatrix}9&6&12\\7&8&16\\4&5&10\end{bmatrix} \] Answer: \(\displaystyle AB=\begin{bmatrix}12&9\\12&15\end{bmatrix}\), \(\displaystyle BA=\begin{bmatrix}9&6&12\\7&8&16\\4&5&10\end{bmatrix}\)

  6. Exercise 16

    Show by an example that for A≠O,B≠O,AB=O\displaystyle \mathrm{A} \neq \mathrm{O}, \mathrm{B} \neq \mathrm{O}, \mathrm{AB}=\mathrm{O}.

    Not cross-checked

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

    Take \[A=\begin{bmatrix}1&1\\1&1\end{bmatrix}\neq O,\qquad B=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\neq O \] \[AB=\begin{bmatrix}1-1&-1+1\\1-1&-1+1\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O \] Answer: \(\displaystyle A=\begin{bmatrix}1&1\\1&1\end{bmatrix}\) and \(\displaystyle B=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\) are both non-zero, yet \(\displaystyle AB=O\).

  7. Exercise 17

    Given A=[240396]\displaystyle \mathrm{A}=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right] and B=[142813]\displaystyle \mathrm{B}=\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right]. Is (AB)′=B′A′\displaystyle (\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime} ?

    Not cross-checked

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

    \[AB=\begin{bmatrix}2&4&0\\3&9&6\end{bmatrix}\begin{bmatrix}1&4\\2&8\\1&3\end{bmatrix}=\begin{bmatrix}2+8+0&8+32+0\\3+18+6&12+72+18\end{bmatrix}=\begin{bmatrix}10&40\\27&102\end{bmatrix} \] \[(AB)'=\begin{bmatrix}10&27\\40&102\end{bmatrix} \] \[B'=\begin{bmatrix}1&2&1\\4&8&3\end{bmatrix},\qquad A'=\begin{bmatrix}2&3\\4&9\\0&6\end{bmatrix} \] \[B'A'=\begin{bmatrix}2+8+0&3+18+6\\8+32+0&12+72+18\end{bmatrix}=\begin{bmatrix}10&27\\40&102\end{bmatrix} \] \[(AB)'=B'A' \] Answer: Yes, \(\displaystyle (AB)'=B'A'\).

  8. Exercise 18

    Solve for x\displaystyle x and y\displaystyle y: x[21]+y[35]+[−8−11]=O.x\left[\begin{array}{l} 2 \\ 1 \end{array}\right]+y\left[\begin{array}{l} 3 \\ 5 \end{array}\right]+\left[\begin{array}{c} -8 \\ -11 \end{array}\right]=\mathrm{O} .

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    NCERT’s answer
    \(\displaystyle x=1, y=2\)
    \[\begin{bmatrix}2x+3y-8\\x+5y-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix} \] \[2x+3y=8 \quad (1) \] \[x+5y=11 \quad (2) \] \[(1)-2\times(2):\quad -7y=-14 \] \[y=2 \] \[x=11-5(2)=1 \] \[\text{Check (1): } 2(1)+3(2)=8 \] Answer: \(\displaystyle x=1,\ y=2\).

  9. Exercise 19

    If X and Y are 2×2\displaystyle 2 \times 2 matrices, then solve the following matrix equations for X and Y 2X+3Y=[2340],3X+2Y=[−221−5].2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right], 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right] .

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    NCERT’s answer
    \(\displaystyle \mathrm{X}=\left[\begin{array}{cc}-2 & 0 \\ -1 & -3\end{array}\right], \mathrm{Y}=\left[\begin{array}{ll}2 & 1 \\ 2 & 2\end{array}\right]\)
    \[3(2X+3Y)-2(3X+2Y)=5Y \] \[5Y=3\begin{bmatrix}2&3\\4&0\end{bmatrix}-2\begin{bmatrix}-2&2\\1&-5\end{bmatrix}=\begin{bmatrix}10&5\\10&10\end{bmatrix} \] \[Y=\begin{bmatrix}2&1\\2&2\end{bmatrix} \] \[3(3X+2Y)-2(2X+3Y)=5X \] \[5X=3\begin{bmatrix}-2&2\\1&-5\end{bmatrix}-2\begin{bmatrix}2&3\\4&0\end{bmatrix}=\begin{bmatrix}-10&0\\-5&-15\end{bmatrix} \] \[X=\begin{bmatrix}-2&0\\-1&-3\end{bmatrix} \] \[\text{Check: } 2X+3Y=\begin{bmatrix}-4&0\\-2&-6\end{bmatrix}+\begin{bmatrix}6&3\\6&6\end{bmatrix}=\begin{bmatrix}2&3\\4&0\end{bmatrix} \] Answer: \(\displaystyle X=\begin{bmatrix}-2&0\\-1&-3\end{bmatrix},\ Y=\begin{bmatrix}2&1\\2&2\end{bmatrix}\).

  10. Exercise 20

    If A=[35],B=[73]\displaystyle \mathrm{A}=\left[\begin{array}{ll}3 & 5\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}7 & 3\end{array}\right], then find a non-zero matrix C such that AC=BC\displaystyle \mathrm{AC}=\mathrm{BC}.

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    NCERT’s answer
    \(\displaystyle \left[\begin{array}{c}k \\ 2 k\end{array}\right],\left[\begin{array}{cc}k & k \\ 2 k & 2 k\end{array}\right]\) etc. where \(\displaystyle k\) is a real number
    \[AC=BC \;\Rightarrow\; (A-B)C=O \] \[A-B=\begin{bmatrix}-4&2\end{bmatrix} \] Take \(\displaystyle C\) as a \(\displaystyle 2\times1\) matrix: \[\begin{bmatrix}-4&2\end{bmatrix}\begin{bmatrix}p\\q\end{bmatrix}=0 \] \[-4p+2q=0 \;\Rightarrow\; q=2p \] \[p=1:\quad C=\begin{bmatrix}1\\2\end{bmatrix} \] \[AC=\begin{bmatrix}3&5\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}=[13],\qquad BC=\begin{bmatrix}7&3\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}=[13] \] Answer: \(\displaystyle C=\begin{bmatrix}1\\2\end{bmatrix}\) (any non-zero \(\displaystyle \begin{bmatrix}p\\2p\end{bmatrix}\) works).