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NCERT Exemplar · Class 12 Mathematics Matrices

101 questions · 101 still being checked

EXERCISE 3.3 1–10 (part 1 of 10)

  1. Exercise 1

    If a matrix has 28\displaystyle 28 elements, what are the possible orders it can have? What if it has 13\displaystyle 13 elements?

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    NCERT’s answer
    \(\displaystyle 28 \times 1, 1 \times 28, 4 \times 7, 7 \times 4, 14 \times 2, 2 \times 14\). If matrix has $\displaystyle 13$ elements then its order will be either \(\displaystyle 13 \times 1\) or \(\displaystyle 1 \times 13\).
    An \(\displaystyle m \times n\) matrix has \(\displaystyle mn\) elements, so the order comes from the factor pairs of the count.\[28 = 1\cdot 28 = 2\cdot 14 = 4\cdot 7 = 7\cdot 4 = 14\cdot 2 = 28\cdot 1 \]\[13 = 1\cdot 13 = 13\cdot 1 \quad (13 \text{ is prime}) \]Answer: $\displaystyle 28$ elements: \(\displaystyle 1\times 28,\ 2\times 14,\ 4\times 7,\ 7\times 4,\ 14\times 2,\ 28\times 1\). $\displaystyle 13$ elements: \(\displaystyle 1\times 13,\ 13\times 1\).

  2. Exercise 2

    In the matrix A=[a1x23x2−y05−25]\displaystyle \mathrm{A}=\left[\begin{array}{ccc}a & 1 & x \\ 2 & \sqrt{3} & x^2-y \\ 0 & 5 & \frac{-2}{5}\end{array}\right], write :
    (i)
    The order of the matrix A
    (ii)
    The number of elements
    (iii)
    Write elements a23,a31,a12\displaystyle a_{23}, a_{31}, a_{12}

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    NCERT’s answer
    (i)
    \(\displaystyle 3 \times 3\)
    (ii)
    $\displaystyle 9$
    (iii)
    \(\displaystyle a_{23}=x^2-y, a_{31}=0, a_{12}=1\)
    (i)
    \[\text{order} = 3\times 3 \quad \text{(3 rows, 3 columns)} \]
    (ii)
    \[3\times 3 = 9 \quad \text{(rows times columns)} \]
    (iii)
    \[a_{23} = x^2 - y \quad \text{(row 2, column 3)} \]
    \[a_{31} = 0 \quad \text{(row 3, column 1)} \]
    \[a_{12} = 1 \quad \text{(row 1, column 2)} \]
    Answer: (i) \(\displaystyle 3\times 3\); (ii) \(\displaystyle 9\); (iii) \(\displaystyle a_{23}=x^2-y,\ a_{31}=0,\ a_{12}=1\).

  3. Exercise 3

    Construct a2×2\displaystyle a_{2 \times 2} matrix where
    (i)
    aij=(i−2j)22\displaystyle a_{i j}=\frac{(i-2 j)^2}{2}
    (ii)
    aij=∣−2i+3j∣\displaystyle a_{i j}=|-2 i+3 j|

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    A \(\displaystyle 2\times 2\) matrix has \(\displaystyle i, j = 1, 2\).
    (i)
    \[a_{11} = \frac{(1-2)^2}{2} = \frac{1}{2}, \qquad a_{12} = \frac{(1-4)^2}{2} = \frac{9}{2} \]
    \[a_{21} = \frac{(2-2)^2}{2} = 0, \qquad a_{22} = \frac{(2-4)^2}{2} = 2 \]
    \[A = \begin{bmatrix} \tfrac{1}{2} & \tfrac{9}{2} \\ 0 & 2 \end{bmatrix} \]
    (ii)
    \[a_{11} = |-2+3| = 1, \qquad a_{12} = |-2+6| = 4 \]
    \[a_{21} = |-4+3| = 1, \qquad a_{22} = |-4+6| = 2 \]
    \[A = \begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix} \]
    Answer: (i) \(\displaystyle \begin{bmatrix} \tfrac12 & \tfrac92 \\ 0 & 2 \end{bmatrix}\); (ii) \(\displaystyle \begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix}\).

  4. Exercise 4

    Construct a 3×2\displaystyle 3 \times 2 matrix whose elements are given by aij=ei.xsin⁡jx\displaystyle a_{i j}=e^{i.x} \sin j x

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    NCERT’s answer
    \(\displaystyle \left[\begin{array}{cc}e^x \sin x & e^x \sin 2 x \\ e^{2 x} \sin x & e^{2 x} \sin 2 x \\ e^{3 x} \sin x & e^{3 x} \sin 2 x\end{array}\right]\)
    Here \(\displaystyle i = 1,2,3\) (rows) and \(\displaystyle j = 1,2\) (columns).\[a_{11} = e^{x}\sin x, \qquad a_{12} = e^{x}\sin 2x \]\[a_{21} = e^{2x}\sin x, \qquad a_{22} = e^{2x}\sin 2x \]\[a_{31} = e^{3x}\sin x, \qquad a_{32} = e^{3x}\sin 2x \]\[A = \begin{bmatrix} e^{x}\sin x & e^{x}\sin 2x \\ e^{2x}\sin x & e^{2x}\sin 2x \\ e^{3x}\sin x & e^{3x}\sin 2x \end{bmatrix} \]Answer: \(\displaystyle A = \begin{bmatrix} e^{x}\sin x & e^{x}\sin 2x \\ e^{2x}\sin x & e^{2x}\sin 2x \\ e^{3x}\sin x & e^{3x}\sin 2x \end{bmatrix}\).

  5. Exercise 5

    Find values of a\displaystyle a and b\displaystyle b if A=B\displaystyle \mathrm{A}=\mathrm{B}, where A=[a+43b8−6],B=[2a+2b2+28b2−5b]\mathrm{A}=\left[\begin{array}{cc} a+4 & 3 b \\ 8 & -6 \end{array}\right], \quad \mathrm{B}=\left[\begin{array}{cc} 2 a+2 & b^2+2 \\ 8 & b^2-5 b \end{array}\right]

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    NCERT’s answer
    \(\displaystyle a=2, b=2\)
    Equal matrices have equal corresponding entries.\[a + 4 = 2a + 2 \Rightarrow a = 2 \]\[3b = b^2 + 2 \Rightarrow b^2 - 3b + 2 = 0 \Rightarrow (b-1)(b-2) = 0 \Rightarrow b = 1 \text{ or } 2 \]\[-6 = b^2 - 5b \Rightarrow b^2 - 5b + 6 = 0 \Rightarrow (b-2)(b-3) = 0 \Rightarrow b = 2 \text{ or } 3 \]Only \(\displaystyle b = 2\) satisfies both. The entry \(\displaystyle 8 = 8\) already matches.Answer: \(\displaystyle a = 2,\ b = 2\).

  6. Exercise 6

    If possible, find the sum of the matrices A and B, where A=[3123]\displaystyle \mathrm{A}=\left[\begin{array}{cc}\sqrt{3} & 1 \\ 2 & 3\end{array}\right], and B=[xyzab6]\displaystyle \mathrm{B}=\left[\begin{array}{lll}x & y & z \\ a & b & 6\end{array}\right]

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

    NCERT’s answer
    Not possible
    Matrices can be added only when their orders are equal.\[A \text{ is } 2\times 2, \qquad B \text{ is } 2\times 3 \]\[2\times 2 \neq 2\times 3 \]Answer: \(\displaystyle A + B\) is not defined, as the orders differ.

  7. Exercise 7

    If X=[31−15−2−3]\displaystyle \mathrm{X}=\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right] and Y=[21−1724]\displaystyle \mathrm{Y}=\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right], find
    (i)
    X+Y\displaystyle \mathrm{X}+\mathrm{Y}
    (ii)
    2X−3Y\displaystyle 2 \mathrm{X}-3 \mathrm{Y}
    (iii)
    A matrix Z such that X+Y+Z\displaystyle \mathrm{X}+\mathrm{Y}+\mathrm{Z} is a zero matrix.

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

    NCERT’s answer
    (i)
    \(\displaystyle \mathrm{X}+\mathrm{Y}=\left[\begin{array}{ccc}5 & 2 & -2 \\ 12 & 0 & 1\end{array}\right]\)
    (ii)
    \(\displaystyle 2 \mathrm{X}-3 \mathrm{Y}=\left[\begin{array}{ccc}0 & -1 & 1 \\ -11 & -10 & -18\end{array}\right]\)
    (iii)
    \(\displaystyle \mathrm{Z}=\left[\begin{array}{ccc}-5 & -2 & 2 \\ -12 & 0 & -1\end{array}\right]\)
    (i)
    \[X + Y = \begin{bmatrix} 3+2 & 1+1 & -1-1 \\ 5+7 & -2+2 & -3+4 \end{bmatrix} = \begin{bmatrix} 5 & 2 & -2 \\ 12 & 0 & 1 \end{bmatrix} \]
    (ii)
    \[2X = \begin{bmatrix} 6 & 2 & -2 \\ 10 & -4 & -6 \end{bmatrix}, \qquad 3Y = \begin{bmatrix} 6 & 3 & -3 \\ 21 & 6 & 12 \end{bmatrix} \]
    \[2X - 3Y = \begin{bmatrix} 6-6 & 2-3 & -2+3 \\ 10-21 & -4-6 & -6-12 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 1 \\ -11 & -10 & -18 \end{bmatrix} \]
    (iii)
    \[X + Y + Z = O \Rightarrow Z = -(X+Y) \]
    \[Z = \begin{bmatrix} -5 & -2 & 2 \\ -12 & 0 & -1 \end{bmatrix} \]
    Answer: (i) \(\displaystyle \begin{bmatrix} 5 & 2 & -2 \\ 12 & 0 & 1 \end{bmatrix}\); (ii) \(\displaystyle \begin{bmatrix} 0 & -1 & 1 \\ -11 & -10 & -18 \end{bmatrix}\); (iii) \(\displaystyle \begin{bmatrix} -5 & -2 & 2 \\ -12 & 0 & -1 \end{bmatrix}\).

  8. Exercise 8

    Find non-zero values of x\displaystyle x satisfying the matrix equation: x[2x23x]+2[85x44x]=2[(x2+8)24(10)6x].x\left[\begin{array}{cc} 2 x & 2 \\ 3 & x \end{array}\right]+2\left[\begin{array}{cc} 8 & 5 x \\ 4 & 4 x \end{array}\right]=2\left[\begin{array}{cc} \left(x^2+8\right) & 24 \\ (10) & 6 x \end{array}\right] .

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    NCERT’s answer
    \(\displaystyle x=4\)
    Multiply the scalars into each matrix, then add.\[\begin{bmatrix} 2x^2 & 2x \\ 3x & x^2 \end{bmatrix} + \begin{bmatrix} 16 & 10x \\ 8 & 8x \end{bmatrix} = \begin{bmatrix} 2x^2+16 & 48 \\ 20 & 12x \end{bmatrix} \]\[\begin{bmatrix} 2x^2+16 & 12x \\ 3x+8 & x^2+8x \end{bmatrix} = \begin{bmatrix} 2x^2+16 & 48 \\ 20 & 12x \end{bmatrix} \]Equate corresponding entries.\[12x = 48 \Rightarrow x = 4 \]\[3x + 8 = 20 \Rightarrow x = 4 \]\[x^2 + 8x = 12x \Rightarrow x(x-4) = 0 \Rightarrow x = 0 \text{ or } 4 \]The non-zero value common to all is \(\displaystyle x = 4\).Answer: \(\displaystyle x = 4\).

  9. Exercise 9

    If A=[0111]\displaystyle \mathrm{A}=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right] and B=[0−110]\displaystyle \mathrm{B}=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right], show that (A+B)(A−B)≠A2−B2\displaystyle (\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B}) \neq \mathrm{A}^2-\mathrm{B}^2.

    Not cross-checked

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

    \[A+B=\begin{bmatrix}0&0\\2&1\end{bmatrix},\qquad A-B=\begin{bmatrix}0&2\\0&1\end{bmatrix} \] \[(A+B)(A-B)=\begin{bmatrix}0&0\\2&1\end{bmatrix}\begin{bmatrix}0&2\\0&1\end{bmatrix}=\begin{bmatrix}0&0\\0&5\end{bmatrix} \] \[A^2=\begin{bmatrix}0&1\\1&1\end{bmatrix}\begin{bmatrix}0&1\\1&1\end{bmatrix}=\begin{bmatrix}1&1\\1&2\end{bmatrix} \] \[B^2=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}=\begin{bmatrix}-1&0\\0&-1\end{bmatrix} \] \[A^2-B^2=\begin{bmatrix}2&1\\1&3\end{bmatrix} \] \[\begin{bmatrix}0&0\\0&5\end{bmatrix}\neq\begin{bmatrix}2&1\\1&3\end{bmatrix} \] Answer: \(\displaystyle (A+B)(A-B)=\begin{bmatrix}0&0\\0&5\end{bmatrix}\neq\begin{bmatrix}2&1\\1&3\end{bmatrix}=A^2-B^2\)

  10. Exercise 10

    Find the value of x\displaystyle x if [1x1][1322511532][12x]=O.\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right] \quad\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=\mathrm{O} .

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

    NCERT’s answer
    \(\displaystyle -2, -14\)
    The row-by-matrix product is \(\displaystyle 1\times3\), so the whole product is a \(\displaystyle 1\times1\) matrix. \[\begin{bmatrix}1&x&1\end{bmatrix}\begin{bmatrix}1&3&2\\2&5&1\\15&3&2\end{bmatrix}=\begin{bmatrix}1+2x+15&\;3+5x+3&\;2+x+2\end{bmatrix}=\begin{bmatrix}16+2x&6+5x&4+x\end{bmatrix} \] \[\begin{bmatrix}16+2x&6+5x&4+x\end{bmatrix}\begin{bmatrix}1\\2\\x\end{bmatrix}=\begin{bmatrix}(16+2x)+2(6+5x)+x(4+x)\end{bmatrix}=\begin{bmatrix}x^2+16x+28\end{bmatrix} \] \[x^2+16x+28=0 \] \[(x+2)(x+14)=0 \] Answer: \(\displaystyle x=-2\) or \(\displaystyle x=-14\)