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NCERT Exemplar · Class 12 Mathematics Matrices

101 questions · 101 still being checked

EXERCISE 3.3 31–40 (part 4 of 10)

  1. Exercise 31

    Show that if A and B are square matrices such that AB=BA\displaystyle \mathrm{AB}=\mathrm{BA}, then (A+B)2=A2+2AB+B2.(\mathrm{A}+\mathrm{B})^2=\mathrm{A}^2+2 \mathrm{AB}+\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.

    \[(A+B)^2 = (A+B)(A+B) \] \[= A^2 + AB + BA + B^2 \quad \text{(distributive law)} \] \[BA = AB \quad \text{(given)} \] \[(A+B)^2 = A^2 + AB + AB + B^2 \] \[(A+B)^2 = A^2 + 2AB + B^2 \] Answer: \(\displaystyle (A+B)^2 = A^2 + 2AB + B^2 \) whenever \(\displaystyle AB = BA \).

  2. Exercise 32

    Let A=[12−13],B=[4015],C=[201−2]\displaystyle \mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right], \quad \mathrm{B}=\left[\begin{array}{cc}4 & 0 \\ 1 & 5\end{array}\right], \quad \mathrm{C}=\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right] and a=4,b=−2\displaystyle a=4, b=-2.
    Show that:
    (a)
    A+(B+C)=(A+B)+C\displaystyle \mathrm{A}+(\mathrm{B}+\mathrm{C})=(\mathrm{A}+\mathrm{B})+\mathrm{C}
    (b)
    A(BC)=(AB)C\displaystyle \mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C}
    (c)
    (a+b)B=a B+b B\displaystyle (a+b) \mathrm{B}=a \mathrm{~B}+b \mathrm{~B}
    (d)
    a(C−A)=aC−a A\displaystyle a(\mathrm{C}-\mathrm{A})=a \mathrm{C}-a \mathrm{~A}
    (e)
    (AT)T=A\displaystyle \left(\mathrm{A}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}
    (f)
    (b A)T=b AT\displaystyle (b \mathrm{~A})^{\mathrm{T}}=b \mathrm{~A}^{\mathrm{T}}
    (g)
    (AB)T=BTAT\displaystyle (\mathrm{AB})^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}
    (h)
    (A−B)C=AC−BC\displaystyle (\mathrm{A}-\mathrm{B}) \mathrm{C}=\mathrm{AC}-\mathrm{BC}
    (i)
    (A−B)T=AT−BT\displaystyle (\mathrm{A}-\mathrm{B})^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}-\mathrm{B}^{\mathrm{T}}

    Not cross-checked

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

    (a) \[A + (B+C) = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 7 & 2 \\ 1 & 6 \end{bmatrix} \] \[(A+B) + C = \begin{bmatrix} 5 & 2 \\ 0 & 8 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 7 & 2 \\ 1 & 6 \end{bmatrix} \] (b) \[BC = \begin{bmatrix} 4\cdot 2 + 0\cdot 1 & 4\cdot 0 + 0(-2) \\ 1\cdot 2 + 5\cdot 1 & 1\cdot 0 + 5(-2) \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 7 & -10 \end{bmatrix} \] \[A(BC) = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} 8 & 0 \\ 7 & -10 \end{bmatrix} = \begin{bmatrix} 22 & -20 \\ 13 & -30 \end{bmatrix} \] \[AB = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} 4 & 0 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 6 & 10 \\ -1 & 15 \end{bmatrix} \] \[(AB)C = \begin{bmatrix} 6 & 10 \\ -1 & 15 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 22 & -20 \\ 13 & -30 \end{bmatrix} \] (c) \[(a+b)B = 2B = \begin{bmatrix} 8 & 0 \\ 2 & 10 \end{bmatrix} \] \[aB + bB = \begin{bmatrix} 16 & 0 \\ 4 & 20 \end{bmatrix} + \begin{bmatrix} -8 & 0 \\ -2 & -10 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 2 & 10 \end{bmatrix} \] (d) \[a(C-A) = 4\begin{bmatrix} 1 & -2 \\ 2 & -5 \end{bmatrix} = \begin{bmatrix} 4 & -8 \\ 8 & -20 \end{bmatrix} \] \[aC - aA = \begin{bmatrix} 8 & 0 \\ 4 & -8 \end{bmatrix} - \begin{bmatrix} 4 & 8 \\ -4 & 12 \end{bmatrix} = \begin{bmatrix} 4 & -8 \\ 8 & -20 \end{bmatrix} \] (e) \[A^T = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \] \[(A^T)^T = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} = A \] (f) \[(bA)^T = (-2A)^T = \begin{bmatrix} -2 & 2 \\ -4 & -6 \end{bmatrix} \] \[bA^T = -2\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ -4 & -6 \end{bmatrix} \] (g) \[(AB)^T = \begin{bmatrix} 6 & -1 \\ 10 & 15 \end{bmatrix} \] \[B^TA^T = \begin{bmatrix} 4 & 1 \\ 0 & 5 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 6 & -1 \\ 10 & 15 \end{bmatrix} \] (h) \[(A-B)C = \begin{bmatrix} -3 & 2 \\ -2 & -2 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} -4 & -4 \\ -6 & 4 \end{bmatrix} \] \[AC - BC = \begin{bmatrix} 4 & -4 \\ 1 & -6 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 7 & -10 \end{bmatrix} = \begin{bmatrix} -4 & -4 \\ -6 & 4 \end{bmatrix} \] (i) \[(A-B)^T = \begin{bmatrix} -3 & -2 \\ 2 & -2 \end{bmatrix} \] \[A^T - B^T = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} - \begin{bmatrix} 4 & 1 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} -3 & -2 \\ 2 & -2 \end{bmatrix} \] Answer: all nine identities hold.

  3. Exercise 33

    If A=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]\displaystyle \mathrm{A}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right], then show that A2=[cos⁡2θsin⁡2θ−sin⁡2θcos⁡2θ]\displaystyle \mathrm{A}^2=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right].

    Not cross-checked

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

    \[A^2 = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix} \] \[A^2 = \begin{bmatrix}\cos^2\theta-\sin^2\theta & 2\sin\theta\cos\theta\\ -2\sin\theta\cos\theta & \cos^2\theta-\sin^2\theta\end{bmatrix} \] \[\cos^2\theta-\sin^2\theta=\cos 2\theta \qquad 2\sin\theta\cos\theta=\sin 2\theta \] \[A^2 = \begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix} \] Answer: \(\displaystyle A^2=\begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}\)

  4. Exercise 34

    If A=[0−xx0],B=[0110]\displaystyle \mathrm{A}=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] and x2=−1\displaystyle x^2=-1, then show that (A+B)2=A2+B2\displaystyle (\mathrm{A}+\mathrm{B})^2=\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.

    \[A^2=\begin{bmatrix}0&-x\\x&0\end{bmatrix}\begin{bmatrix}0&-x\\x&0\end{bmatrix}=\begin{bmatrix}-x^2&0\\0&-x^2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix} \quad (x^2=-1) \] \[B^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix} \] \[A^2+B^2=\begin{bmatrix}2&0\\0&2\end{bmatrix} \] \[A+B=\begin{bmatrix}0&1-x\\1+x&0\end{bmatrix} \] \[(A+B)^2=\begin{bmatrix}(1-x)(1+x)&0\\0&(1+x)(1-x)\end{bmatrix}=\begin{bmatrix}1-x^2&0\\0&1-x^2\end{bmatrix}=\begin{bmatrix}2&0\\0&2\end{bmatrix} \quad (x^2=-1) \] Answer: \(\displaystyle (A+B)^2=A^2+B^2=\begin{bmatrix}2&0\\0&2\end{bmatrix}\)

  5. Exercise 35

    Verify that A2=I\displaystyle \mathrm{A}^2=\mathrm{I} when A=[01−14−343−34]\displaystyle \mathrm{A}=\left[\begin{array}{ccc}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right].

    Not cross-checked

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

    \[A^2=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix} \] \[A^2=\begin{bmatrix}0+4-3 & 0-3+3 & 0+4-4\\ 0-12+12 & 4+9-12 & -4-12+16\\ 0-12+12 & 3+9-12 & -3-12+16\end{bmatrix} \] \[A^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I \] Answer: \(\displaystyle A^2=I\)

  6. Exercise 36

    Prove by Mathematical Induction that (A′)n=(An)′\displaystyle \left(\mathrm{A}^{\prime}\right)^n=\left(\mathrm{A}^n\right)^{\prime}, where n∈N\displaystyle n \in \mathbf{N} for any square matrix A.

    Not cross-checked

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

    \[P(n):\ (A')^n=(A^n)' \] \[n=1:\ (A')^1=A'=(A^1)' \quad \text{(true)} \] \[\text{Assume } P(k):\ (A')^k=(A^k)' \] \[(A')^{k+1}=(A')^k A' \] \[(A')^{k+1}=(A^k)'A' \quad \text{(assumption)} \] \[(A')^{k+1}=(A\,A^k)' \quad \text{(reversal law: } (XY)'=Y'X' \text{)} \] \[(A')^{k+1}=(A^{k+1})' \] \(\displaystyle P(1)\) is true and \(\displaystyle P(k)\Rightarrow P(k+1)\), so \(\displaystyle P(n)\) holds for every \(\displaystyle n\in\mathbf N\).Answer: \(\displaystyle (A')^n=(A^n)'\) for all \(\displaystyle n\in\mathbf N\).

  7. Exercise 37

    Find inverse, by elementary row operations (if possible), of the following matrices
    (i)
    [13−57]\displaystyle \left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right]
    (ii)
    [1−3−26]\displaystyle \left[\begin{array}{cc}1 & -3 \\ -2 & 6\end{array}\right].

    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
    (i)
    \(\displaystyle \frac{1}{22}\left[\begin{array}{cc}7 & -3 \\ 5 & 1\end{array}\right]\)
    (ii)
    not possible
    (i) Write \(\displaystyle [A\mid I]\) and reduce. \[\left[\begin{array}{cc|cc}1&3&1&0\\-5&7&0&1\end{array}\right] \] \[\left[\begin{array}{cc|cc}1&3&1&0\\0&22&5&1\end{array}\right] \quad (R_2\to R_2+5R_1) \] \[\left[\begin{array}{cc|cc}1&3&1&0\\0&1&\frac{5}{22}&\frac{1}{22}\end{array}\right] \quad (R_2\to \tfrac{1}{22}R_2) \] \[\left[\begin{array}{cc|cc}1&0&\frac{7}{22}&-\frac{3}{22}\\0&1&\frac{5}{22}&\frac{1}{22}\end{array}\right] \quad (R_1\to R_1-3R_2) \] \[A^{-1}=\frac{1}{22}\begin{bmatrix}7&-3\\5&1\end{bmatrix} \](ii) \[\left[\begin{array}{cc|cc}1&-3&1&0\\-2&6&0&1\end{array}\right] \] \[\left[\begin{array}{cc|cc}1&-3&1&0\\0&0&2&1\end{array}\right] \quad (R_2\to R_2+2R_1) \] The left half has a row of zeros, so the inverse does not exist.Answer: (i) \(\displaystyle \frac{1}{22}\begin{bmatrix}7&-3\\5&1\end{bmatrix}\); (ii) inverse does not exist.

  8. Exercise 38

    If [xy4z+6x+y]=[8w06]\displaystyle \left[\begin{array}{cc}x y & 4 \\ z+6 & x+y\end{array}\right]=\left[\begin{array}{cc}8 & w \\ 0 & 6\end{array}\right], then find values of x,y,z\displaystyle x, y, z and w\displaystyle w.

    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
    \(\displaystyle x=2, y=4\) or \(\displaystyle x=4, y=2, z=-6, w=4\)
    Equal matrices have equal corresponding entries. \[xy=8,\qquad w=4,\qquad z+6=0,\qquad x+y=6 \] \[z=-6 \] \[y=6-x \Rightarrow x(6-x)=8 \] \[x^2-6x+8=0 \Rightarrow (x-2)(x-4)=0 \] \[x=2,\ y=4 \quad\text{or}\quad x=4,\ y=2 \] Answer: \(\displaystyle x=2,\ y=4\) (or \(\displaystyle x=4,\ y=2\)), \(\displaystyle z=-6\), \(\displaystyle w=4\).

  9. Exercise 39

    If A=[15712]\displaystyle \mathrm{A}=\left[\begin{array}{cc}1 & 5 \\ 7 & 12\end{array}\right] and B=[9178]\displaystyle \mathrm{B}=\left[\begin{array}{ll}9 & 1 \\ 7 & 8\end{array}\right], find a matrix C such that 3 A+5 B+2C\displaystyle 3 \mathrm{~A}+5 \mathrm{~B}+2 \mathrm{C} is a null matrix.

    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
    \(\displaystyle \left[\begin{array}{ll}-24 & -10 \\ -28 & -38\end{array}\right]\)
    \[3A+5B+2C=O \Rightarrow C=-\tfrac12(3A+5B) \] \[3A=\begin{bmatrix}3&15\\21&36\end{bmatrix},\qquad 5B=\begin{bmatrix}45&5\\35&40\end{bmatrix} \] \[3A+5B=\begin{bmatrix}48&20\\56&76\end{bmatrix} \] \[C=-\tfrac12\begin{bmatrix}48&20\\56&76\end{bmatrix}=\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix} \] Answer: \(\displaystyle C=\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}\)

  10. Exercise 40

    If A=[3−5−42]\displaystyle \mathrm{A}=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right], then find A2−5 A−14I\displaystyle \mathrm{A}^2-5 \mathrm{~A}-14 \mathrm{I}. Hence, obtain A3\displaystyle \mathrm{A}^3.

    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
    \(\displaystyle \mathrm{A}^3=\left[\begin{array}{cc}187 & -195 \\ -156 & 148\end{array}\right]\)
    \[A^2=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}=\begin{bmatrix}29&-25\\-20&24\end{bmatrix} \] \[5A=\begin{bmatrix}15&-25\\-20&10\end{bmatrix},\qquad 14I=\begin{bmatrix}14&0\\0&14\end{bmatrix} \] \[A^2-5A-14I=\begin{bmatrix}29-15-14&-25+25\\-20+20&24-10-14\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix} \] Hence \(\displaystyle A^2=5A+14I\). Multiply by \(\displaystyle A\): \[A^3=5A^2+14A=5(5A+14I)+14A=39A+70I \] \[A^3=\begin{bmatrix}117+70&-195\\-156&78+70\end{bmatrix}=\begin{bmatrix}187&-195\\-156&148\end{bmatrix} \] Answer: \(\displaystyle A^2-5A-14I=O\); \(\displaystyle A^3=\begin{bmatrix}187&-195\\-156&148\end{bmatrix}\)