SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 12 Mathematics Matrices

101 questions · 101 still being checked

EXERCISE 3.3 41–50 (part 5 of 10)

  1. Exercise 41

    Find the values of a,b,c\displaystyle a, b, c and d\displaystyle d, if 3[abcd]=[a6−12d]+[4a+bc+d3].3\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} a & 6 \\ -1 & 2 d \end{array}\right]+\left[\begin{array}{cc} 4 & a+b \\ c+d & 3 \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
    \(\displaystyle a=2, b=4, c=1, d=3\)
    Equate corresponding entries.\[3a = a + 4 \Rightarrow a = 2 \]\[3b = 6 + a + b \Rightarrow 2b = 6 + 2 \Rightarrow b = 4 \]\[3d = 2d + 3 \Rightarrow d = 3 \]\[3c = -1 + c + d \Rightarrow 2c = -1 + 3 \Rightarrow c = 1 \]Answer: \(\displaystyle a = 2,\ b = 4,\ c = 1,\ d = 3\)

  2. Exercise 42

    Find the matrix A such that [2−110−34]A=[−1−8−101−2−592215].\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right] \quad \mathrm{A}=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \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
    \(\displaystyle \left[\begin{array}{ccc}1 & -2 & -5 \\ 3 & 4 & 0\end{array}\right]\)
    The left factor is \(\displaystyle 3\times 2\) and the product is \(\displaystyle 3\times 3\), so A is \(\displaystyle 2\times 3\). Write the rows of A as \(\displaystyle (p,q,r)\) and \(\displaystyle (s,t,u)\).\[\begin{bmatrix}1 & 0\end{bmatrix} A = \begin{bmatrix}1 & -2 & -5\end{bmatrix} \Rightarrow (p,q,r) = (1,-2,-5) \]\[2(p,q,r) - (s,t,u) = (-1,-8,-10) \]\[(s,t,u) = 2(1,-2,-5) - (-1,-8,-10) = (3,\,4,\,0) \]Check with row $\displaystyle 3$:\[-3(1,-2,-5) + 4(3,4,0) = (9,\,22,\,15) \quad \checkmark \]Answer: \(\displaystyle A = \begin{bmatrix}1 & -2 & -5\\ 3 & 4 & 0\end{bmatrix}\)

  3. Exercise 43

    If A=[1241]\displaystyle \mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right], find A2+2 A+7I\displaystyle \mathrm{A}^2+2 \mathrm{~A}+7 \mathrm{I}.

    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
    \(\displaystyle \left[\begin{array}{cc}18 & 8 \\ 16 & 18\end{array}\right]\)
    \[A^2 = \begin{bmatrix}1 & 2\\ 4 & 1\end{bmatrix}\begin{bmatrix}1 & 2\\ 4 & 1\end{bmatrix} = \begin{bmatrix}1+8 & 2+2\\ 4+4 & 8+1\end{bmatrix} = \begin{bmatrix}9 & 4\\ 8 & 9\end{bmatrix} \]\[2A = \begin{bmatrix}2 & 4\\ 8 & 2\end{bmatrix}, \qquad 7I = \begin{bmatrix}7 & 0\\ 0 & 7\end{bmatrix} \]\[A^2 + 2A + 7I = \begin{bmatrix}9+2+7 & 4+4+0\\ 8+8+0 & 9+2+7\end{bmatrix} \]Answer: \(\displaystyle \begin{bmatrix}18 & 8\\ 16 & 18\end{bmatrix}\)

  4. Exercise 44

    If 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], and A−1=A′\displaystyle \mathrm{A}^{-1}=\mathrm{A}^{\prime}, find value of α\displaystyle \alpha.

    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 for all real values of \(\displaystyle \alpha\)
    \[AA' = \begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix} \]\[= \begin{bmatrix}\cos^2\alpha + \sin^2\alpha & -\cos\alpha\sin\alpha + \sin\alpha\cos\alpha\\ -\sin\alpha\cos\alpha + \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha\end{bmatrix} = \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} = I \]\[AA' = I \Rightarrow A^{-1} = A' \]This holds for every value of \(\displaystyle \alpha\); the condition puts no restriction on it.Answer: \(\displaystyle \alpha\) can be any real number

  5. Exercise 45

    If the matrix [0a32b−1c10]\displaystyle \left[\begin{array}{ccc}0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0\end{array}\right] is a skew symmetric matrix, find the values of a,b\displaystyle a, b and c\displaystyle c.

    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
    \(\displaystyle a=-2, b=0, c=-3\)
    For a skew-symmetric matrix \(\displaystyle a_{ij} = -a_{ji}\).\[a_{22} = -a_{22} \Rightarrow b = 0 \]\[a_{12} = -a_{21} \Rightarrow a = -2 \]\[a_{13} = -a_{31} \Rightarrow 3 = -c \Rightarrow c = -3 \]\[a_{23} = -a_{32} \Rightarrow -1 = -1 \quad \checkmark \]Answer: \(\displaystyle a = -2,\ b = 0,\ c = -3\)

  6. Exercise 46

    If P(x)=[cos⁡xsin⁡x−sin⁡xcos⁡x]\displaystyle \mathrm{P}(x)=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right], then show that P(x).P(y)=P(x+y)=P(y).P(x).\mathrm{P}(x) . \mathrm{P}(y)=\mathrm{P}(x+y)=\mathrm{P}(y) . \mathrm{P}(x) .

    Not cross-checked

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

    \[P(x)P(y) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}\begin{bmatrix}\cos y & \sin y\\ -\sin y & \cos y\end{bmatrix} \]\[= \begin{bmatrix}\cos x\cos y - \sin x\sin y & \cos x\sin y + \sin x\cos y\\ -\sin x\cos y - \cos x\sin y & -\sin x\sin y + \cos x\cos y\end{bmatrix} \]\[= \begin{bmatrix}\cos(x+y) & \sin(x+y)\\ -\sin(x+y) & \cos(x+y)\end{bmatrix} = P(x+y) \]Interchanging \(\displaystyle x\) and \(\displaystyle y\):\[P(y)P(x) = P(y+x) = P(x+y) \]Answer: \(\displaystyle P(x)P(y) = P(x+y) = P(y)P(x)\)

  7. Exercise 47

    If A is square matrix such that A2=A\displaystyle \mathrm{A}^2=\mathrm{A}, show that (I+A)3=7 A+I\displaystyle (\mathrm{I}+\mathrm{A})^3=7 \mathrm{~A}+\mathrm{I}.

    Not cross-checked

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

    I and A commute, so the binomial expansion applies:\[(I+A)^3 = I + 3A + 3A^2 + A^3 \]\[A^2 = A \Rightarrow A^3 = A^2 A = A\,A = A \]\[(I+A)^3 = I + 3A + 3A + A = I + 7A \]Answer: \(\displaystyle (I+A)^3 = 7A + I\)

  8. Exercise 48

    If A, B are square matrices of same order and B is a skew-symmetric matrix, show that A′BA\displaystyle \mathrm{A}^{\prime} \mathrm{BA} is skew symmetric.

    Not cross-checked

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

    \[(A'BA)' = A'B'(A')' \quad \text{(reversal law)} \]\[= A'B'A \]\[B' = -B \quad \text{(B is skew-symmetric)} \]\[(A'BA)' = A'(-B)A = -A'BA \]Answer: \(\displaystyle (A'BA)' = -A'BA\), so \(\displaystyle A'BA\) is skew-symmetric

  9. Exercise 49

    If AB=BA\displaystyle \mathrm{AB}=\mathrm{BA} for any two sqaure matrices, prove by mathematical induction that (AB)n=An Bn\displaystyle (\mathrm{AB})^n=\mathrm{A}^n \mathrm{~B}^n.

    Not cross-checked

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

    Let \(\displaystyle P(n):\ (AB)^n=A^nB^n\), given \(\displaystyle AB=BA\). \[P(1):\ (AB)^1=A^1B^1 \quad \text{(true)} \] Assume \(\displaystyle P(k)\): \[(AB)^k=A^kB^k \] Using \(\displaystyle BA=AB\) repeatedly, \(\displaystyle B\) moves past \(\displaystyle A\): \[B^kA=B^{k-1}(BA)=B^{k-1}AB=\cdots=AB^k \] Then \[(AB)^{k+1}=(AB)^k(AB) \] \[=A^kB^k\,AB \quad \text{(assumption)} \] \[=A^k(B^kA)B \] \[=A^k(AB^k)B \quad (B^kA=AB^k) \] \[=A^{k+1}B^{k+1} \] So \(\displaystyle P(k)\Rightarrow P(k+1)\), and \(\displaystyle P(1)\) is true.Answer: \(\displaystyle (AB)^n=A^nB^n\) for every natural number \(\displaystyle n\).

  10. Exercise 50

    Find x,y,z\displaystyle x, y, z if A=[02yzxy−zx−yz]\displaystyle \mathrm{A}=\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right] satisfies A′=A−1\displaystyle \mathrm{A}^{\prime}=\mathrm{A}^{-1}.

    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
    \(\displaystyle x= \pm \frac{1}{\sqrt{2}}, y= \pm \frac{1}{\sqrt{6}}, z= \pm \frac{1}{\sqrt{3}}\)
    \(\displaystyle A^{-1}=A'\) means \(\displaystyle A'A=I\). \[A'A=\begin{bmatrix}0&x&x\\2y&y&-y\\z&-z&z\end{bmatrix}\begin{bmatrix}0&2y&z\\x&y&-z\\x&-y&z\end{bmatrix} \] \[=\begin{bmatrix}2x^2&0&0\\0&6y^2&0\\0&0&3z^2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \] \[2x^2=1,\quad 6y^2=1,\quad 3z^2=1 \] \[x=\pm\frac{1}{\sqrt2},\quad y=\pm\frac{1}{\sqrt6},\quad z=\pm\frac{1}{\sqrt3} \]Answer: \(\displaystyle x=\pm\dfrac{1}{\sqrt2},\ y=\pm\dfrac{1}{\sqrt6},\ z=\pm\dfrac{1}{\sqrt3}\)