RD Sharma Class 12 Exercise 5.1 Determinant Solutions Maths

RD Sharma Class 12 Exercise 5.1 Determinant Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 07:12 PM IST

The CBSE board students are gifted to access the best solutions book, the RD Sharma solutions, to clear their doubts. It is indeed a fact that a teacher cannot be present 24 x 7 to clear the doubts of the students. Hence, a better alternative is required. As most of the 12th standard students find mathematics a complex subject, even a simple chapter like the Determinants becomes difficult for them to solve. This is where the RD Sharma Class 12th exercise 5.1 comes to the rescue.

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RD Sharma Class 12 Solutions Chapter 5 Determinants - Other Exercise

Determinants Excercise: 5.1

Determinants exercise 5.1 question 1 (i)

Answer:
$\mathrm{M}_{11}=-1, \mathrm{M}_{21}=20, \mathrm{C}_{11}=-1, \mathrm{C}_{21}=-20$
Hint:
Let M_ij, C_ij represents minor and cofactor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present.
Given:
$A=\begin{bmatrix} 5 &20 \\ 0 &-1 \end{bmatrix}$
Solution:
$\begin{aligned} &\mathrm{M}_{11}=-1 \\ &\mathrm{M}_{21}=20 \\ &\mathrm{C}_{\mathrm{ij}}=(-1)^{\mathrm{i}+\mathrm{j}} \mathrm{M}_{\mathrm{ij}} \\ &\mathrm{C}_{11}=(-1)^{2}(-1)=-1 \\ &\mathrm{C}_{21}=(-1)^{3}(20)=-20 \\ &\mathrm{D}=(-1 \times 5)-(20 \times 0)=-5 \end{aligned}$

Determinants exercise 5.1 question 1 (ii)

Answer:
$\mathrm{M}_{11}=3, \mathrm{M}_{21}=4, \mathrm{C}_{11}=3, \mathrm{C}_{21}=-4$
Hint:
Let M_ij, C_ij represents minor and cofactor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present.
Given:
$A=\begin{bmatrix} -1 &4 \\ 2 &3 \end{bmatrix}$
Solution:
$\begin{aligned} &\mathrm{M}_{11}=3 \\ &\mathrm{M}_{21}=4 \\ &\mathrm{C}_{\mathrm{ij}}=(-1)^{\mathrm{i}+j} \mathrm{M}_{\mathrm{ij}} \\ &\mathrm{C}_{11}=(-1)^{2} 3=3 \\ &\mathrm{C}_{21}=(-1)^{3} 4=-4 \\ &\mathrm{D}=(-1 \times 3)-(2 \times 4)=-3-8=-11 \end{aligned}$

Determinants exercise 5.1 question 1 (iii)

Answer:
$\mathrm{M}_{11}=-12, \mathrm{M}_{21}=-16, \mathrm{M}_{31}=-4, \\ \mathrm{C}_{11}=-12, \mathrm{C}_{21}=16, \mathrm{C}_{31}=-4,$
Hint:
Let M_ij, C_ijrepresents minor and cofactor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present.
Given:
$\begin{gathered} \mathrm{A}=\left[\begin{array}{ccc} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right] \\ \end{gathered}$
Solution:
$\begin{gathered} \mathrm{A}=\left[\begin{array}{ccc} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right] \\ \mathrm{M}_{11}=\left|\begin{array}{cc} -1 & 2 \\ 5 & 2 \end{array}\right|=-2-10=-12 \\ \mathrm{M}_{21}=\left|\begin{array}{cc} -3 & 2 \\ 5 & 2 \end{array}\right|=-6-10=-16 \\ \mathrm{M}_{31}=\left|\begin{array}{ll} -3 & 2 \\ -1 & 2 \end{array}\right|=-6-(-2)=-4 \end{gathered}$
$\begin{aligned} &\mathrm{C}_{\mathrm{ij}}=(-1)^{\mathrm{i}+\mathrm{j}} \mathrm{M}_{\mathrm{ij}} \\ &\mathrm{C}_{11}=(-1)^{2}(-12)=-12 \\ &\mathrm{C}_{21}=(-1)^{3}(-16)=16 \\ &\mathrm{C}_{31}=(-1)^{4}(-4)=-4 \\ &\mathrm{D}=1(-12)+3(8-6)+2(20+3) \\ &=-12+3(2)+2(23) \\ &\qquad \begin{array}{l} = -12+6+46 \\ =40 \end{array} \\ & \end{aligned}$

Determinants exercise 5.1 question 1 (iv)

Answer:
$\begin{aligned} &\mathrm{M}_{11}=\mathrm{a}\left(\mathrm{b}^{2}-\mathrm{c}^{2}\right), \mathrm{M}_{21}=\mathrm{b}\left(\mathrm{a}^{2}-\mathrm{c}^{2}\right), \mathrm{M}_{31}=\mathrm{c}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \\ &\mathrm{C}_{11}=\mathrm{a}\left(\mathrm{b}^{2}-\mathrm{c}^{2}\right), \mathrm{C}_{21}=\mathrm{b}\left(\mathrm{a}^{2}-\mathrm{c}^{2}\right), \mathrm{C}_{31}=\mathrm{c}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \end{aligned}$
Hint:
Let M_ij, C_ij represents minor and cofactor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present.
Given:
$\begin{aligned} &\mathrm{A}=\left[\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right] \\ \end{aligned}$
Solution:
$\begin{aligned} &\mathrm{A}=\left[\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right] \\ &\mathrm{M}_{11}=\left|\begin{array}{ll} b & c a \\ c & a b \end{array}\right|=\mathrm{ab}^{2}-\mathrm{ac}^{2}=\mathrm{a}\left(\mathrm{b}^{2}-\mathrm{c}^{2}\right) \\ &\mathrm{M}_{21}=\left|\begin{array}{ll} a & b c \\ c & a b \end{array}\right|=\mathrm{a}^{2} \mathrm{~b}-\mathrm{bc}^{2}=\mathrm{b}\left(\mathrm{a}^{2}-\mathrm{c}^{2}\right) \\ &\mathrm{M}_{31}=\left|\begin{array}{ll} a & b c \\ b & c a \end{array}\right|=\mathrm{a}^{2} \mathrm{c}-\mathrm{b}^{2} \mathrm{c}=\mathrm{c}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \\ &\mathrm{C}_{\mathrm{ij}}=(-1)^{\mathrm{i}+\mathrm{j}} \mathrm{M}_{\mathrm{ij}} \end{aligned}$

Determinants exercise 5.1 question 1 (v)

Answer:
$\mathrm{M}_{11}=5, \mathrm{M}_{21}=-40, \mathrm{M}_{31}=-30 \\ \mathrm{C}_{11}=5, \mathrm{C}_{21}=40, \mathrm{C}_{31}=-30$
Hint:
Let M_ij, C_ij represents minor and cofactor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present.
Given:
$\begin{aligned} &\mathbf{A}=\left[\begin{array}{lll} 0 & 2 & 6 \\ 1 & 5 & 0 \\ 3 & 7 & 1 \end{array}\right] \\ \end{aligned}$
Solution:
$\begin{aligned} &\mathbf{A}=\left[\begin{array}{lll} 0 & 2 & 6 \\ 1 & 5 & 0 \\ 3 & 7 & 1 \end{array}\right] \\ &\mathrm{M}_{11}=\left|\begin{array}{ll} 5 & 0 \\ 7 & 1 \end{array}\right|=5-0=5 \\ &\mathrm{M}_{21}=\left|\begin{array}{ll} 2 & 6 \\ 7 & 1 \end{array}\right|=2-42=-40 \\ &\mathrm{M}_{31}=\left|\begin{array}{ll} 2 & 6 \\ 5 & 0 \end{array}\right|=0-(30)=-30 \end{aligned}$
$\begin{aligned} &C_{i j}=(-1)^{\mathrm{i}+j} \mathrm{M}_{\mathrm{ij}} \\ &\mathrm{C}_{11}=(-1)^{2} 5=5 \\ &\mathrm{C}_{21}=(-1)^{3}(-40)=40 \\ &\mathrm{C}_{31}=(-1)^{4}(-30)=-30 \\ &\mathrm{D}=0(5)-2(1)+6(7-5) \\ &=0-2-48 \\ &=-50 \end{aligned}$

Determinants exercise 5.1 question 1 (vi)

Answer:
$\begin{aligned} &\mathrm{M}_{11}=\mathrm{bc}-\mathrm{f}^{2}, \mathrm{M}_{21}=\mathrm{hc}-\mathrm{fg}, \mathrm{M}_{31}=\mathrm{hf}-\mathrm{gb} \\ &\mathrm{C}_{11}=\mathrm{bc}-\mathrm{f}^{2}, \mathrm{C}_{21}=-(\mathrm{hc}-\mathrm{fg}), \mathrm{C}_{31}=\mathrm{hf}-\mathrm{gb} \end{aligned}$
Hint:
Let M_ij, C_ij represents minor and cofactor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present.
Given:
$\begin{aligned} &\mathrm{A}=\left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right] \\ \end{aligned}$
Solution:
$\begin{aligned} &\mathrm{A}=\left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right] \\ &\mathrm{M}_{11}=\left|\begin{array}{ll} b & f \\ f & c \end{array}\right|=\mathrm{bc}-\mathrm{f}^{2} \\ &\mathrm{M}_{21}=\left|\begin{array}{ll} h & g \\ f & c \end{array}\right|=\mathrm{hc}-\mathrm{fg} \\ &\mathrm{M}_{31}=\left|\begin{array}{ll} h & g \\ b & f \end{array}\right|=\mathrm{hf}-\mathrm{gb} \end{aligned}$
$\begin{aligned} &C_{i j}=(-1)^{i+j} M_{i j} \\ &C_{11}=(-1)^{2} b c-f^{2}=b c-f^{2} \\ &C_{21}=(-1)^{3} \mathrm{hc}-f g=-(h c-f g) \\ &C_{31}=(-1)^{4} \mathrm{hf}-g b=h f-g b \end{aligned}$
$\begin{aligned} \mathrm{D} &=\mathrm{a}\left(\mathrm{b} \mathrm{c}-\mathrm{f}^{2}\right)-\mathrm{h}(\mathrm{hc}-\mathrm{fg})+\mathrm{g}(\mathrm{hf}-\mathrm{g} \mathrm{b}) \\ &=\mathrm{abc}-\mathrm{af}^{2}-\mathrm{h}^{2} \mathrm{c}+\mathrm{hfg}+\mathrm{ghf}-\mathrm{g}^{2} \mathrm{~b} \\ &=\mathrm{abc}-\mathrm{af}^{2}-\mathrm{h}^{2} \mathrm{c}+2 \mathrm{hfg}-\mathrm{g}^{2} \mathrm{~b} \end{aligned}$

Determinants exercise 5.1 question 1 (vii)

Answer:
$\begin{aligned} &\mathrm{M}_{11}=-9, \mathrm{M}_{21}=9, \mathrm{M}_{31}=-9, \mathrm{M}_{41}=0 \\ &\mathrm{C}_{11}=-9, \mathrm{C}_{21}=-9, \mathrm{C}_{31}=-9, \mathrm{C}_{41}=0 \end{aligned}$
Hint:
Let M_ij, C_ij represents minor and cofactor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present.
Given:
$\begin{aligned} &\mathrm{A}=\left[\begin{array}{cccc} 2 & -1 & 0 & 1 \\ -3 & 0 & 1 & -2 \\ 1 & 1 & -1 & 1 \\ 2 & -1 & 5 & 0 \end{array}\right] \\ \end{aligned}$
Solution:
$\begin{aligned} &\mathrm{A}=\left[\begin{array}{cccc} 2 & -1 & 0 & 1 \\ -3 & 0 & 1 & -2 \\ 1 & 1 & -1 & 1 \\ 2 & -1 & 5 & 0 \end{array}\right] \\ &\begin{aligned} M_{11}=\left|\begin{array}{ccc} 0 & 1 & -2 \\ 1 & -1 & 1 \\ -1 & 5 & 0 \end{array}\right| &=0(0-5)-1(0+1)-2(5-1) \\ &=-1-8 \\ &=-9 \end{aligned} \end{aligned}$
$\begin{aligned} \mathrm{M}_{21}=\left|\begin{array}{ccc} -1 & 0 & 1 \\ 1 & -1 & 1 \\ -1 & 5 & 0 \end{array}\right| &=-1(0-5)+1(5-1) \\ &=5+4 \\ &=9 \end{aligned}$
$\begin{aligned} &\mathrm{M}_{31}=\left|\begin{array}{ccc} -1 & 0 & 1 \\ 0 & 1 & -2 \\ -1 & 5 & 0 \end{array}\right| \\ & =-1(0+10)+1(0+1) \\ & =-10+1 \\ & =-9 \\ & \begin{array}{rll} M_{41}=\left|\begin{array}{ccc} -1 & 0 & 1 \\ 0 & 1 & -2 \\ 1 & -1 & 1 \end{array}\right| & =-1(1-2)+1(0-1) \\ & =1-1 \\ & =0 \end{array} \end{aligned}$
$\begin{aligned} &C_{i j}=(-1)^{\mathrm{i}+j} M_{i j} \\ &C_{11}=(-1)^{2}(-9)=-9 \\ &C_{21}=(-1)^{3}(9)=-9 \\ &C_{31}=(-1)^{4}(-9)=-9 \\ &C_{41}=(-1)^{5}(0)=0 \end{aligned}$
$\begin{aligned} \mathrm{D} &=2\left|\begin{array}{ccc} 0 & 1 & -2 \\ 1 & -1 & 1 \\ -1 & 5 & 0 \end{array}\right|+1\left|\begin{array}{ccc} -3 & 1 & -2 \\ 1 & -1 & 1 \\ 2 & 5 & 0 \end{array}\right|-1\left|\begin{array}{ccc} -3 & 0 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 5 \end{array}\right| \\ &=-18-27+15 \\ &=-30 \end{aligned}$

Determinants exercise 5.1 question 2 (i)

Answer:
$D=5x^{2}+8x$
Hint:
Determinant matrix must be square (i.e. same number of rows and columns)
Given:
$\begin{vmatrix} x &-7 \\ x &5x+1 \end{vmatrix}$
Solution: $\begin{aligned} &\mathrm{D}=\mathrm{a}_{11} \times \mathrm{C}_{11}+\mathrm{a}_{21} \times \mathrm{C}_{21} \\ &=\mathrm{x}(5 \mathrm{x}+1)+(-7 \mathrm{x}) \\ &=5 \mathrm{x}^{2}+\mathrm{x}+7 \mathrm{x} \\ &=5 \mathrm{x}^{2}+8 \mathrm{x} \end{aligned}$

Determinants exercise 5.1 question 2 (ii)

Answer:
$cos^{2}\theta +sin^{2} \theta =1$
Hint:
Determinant matrix must be square (i.e. same number of rows and columns)
Given:
$\begin{aligned} \left|\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right|\\ \end{aligned}$
Solution:
$\begin{aligned} &\begin{aligned} &\mathrm{D}=\mathrm{a}_{11} \times \mathrm{C}_{11}+\mathrm{a}_{21} \times \mathrm{C}_{21} \\ &=\cos \theta(\cos \theta)+(-\sin \theta \times(-\sin \theta)) \\ &=\cos ^{2} \theta+\sin ^{2} \theta \\ &=1 \end{aligned} \end{aligned}$

Determinants exercise 5.1 question 2 (iii)

Answer:
$0$
Hint:
Determinant matrix must be square (i.e. same number of rows and columns)
Given:
$\begin{aligned} \left|\begin{array}{cc} \cos 15^{\circ} & \sin 15^{\circ} \\ -\sin 75^{\circ} & \cos 75^{\circ} \end{array}\right|\\ \end{aligned}$
Solution:
$\begin{aligned} &\begin{aligned} \mathrm{D} &=\mathrm{a}_{11} \times \mathrm{C}_{11}+\mathrm{a}_{21} \times \mathrm{C}_{21} \\ &=\cos 15^{\circ}\left(\cos 75^{\circ}\right)+\left(\sin 15^{\circ}\left(-\sin 75^{\circ}\right)\right) \\ &=\cos 15^{\circ} \cos 75^{\circ}-\sin \left(90^{\circ}-15^{\circ}\right) \sin \left(90^{\circ}-75^{\circ}\right) \quad\left[\sin \left(90^{\circ}-\theta\right)=\cos \theta\right] \\ &=\cos 15^{\circ} \cos 75^{\circ}-\cos 75^{\circ} \cos 15^{\circ} \\ &=0 \end{aligned} \end{aligned}$

Determinants exercise 5.1 question 2 (iv)

Answer:
$a^{2}+b^{2}+c^{2}+d^{2}$
Hint:
Determinant matrix must be square (i.e. same number of rows and columns)
Given:
$\begin{aligned} \left|\begin{array}{cc} \mathrm{a}+\mathrm{ib} & \mathrm{c}+\mathrm{id} \\ -\mathrm{c}+\mathrm{id} & \mathrm{a}-\mathrm{ib} \end{array}\right|\\ \end{aligned}$
Solution:
$\begin{aligned} &\begin{aligned} &\mathrm{D}=\mathrm{a}_{11} \times \mathrm{C}_{11}+\mathrm{a}_{21} \times \mathrm{C}_{21} \\ &\quad=\mathrm{a}+\mathrm{ib}(\mathrm{a}-\mathrm{ib})+\mathrm{c}+\mathrm{id}(-\mathrm{c}+\mathrm{id}) \\ &\quad=\mathrm{a}^{2}-\mathrm{i}^{2} \mathrm{~b}^{2}-\mathrm{i}^{2} \mathrm{~d}^{2}+\mathrm{c}^{2} \\ &\quad=\mathrm{a}^{2}+\mathrm{c}^{2}+\mathrm{b}^{2}+\mathrm{d}^{2} \end{aligned} \end{aligned}$

Determinants exercise 5.1 question 4

Answer:
$D=1$
Hint:
Determinant matrix must be square (i.e. same number of rows and columns)
Given:
$\begin{aligned} \left|\begin{array}{cc} \sin 10^{\circ} & -\cos 10^{\circ} \\ \sin 80^{\circ} & \cos 80^{\circ} \end{array}\right|\\ \end{aligned}$
Solution:
$\begin{aligned} &\Delta=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21}\\ &=\sin 10^{\circ} \cos 80^{\circ}+\cos 10^{\circ} \sin 80^{\circ}\\ &=\sin 10^{\circ} \cos \left(90^{\circ}-10^{\circ}\right)+\cos 10^{\circ} \sin \left(90^{\circ}-10^{\circ}\right) \quad\quad\quad \quad\left[\sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\\ &=\sin 10^{\circ} \sin 10^{\circ}+\cos 10^{\circ} \cos 10^{\circ}\\ &=\sin ^{2} 10^{\circ}+\cos ^{2} 10^{\circ} \quad \quad \quad \quad \quad\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right]\\ &=1 \end{aligned}$

Determinants exercise 5.1 question 5

Answer:
$-140$
Hint:
Determinant matrix must be square (i.e. same number of rows and columns)
Given:
$\begin{aligned} \left|\begin{array}{ccc} 2 & 3 & -5 \\ 7 & 1 & -2 \\ -3 & 4 & 1 \end{array}\right|\\ \end{aligned}$
Solution:
$\begin{aligned} &\Delta=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21+} \mathrm{a}_{31} \mathrm{C}_{31} \end{aligned}$
$\begin{aligned} &=(-1)^{1+1} 2(1+8)+(-1)^{1+2} 3(7-6)+(-1)^{1+3}(-5)(28+3) \\ &=2(1+8)-3(7-6)-5(28+3) \\ &=18-3-155 \\ &\Delta=-140 \end{aligned}$
Second method is the Sarus method where we adjoin the first two columns to the right to get:
$\left|\begin{array}{ccccc} 2 & 3 & -5 & 2 & 3 \\ 7 & 1 & -2 & 7 & 1 \\ -3 & 4 & 1 & -3 & 4 \end{array}\right|$
$\begin{aligned} \Delta &=\{((2 \times 1 \times 1)+(3 \times-2 \times-3)-(5 \times 7 \times 4))-((-5 \times 1 \times-3)+(2 \times-2 \times 4)+(3 \times 7 \times 1))\} \\ &=(2+18-140)-(15-16+21) \\ &=-120-20 \\ \Delta &=-140 \end{aligned}$

Determinants exercise 5.1 question 6

Answer:
$0$
Hint:
Determinant matrix must be square (i.e. same number of rows and columns)
Given:
$\left|\begin{array}{ccc} 0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0 \end{array}\right|$
Solution:
$\begin{aligned} &\Delta=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21+} \mathrm{a}_{31} \mathrm{C}_{31}\\ &=(-1)^{1+1} 0\left(0+\sin ^{2} \beta\right)+(-1)^{1+2} \sin \alpha(0-\sin \beta \cos \alpha)+(-1)^{1+3}(-\cos \alpha)(\sin \alpha \sin \beta-0)\\ &=0\left(0+\sin ^{2} \beta\right)-\sin \alpha(0-\sin \beta \cos \alpha)-\cos \alpha(\sin \alpha \sin \beta-0)\\ &=\sin \alpha \sin \beta \cos \alpha-\cos \alpha \sin \alpha \sin \beta\\ &\Delta=0 \end{aligned}$

Determinants exercise 5.1 question 7

Answer:
$1$
Hint:
Determinant matrix must be square (i.e. same number of rows and columns)
Given:
$\left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right|$
Solution:
$\begin{aligned} &\Delta=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21}+\mathrm{a}_{31} \mathrm{C}_{31} \\ &=(-1)^{1+1} \cos \alpha \cos \beta(\cos \alpha \cos \beta-0)+(-1)^{1+2} \cos \alpha \sin \beta(-\sin \beta \cos \alpha-0)+(-\text { 1) }^{1+3}(-\sin \alpha)\left(-\sin \alpha \sin ^{2} \beta-\sin \alpha \cos ^{2} \beta\right) \\ &=\cos \alpha \cos \beta(\cos \alpha \cos \beta-0)-\cos \alpha \sin \beta(-\sin \beta \cos \alpha-0)+(-\sin \alpha)\left(-\sin \alpha \sin ^{2} \beta-\right. \left.\sin \alpha \cos ^{2} \beta\right) \\ &=\cos ^{2} \alpha \cos ^{2} \beta+\cos ^{2} \alpha \sin ^{2} \beta+\sin ^{2} \alpha \sin ^{2} \beta+\sin ^{2} \alpha \cos ^{2} \beta \\ &=\cos ^{2} \alpha\left(\cos ^{2} \beta+\sin ^{2} \beta\right)+\sin ^{2} \alpha\left(\sin ^{2} \beta+\cos ^{2} \beta\right) \quad \quad \quad \quad \quad \quad \quad\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right] \\ &=\cos ^{2} \alpha+\sin ^{2} \alpha \\ &\Delta=1 \end{aligned}$

Determinants exercise 5.1 question 8

Answer:
$\begin{aligned} |A B|=|A||B| \\ \end{aligned}$
Given:

$\begin{aligned} & A=\left[\begin{array}{ll} 2 & 5 \\ 2 & 1 \end{array}\right], B=\left[\begin{array}{cc} 4 & -3 \\ 2 & 5 \end{array}\right] \\ & \text { To prove: }|A B|=|A||B| \end{aligned}$

Solution:
Consider AB in LHS:
$\begin{aligned} \mathrm{AB} &=\left[\begin{array}{ll} 2 & 5 \\ 2 & 1 \end{array}\right]\left[\begin{array}{cc} 4 & -3 \\ 2 & 5 \end{array}\right] \\ &=\left[\begin{array}{cc} 8+10 & -6+25 \\ 8+2 & -6+5 \end{array}\right] \\ &=\left[\begin{array}{cc} 18 & 19 \\ 10 & -1 \end{array}\right] \\ |A B| &=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21} \\ &=-18-190 \\ |A B| &=-208 \end{aligned}$
Consider RHS:
$\begin{aligned} &|A|=2-10=-8 \\ &|B|=20+6=26 \\ &\begin{array}{l} |A||B|=|-8||26| \\ =-208 \\ \therefore \text { LHS }=\text { RHS } \\ |A B|=|A||B| \text { is proved. } \end{array} \end{aligned}$

Determinants exercise 5.1 question 9

Answer:
$\begin{aligned} |3 A|=27|A| \\ \end{aligned}$
Given:
$\begin{aligned} \mathrm{A}=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right] \\ \text { To prove: }|3 A|=27|A| \end{aligned}$
Solution:
$\begin{aligned} &\mathrm{A}=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}\right], 3 \mathrm{~A}=\left[\begin{array}{llc} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array}\right] \\ &|3 A|=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21}+\mathrm{a}_{31} \mathrm{C}_{31} \\ &\quad=(-1)^{1+1} 3(36-0)+(-1)^{1+2} 0(0-0)+(-1)^{1+3} 3(0-0) \\ &\quad=3(36)+0+0 \\ &\quad=108 \quad \quad \quad \quad \quad \quad.....(1) \end{aligned}$
$\begin{aligned} &|A|=(-1)^{1+1} 1(4-0)+(-1)^{1+2} 0(0-0)+(-1)^{1+3} 1(0-0) \\ &27|A|=27(4+0+0) \\ &\quad=108 \quad \quad \quad \quad.....(2)\end{aligned} \\ F\! rom\; (1)\; and\; (2),\\ |3 A|=27|A| is proved.$

Determinants exercise 5.1 question 10 (i)

Answer:
$x=\pm \sqrt{3}$
Hint:
Take determinant of matrix on both side and equate it.
Given:
$\begin{aligned} &\left|\begin{array}{ll} 2 & 4 \\ 5 & 1 \end{array}\right|=\left|\begin{array}{ll} 2 x & 4 \\ 6 x & x \end{array}\right| \end{aligned}$
Solution:
$\begin{aligned} &\left|\begin{array}{ll} 2 & 4 \\ 5 & 1 \end{array}\right|=\left|\begin{array}{ll} 2 x & 4 \\ 6 x & x \end{array}\right| \\ &2-20=2 \mathrm{x}^{2}-24 \\ &-18=2 \mathrm{x}^{2}-24 \\ &2 \mathrm{x}^{2}=6 \\ &\mathrm{x}^{2}=3 \\ &\mathrm{x}=\pm \sqrt{3} \end{aligned}$

Determinants exercise 5.1 question 10 (ii)

Answer:
$x=2$
Hint:
Take determinant of matrix on both side and equate it.
Given:
$\begin{aligned} &\left|\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right|=\left|\begin{array}{cc} x & 3 \\ 2 x & 5 \end{array}\right| \end{aligned}$
Solution:
$\begin{aligned} &\left|\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}\right|=\left|\begin{array}{cc} x & 3 \\ 2 x & 5 \end{array}\right| \\ &10-12=5 x-6 x \\ &-2=-x \\ &x=2 \end{aligned}$

Determinants exercise 5.1 question 10 (iii)

Answer:
$x=\pm 2\sqrt{2}$
Hint:
Take determinant of matrix on both side and equate it.
Given:
$\begin{aligned} &\left|\begin{array}{ll} 3 & x \\ x & 1 \end{array}\right|=\left|\begin{array}{ll} 3 & 2 \\ 4 & 1 \end{array}\right| \\ \end{aligned}$
Solution:
$\begin{aligned} &\left|\begin{array}{ll} 3 & x \\ x & 1 \end{array}\right|=\left|\begin{array}{ll} 3 & 2 \\ 4 & 1 \end{array}\right| \\ &3-x^{2}=3-8 \\ &-x^{2}=-5-3 \\ &x^{2}=8 \\ &x=\pm 2 \sqrt{2} \end{aligned}$

Determinants exercise 5.1 question 10 (iv)

Answer:
$x=2$
Hint:
Take determinant of matrix on both side and equate it.
Given:
$\begin{aligned} &\left|\begin{array}{cc} 3 x & 7 \\ 2 & 4 \end{array}\right|=10 \\ \end{aligned}$
Solution:
$\begin{aligned} &\left|\begin{array}{cc} 3 x & 7 \\ 2 & 4 \end{array}\right|=10 \\ &12 \mathrm{x}-14=10 \\ &12 \mathrm{x}=24 \\ &\mathrm{x}=2 \end{aligned}$

Determinants exercise 5.1 question 10 (v)

Answer:
$x=2$
Hint:
Take determinant of matrix on both side and equate it.
Given:
$\begin{aligned} &\left|\begin{array}{ll} x+1 & x-1 \\ x-3 & x+2 \end{array}\right|=\left|\begin{array}{cc} 4 & -1 \\ 1 & 3 \end{array}\right| \end{aligned}$
Solution:
$\begin{aligned} &\left|\begin{array}{ll} x+1 & x-1 \\ x-3 & x+2 \end{array}\right|=\left|\begin{array}{cc} 4 & -1 \\ 1 & 3 \end{array}\right| \\ &(x+1)(x+2)-(x-3)(x-1)=12+1 \\ &x^{2}+3 x+2-x^{2}+4 x-3=13 \\ &7 x-1=13 \\ &7 x=14 \\ &x=2 \end{aligned}$

Determinants exercise 5.1 question 10 (vi)

Answer:
$x=\pm 3$
Hint:
Take determinant of matrix on both side and equate it.
Given:
$\begin{aligned} &\left|\begin{array}{ll} 2 x & 5 \\ 8 & x \end{array}\right|=\left|\begin{array}{ll} 6 & 5 \\ 8 & 3 \end{array}\right| \\ \end{aligned}$
Solution:
$\begin{aligned} &2 \mathrm{x}^{2}-40=18-40 \\ &2 \mathrm{x}^{2}=18 \\ &\mathrm{x}^{2}=9 \\ &\mathrm{x}=\pm 3 \end{aligned}$

Determinants exercise 5.1 question 11

Answer:
$x=2$
Hint:
Take determinant of matrix on both side and equate it.
Given:
$\left|\begin{array}{lll} x^{2} & x & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 4 \end{array}\right|=28$
Solution:
$\begin{aligned} &x^{2}(8-1)-x(0-3)+1(0-6)=28\\ &8 x^{2}-x^{2}+3 x-6=28\\ &7 x^{2}+3 x-34=0\\ &(7 x+17)(x-2)=0\\ &(\mathrm{x}=2) \text { or }\left(\mathrm{x}=\frac{-17}{7}\right)\\ &\mathrm{x}=\frac{-17}{7} \text { is not integer. So, } \mathrm{x}=2 \end{aligned}$

Determinants exercise 5.1 question 12 (i)

Answer:
$x=\frac{13}{15}$
Hint:
If matrix is singular, then |A| = 0
Given:
$\begin{bmatrix} 1+x &7 \\ 3-x &8 \end{bmatrix}$
Solution:
$\begin{aligned} &|\mathrm{A}|=0 \\ &8+8 \mathrm{x}-21+7 \mathrm{x}=0 \\ &15 \mathrm{x}-13=0 \\ &15 \mathrm{x}=13 \\ &\mathrm{x}=\frac{13}{15} \end{aligned}$

Determinants exercise 5.1 question 12 (ii)

Answer:
$x=2\; or\; x=-1$
Hint:
If matrix is singular, then |A| = 0
Given:
$\mathrm{A}=\left[\begin{array}{ccc} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{array}\right]$
Solution:
$\begin{aligned} &|\mathbf{A}|=\mathbf{0} \\ &(\mathrm{x}-1)\left(\mathrm{x}^{2}-2 \mathrm{x}\right)-1(\mathrm{x}-2)+1(2-\mathrm{x})=0 \\ &\mathrm{x}^{3}-2 \mathrm{x}^{2}-\mathrm{x}^{2}+2 \mathrm{x}-\mathrm{x}+2-\mathrm{x}+2=0 \\ &\mathrm{x}^{3}-3 \mathrm{x}^{2}+4=0 \\ &(\mathrm{x}-2)^{2}(\mathrm{x}+1)=0 \\ &\Rightarrow[\mathrm{x}=2] \text { or }[\mathrm{x}=-1] \\ &x=-1,2 \end{aligned}$

The 5th chapter, Determinants of Class 12 mathematics, is one of the most uncomplicated chapters to score good marks. Hence, a student needs to concentrate more on this chapter to use the opportunity to the fullest. The concepts in the first exercise, ex 5.1, are Minors and Cofactors of determinants, the Integral value of determinants; if the matrix is singular, how do we find the value of x and many more. There are about twenty-seven questions, including the subparts in this exercise. The RD Sharma Class 12 Chapter 5 Exercise 5.1 solutions book will make solving the problems more manageable.

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Chapter-wise RD Sharma Class 12 Solutions

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