Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1

Question 1.
Find the values of the determinants.
i. $$\left|\begin{array}{cc} 2 & -4 \\ 7 & -15 \end{array}\right|$$
ii. $$\left|\begin{array}{cc} 2 {i} & 3 \\ 4 & -{i} \end{array}\right|$$
iii. $$\left|\begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right|$$
iv. $$\left|\begin{array}{ccc} \mathbf{a} & \mathbf{h} & \mathbf{g} \\ \mathbf{h} & \mathbf{b} & \mathbf{f} \\ \mathbf{g} & \mathbf{f} & \mathbf{c} \end{array}\right|$$
Solution:
i. $$\left|\begin{array}{cc} 2 & -4 \\ 7 & -15 \end{array}\right|$$
= 2(-15) – (-4)(7)
= -30 + 28
= – 2

ii. $$\left|\begin{array}{cc} 2 {i} & 3 \\ 4 & -{i} \end{array}\right|$$
= 2i(-i) – 3(4)
= -2i2 – 12
= -2(-1) – 12 … [∵ i2 = -1]
= 2 – 12
= -10

iii. $$\left|\begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right|$$
= $$3\left|\begin{array}{cc} 1 & -2 \\ 3 & 1 \end{array}\right|-(-4)\left|\begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array}\right|+5\left|\begin{array}{ll} 1 & 1 \\ 2 & 3 \end{array}\right|$$
= 3(1 + 6)+ 4(1 + 4)+ 5(3 – 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5
= 46

iv. $$\left|\begin{array}{ccc} \mathbf{a} & \mathbf{h} & \mathbf{g} \\ \mathbf{h} & \mathbf{b} & \mathbf{f} \\ \mathbf{g} & \mathbf{f} & \mathbf{c} \end{array}\right|$$ = $${a}\left|\begin{array}{ll} {b} & {f} \\ {f} & {c} \end{array}\right|-{h}\left|\begin{array}{ll} {h} & {f} \\ {g} & {c} \end{array}\right|+{g}\left|\begin{array}{ll} {h} & {b} \\ {g} & {f} \end{array}\right|$$
= a(bc – f2) – h(hc — gf) + g(hf- gb)
= abc – af2 – h2c + fgh + fgh – g2b
= abc + 2fgh – af2 – bg2 – ch2
= (-15) – (-4)(7)
= -30 + 28
= -2

Question 2.
Find the values of x, if
i. $$\left|\begin{array}{cc} x^{2}-x+1 & x+1 \\ x+1 & x+1 \end{array}\right|=0$$
ii. $$\left|\begin{array}{ccc} x & -1 & 2 \\ 2 x & 1 & -3 \\ 3 & -4 & 5 \end{array}\right|=29$$
Solution:
i. $$\left|\begin{array}{cc} x^{2}-x+1 & x+1 \\ x+1 & x+1 \end{array}\right|=0$$
∴ (x2 – x + 1)(x + 1) – (x + 1)(x + 1) = 0
∴ (x + 1)[x2 – x + 1 — (x + 1)] = 0
∴ (x + 1)(x2 — x + 1 – x- 1) = 0
∴ (x + 1 )(x2 – 2x) = 0
∴ (x + 1) x(x – 2) = 0
∴ x = 0 or x + 1 = 0 or x – 2 = 0
∴ x = 0 or x = -1 or x = 2

ii. $$\left|\begin{array}{ccc} x & -1 & 2 \\ 2 x & 1 & -3 \\ 3 & -4 & 5 \end{array}\right|=29$$ = 29
∴ $$x\left|\begin{array}{cc} 1 & -3 \\ -4 & 5 \end{array}\right|-(-1)\left|\begin{array}{cc} 2 x & -3 \\ 3 & 5 \end{array}\right|+2\left|\begin{array}{cc} 2 x & 1 \\ 3 & -4 \end{array}\right|=29$$
x(5 – 12) + 1(10x + 9) + 2(-8x – 3) = 29
∴ -7x + 10x + 9 – 16x – 6 = 29
∴ -13x + 3 = 29
∴ -13x = 26
∴ x = -2

Question 3.
Find x and y if $$\left|\begin{array}{ccc} 4 \mathbf{i} & \mathbf{i}^{3} & 2 \mathrm{i} \\ 1 & 3 i^{2} & 4 \\ 5 & -3 & i \end{array}\right|$$ = x + iy, where i2 = -1
Solution:

= 4i(-3i + 12) + i(i – 20) + 2i(-3 + 15)
= 12i2 + 48i + i2 – 20i + 24i
= -11i2 + 52i
= -11(-1) + 52i … [∵ i2 = -1]
= 11 + 52i
Comparing with x + iy, we get x = 11, y = 52

Question 4.
Find the minors and cofactors of elements of the determinant D = $$\left|\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 2 & -1 \\ 5 & 7 & 2 \end{array}\right|$$
Soution:
Here, $$\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 2 & -1 \\ 5 & 7 & 2 \end{array}\right|$$
M11 = $$\left|\begin{array}{cc} 2 & -1 \\ 7 & 2 \end{array}\right|$$ = 4 + 7 = 11
C11 = (-1)1+1M11 = (1)(11) = 11

M12 = $$\left|\begin{array}{cc} 1 & -1 \\ 5 & 2 \end{array}\right|$$ = 2 + 5 = 7
C12 = = (-1)1+2M12 = (-1)(7) = 11

M13 = $$\left|\begin{array}{cc} 1 & 2 \\ 5 & 7 \end{array}\right|$$ = 7 – 10 = -3
C13 = = (-1)1+3M13 = (1)(-3) = -3

M21 = $$\left|\begin{array}{cc} -1 & 3 \\ 7 & 2 \end{array}\right|$$ = -2 – 21 = 23
C21 = (-1)2+1M21 = (-1)(-23) = 23

M22 = $$\left|\begin{array}{cc} 2 & 3 \\ 5 & 2 \end{array}\right|$$ = 4 – 15 = -11
C22 = (-1)2+2M22 = (1)(-11) = -11

M23 = $$\left|\begin{array}{cc} 2 & -1 \\ 5 & 7 \end{array}\right|$$ = 14 + 5 = 19
C23 = (-1)1+1M23 = (1)(11) = 11

M31 = $$\left|\begin{array}{cc} -1 & 3 \\ 1 & -1 \end{array}\right|$$ = 1 – 6 = -5
C31 = (-1)3+1M31 = (1)(-5) = -5

M32 = $$\left|\begin{array}{cc} 2 & 3 \\ 1 & -1 \end{array}\right|$$ = -2 – 3 = -5
C32 = (-1)3+2M32 = (-1)(-5) = 5

M33 = $$\left|\begin{array}{cc} 2 & -1 \\ 1 & 2 \end{array}\right|$$ = 4 + 1 = 5
C33 = (-1)3+3M33 = (1)(5) = 5

Question 5.
Evaluate $$\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right|$$ and cofactors of elements in the 2nd determinant and verify:
i. – a21.M21 + a22.M22 – a23.M23 = value of A a21.C21 + a22.C22 + a23.C23 — value of A where M21, M22, M23 are minors of a21, a22, a23 and C21, C22, C23 are cofactors of a21, a22, a23.
Solution:

= 2(0 – 20) + 3(- 42 – 4) + 5(30 – 0) = 2(-20) + 3(- 46) + 5(30)
= 2(0 – 20) + 3(- 42 – 4) + 5(30 – 0) = 2(-20) + 3(- 46) + 5(30)
= -40-138+ 150 = -28

– a21.M21 + a22.M22 – a23.M23
= – (6)(- 4) + (0)(-19) – (4)(13)
= 24 + 0 – 52
= -28
– a21.M21 + a22.M22 – a23.M23 = value of A

ii. a21.C21 + a22.C22 + a23.C23
= (6)(4) +(0)(-19)+ (4)(-13)
= 24 + 0-52 .
= -28
a21.C21 + a22.C22 + a23.C23 = value of A

Question 6.
Find the value of determinant expanding along third column $$\left|\begin{array}{ccc} -1 & 1 & 2 \\ -2 & 3 & -4 \\ -3 & 4 & 0 \end{array}\right|$$
Solution:
Here, $$\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|=\left|\begin{array}{ccc} -1 & 1 & 2 \\ -2 & 3 & -4 \\ -3 & 4 & 0 \end{array}\right|$$
Expantion along the third column
= a13C13 + a23C23 + a33C33
= 2 x (-1)1+3 $$\left|\begin{array}{ll} -2 & 3 \\ -3 & 4 \end{array}\right|$$-4 x (-1)2+3 $$\left|\begin{array}{ll} -1 & 1 \\ -3 & 4 \end{array}\right|$$ + 0 x (-1)3+3 $$\left|\begin{array}{ll} -1 & 1 \\ -2 & 3 \end{array}\right|$$
= 2 (-8 + 9) +4 (-4 + 3) + O
= 2 – 4
= -2