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Class 12 NCERT Solutions- Mathematics Part I – Chapter 4 Determinants – Miscellaneous Exercises on Chapter 4
  • Last Updated : 30 Apr, 2021

Question 1. Prove that the determinant \begin{vmatrix} x & sin\theta  & cos\theta \\ -sin\theta  & -x & 1\\ cos\theta  & 1 & x \end{vmatrix}      is independent of θ.

Solution: 

A = \begin{vmatrix} x & sin\theta  & cos\theta \\ -sin\theta  & -x & 1\\ cos\theta  & 1 & x \end{vmatrix}

A = x(x2 – 1) – sinθ(-x sinθ – cosθ) + cosθ(-sinθ + x cosθ)

A = x3 – x + x sin2θ + sinθcosθ – sinθcosθ + x cos2θ

A = x3 – x + x(sin2θ + cos2θ)



A = x3 – x + x

A = x3(Independent of θ).

Hence, it is independent of θ

Question 2. Without expanding the determinant, prove that

\begin{vmatrix} a & a^{2} & bc\\ b & b^{2} & ca\\ c & c^{2} & ab \end{vmatrix}      \begin{vmatrix} 1 & a^{2} & a^{3}\\ 1 & b^{2} & b^{3}\\ 1 & c^{2} & c^{3} \end{vmatrix}

Solution: 

L.H.S. = \begin{vmatrix} a & a^{2} & bc\\ b & b^{2} & ca\\ c & c^{2} & ab \end{vmatrix}

\frac{1}{abc}\begin{vmatrix} a^{2} & a^{3} & abc\\ b^{2} & b^{3} & bca\\ c^{2} & c^{3} & cab \end{vmatrix}

\frac{1}{abc}.abc\begin{vmatrix} a^{2} & a^{3} & 1\\ b^{2} & b^{3} & 1\\ c^{2} & c^{3} & 1 \end{vmatrix}

(Taking abc out from C3)



\begin{vmatrix} a^{2} & a^{3} & 1\\ b^{2} & b^{3} & 1\\ c^{2} & c^{3} & 1 \end{vmatrix}

\begin{vmatrix} 1 & a^{2} & a^{3}\\ 1 & b^{2} & a^{3}\\ 1 & c^{2} & a^{3} \end{vmatrix}

(Applying column transformation between C1 and C3 and between C2 and C3)

= R.H.S.

Hence, it is proved that \begin{vmatrix} a & a^{2} & bc\\ b & b^{2} & ca\\ c & c^{2} & ab \end{vmatrix}      \begin{vmatrix} 1 & a^{2} & a^{3}\\ 1 & b^{2} & b^{3}\\ 1 & c^{2} & c^{3} \end{vmatrix}

Question 3. Evaluate \begin{vmatrix} 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{vmatrix}

Solution: 

A = \begin{vmatrix} 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{vmatrix}

Expanding along C3 

A = -sinα(-sinα sin2β – cos2β sinα) + cosα(cosα cos2β + cosα sin2β)

A = sin2α(sin2β + cos2β) + cos2α(cos2β + sin2β)



A = sin2(1) + cos2(1)

A = 1

Question 4. If a, b and c are real numbers, and Δ = \begin{vmatrix} b+c & c+a & a+b\\ c+a & a+b & b+c\\ a+b & b+c & c+a \end{vmatrix}     = 0

Show that either a + b + c = 0 or a = b = c

Solution:

 Δ = \begin{vmatrix} b+c & c+a & a+b\\ c+a & a+b & b+c\\ a+b & b+c & c+a \end{vmatrix}

Applying R1 ⇢ R1 + R2 + R3

 Δ = \begin{vmatrix} 2(a+b+c) & 2(a+b+c) & 2(a+b+c)\\ c+a & a+b & b+c\\ a+b & b+c & c+a \end{vmatrix}

= 2(a + b + c) \begin{vmatrix} 1 & 1 & 1\\ c+a & a+b & b+c\\ a+b & b+c & c+a \end{vmatrix}

Applying C2 ⇢ C2-C1 and C3 ⇢ C3 – C1

Δ = 2(a + b + c) \begin{vmatrix} 1 & 0 & 0\\ c+a & b-c & b-a\\ a+b & c-a & c-b \end{vmatrix}

Expanding along R1

Δ = 2(a + b + c)(1)[(b – c)(c – b) – (b – a)(c – a)]

= 2(a + b + c)[-b2 – c2 + 2bc – bc + ba + ac – a2]

= 2(a + b + c)[ab + bc + ca – a2 – b2 – c2]

According to the question Δ = 0

2(a + b + c)[ab + bc + ca – a2 – b2 – c2] = 0

From above, you can see that either a + b + c =0 or ab + bc + ca – a2 – b2 – c2 = 0

Now,

ab + bc + ca – a2 – b2 – c2 = 0

-2ab – 2bc – 2ac + 2a2 + 2b2 + 2c2 = 0

(a – b)2 + (b – c)2 + (c – a)2 = 0

(a – b)2 = (b – c)2 = (c – a)2 = 0  (because (a – b)2, (b – c)2, (c – a)2 are non negative)

(a – b) = (b – c) = (c – a) = 0

a = b = c

Hence, it is proved that if Δ = 0 then either a + b + c = 0 or a = b = c.

Question 5. Solve the equations \begin{vmatrix} x+a & x & x\\ x & x+a & x\\ x & x & x+a \end{vmatrix}     = 0, a ≠ 0

Solution:

\begin{vmatrix} x+a & x & x\\ x & x+a & x\\ x & x & x+a \end{vmatrix}     = 0

Applying R1 ⇢ R1 + R2 + R3

\begin{vmatrix} 3x+a & 3x+a & 3x+a\\ x & x+a & x\\ x & x & x+a \end{vmatrix}     = 0

(3x + a)\begin{vmatrix} 1 & 1 & 1\\ x & x+a & x\\ x & x & x+a \end{vmatrix}     = 0

Applying C2 ⇢ C2-C1 and C3 ⇢ C3 – C1

(3x + a)\begin{vmatrix} 1 & 0 & 0\\ x & a & 0\\ x & 0 & a \end{vmatrix}     = 0

Expanding along R1

(3x + a)[a2] = 0

a2(3x + a) = 0

But a ≠ 0

Therefore,

3x + a = 0

x = a/3

Question 6. Prove that \begin{vmatrix} a^{2} & bc & ac+c^{2}\\ a^{2}+ab & b^{2} &ac\\ ab & b^{2}+bc & c^{2} \end{vmatrix}     = 4a2b2c2

Solution:

A = \begin{vmatrix} a^{2} & bc & ac+c^{2}\\ a^{2}+ab & b^{2} &ac\\ ab & b^{2}+bc & c^{2} \end{vmatrix}

Taking out common factors a, b and c from C1, C2 and C3

A = abc \begin{vmatrix} a & c & a+c\\ a+b & b &c\\ b & b+c & c \end{vmatrix}

Applying R2 ⇢ R2 – R1 and R3 ⇢ R3 – R1

A = abc \begin{vmatrix} a & c & a+c\\ b & b-c &-c\\ b-a & b & -a \end{vmatrix}      

Applying R2 ⇢ R2 + R1

A = abc \begin{vmatrix} a & c & a+c\\ a+b & b &a\\ 2b & 2b & 0 \end{vmatrix}

A = 2ab2\begin{vmatrix} a & c & a+c\\ a+b & b &a\\ 1 & 1 & 0 \end{vmatrix}      

Applying C2 ⇢ C2 – C1

A = 2ab2\begin{vmatrix} a & c-a & a+c\\ a+b & -a &a\\ 1 & 0 & 0 \end{vmatrix}

Expanding along R3

A = 2ab2c[a(c – a) + a(a + c)]

= 2ab2c[ac – a2 + a2 + ac]

= 2ab2c(2ac)

= 4a2b2c2

Hence, it is proved.

Question 7. If A-1 =\begin{vmatrix} 3 & -1 & 1\\ -15 & 6 &-5\\ 5 & -2 & 2 \end{vmatrix}     and B =\begin{vmatrix} 1 & 2 & -2\\ -1 & 3 &0\\ 0 & -2 & 1 \end{vmatrix}     . Find (AB)-1

Solution:

|B| = 1(3 – 0) + 1(2 – 4) = 1

B11 = 3 – 0 = 3

B12 = 1

B13 = 2 – 0 = 2

B21 = -(2 – 4) = 2

B22 = 1 – 0 = 1

B23 = 2

B31 = 0 + 6 = 6

B32 = -(0 – 2) = 2

B33 = 3 + 2 = 5

adj B = \begin{bmatrix} 3 & 2 & 6\\ 1 & 1 & 2\\ 2 & 2 & 5 \end{bmatrix}

B-1 = (adj B)/|B|

B-1\begin{bmatrix} 3 & 2 & 6\\ 1 & 1 & 2\\ 2 & 2 & 5 \end{bmatrix}

Now,

(AB)-1 = B-1A-1

(AB)-1 \begin{bmatrix} 3 & 2 & 6\\ 1 & 1 & 2\\ 1 & 2 &5 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1\\ -15 & 6 & -5\\ 5 & -2 & 2 \end{bmatrix}

\begin{bmatrix} 9-30+30 & -3+12-12 & 3-10+12\\ 3-15+10 & -1+6-4 & 1-5+4\\ 6+12-10 & -2+12-10 & 2-10+10 \end{bmatrix}

(AB)-1\begin{bmatrix} 9 & -3 & 5\\ -2 & 1 & 0\\ 8 & 0 & 2 \end{bmatrix}

Question 8. Let A =\begin{bmatrix} 1 & -2 & 1\\ -2 & 3 & 1\\ 1 & 1 & 5 \end{bmatrix}     verify that

(i) [adj A]-1 = adj(A-1)

(ii) (A-1)-1 = A

Solution: 

A = \begin{bmatrix} 1 & -2 & 1\\ -2 & 3 & 1\\ 1 & 1 & 5 \end{bmatrix}

|A| = 1(15 – 1) + 2(-10 – 1) + 1(-2 – 3) = 14 – 27 = -13

A11 = 14

A12 = 11

A13 = -5

A21 = 11

A22 = 4

A23 = -3

A31 = -5

A32 = -3

A33 = -1

adj A = \begin{bmatrix} 14 & 11 & -5\\ 11 & 4 & -3\\ -5 & -3 & -1 \end{bmatrix}

Arrr-1 = (adj A)/|A|

-1/13\begin{bmatrix} 14 & 11 & -5\\ 11 & 4 & -3\\ -5 & -3 & -1 \end{bmatrix}

1/13\begin{bmatrix} -14 & -11 & 5\\ -11 & -4 & 3\\ 5 & 3 & 1 \end{bmatrix}

(i). |adj A| = 14(-4 – 9) – 11(-11 – 15) – 5(-33 + 20)

= 14(-13) – 11(-26) – 5(-13)

= -182 + 286 + 65 = 169

adj(adj A) = \begin{bmatrix} -13 & 26 & -13\\ 26 & -39 & -13\\ -13 & -13 & -65 \end{bmatrix}

[adj A]-1 = (adj(adj A))/|adj A|

1/169\begin{bmatrix} -13 & 26 & -13\\ 26 & -39 & -13\\ -13 & -13 & -65 \end{bmatrix}

1/13\begin{bmatrix} -1 & 2 & -1\\ 2 & -3 & -1\\ -1 & -1 & -5 \end{bmatrix}

Now, A-11/13\begin{bmatrix} -14 & -11 & 5\\ -11 & -4 & 3\\ 5 & 3 & 1 \end{bmatrix}

\begin{bmatrix} \frac{-14}{13} & \frac{-11}{13} & \frac{5}{13}\\ \frac{-11}{13} & \frac{-4}{13} & \frac{3}{13}\\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{bmatrix}

adj(A-1) = \begin{bmatrix} \frac{-4}{169}-\frac{9}{169} & -(\frac{-11}{169}-\frac{15}{169}) & \frac{-33}{169}+\frac{20}{169}\\ -(\frac{-11}{169}-\frac{-15}{169}) & \frac{-14}{169}-\frac{25}{169} &-(\frac{-42}{169}+\frac{55}{169}) \\ \frac{-33}{169}+\frac{20}{169} & -(\frac{-42}{169}+\frac{55}{169}) & \frac{56}{169}-\frac{121}{169} \end{bmatrix}

1/169\begin{bmatrix} -13 & 26 & -13\\ 26 & -39 & -13\\ -13 & -13 & -65 \end{bmatrix}

1/13\begin{bmatrix} -1 & 2 & -1\\ 2 & -3 & -1\\ -1 & -1 & -5 \end{bmatrix}

Hence, [adj A]-1 = adj(A-1)

(ii). A-11/13\begin{bmatrix} -14 & -11 & 5\\ -11 & -4 & 3\\ 5 & 3 & 1 \end{bmatrix}

adj A-11/13\begin{bmatrix} -1 & 2 & -1\\ 2 & -3 & -1\\ -1 & -1 & -5 \end{bmatrix}

|A-1| = (1/13)3[-14 × (-13) +11 × (-26) + 5 × (-13)]

= (1/13)3 × (-169)

= -1/13

Now, (A-1)-1 = (adj A-1)/|A-1|

\frac{1}{(\frac{-1}{13})}*\frac{1}{13}\begin{bmatrix} -1 & 2 & -1\\ 2 & -3 & -1\\ -1 & -1 & -5 \end{bmatrix}

\begin{bmatrix} 1 & -2 & 1\\ -2 & 3 & 1\\ 1 & 1 & 5 \end{bmatrix}

= A

Hence, it is proved that (A-1)-1 = A

Question 9. Evaluate \begin{vmatrix} x & y & x+y\\ y & x+y & x\\ x+y & x &y \end{vmatrix}

Solution: 

A = \begin{vmatrix} x & y & x+y\\ y & x+y & x\\ x+y & x &y \end{vmatrix}

Applying R1 -> R1+R2+R3

A = \begin{vmatrix} 2(x+y) & 2(x+y) & 2(x+y)\\ y & x+y & x\\ x+y & x &y \end{vmatrix}

= 2(x+y)\begin{vmatrix} 1 & 1 & 1\\ y & x+y & x\\ x+y & x &y \end{vmatrix}

Applying C2-> C2 – C1 and C3-> C3 – C1

A = 2(x + y)\begin{vmatrix} 1 & 0 & 0\\ y & x & x-y\\ x+y & -y &-x \end{vmatrix}

Expanding along R1

A = 2(x + y)[-x2 + y(x – y)]

= -2(x + y)(x2 + y2 – yx)

A = -2(x3 + y3)

Question 10. Evaluate \begin{vmatrix} 1 & x & y\\ 1 & x+y & y\\ 1 & x &x+y \end{vmatrix}

Solution:

A = \begin{vmatrix} 1 & x & y\\ 1 & x+y & y\\ 1 & x &x+y \end{vmatrix}

Applying R2->R2 – R1 and R3->R3 – R1

A = \begin{vmatrix} 1 & x & y\\ 0 & y & 0\\ 0 & 0 &x \end{vmatrix}

Expanding along C1

A = 1(xy – 0)

A = xy

Question 11. Using properties of determinants, prove that:

\begin{vmatrix} \alpha  & \alpha ^{2} & \beta +\gamma  \\ \beta  & \beta ^{2} & \gamma +\alpha \\ \gamma  & \gamma ^{2} &\alpha  +\beta   \end{vmatrix}   = (β – γ)(γ – α)(α – β)(α + β + γ)

Solution:

A = \begin{vmatrix} \alpha  & \alpha ^{2} & \beta +\gamma  \\ \beta  & \beta ^{2} & \gamma +\alpha \\ \gamma  & \gamma ^{2} &\alpha  +\beta   \end{vmatrix}

Applying R2->R2 – R1 and R3->R3 – R1

A = \begin{vmatrix} \alpha  & \alpha ^{2} & \beta +\gamma  \\ \beta-\alpha   & \beta ^{2}-\alpha ^{2} & \alpha-\beta \\ \gamma-\alpha   & \gamma ^{2}-\alpha ^{2} &\alpha-\gamma   \end{vmatrix}

A = (γ – α)(β – α)\begin{vmatrix} \alpha  & \alpha ^{2} & \beta +\gamma  \\ 1   & \beta +\alpha & -1 \\ 1   & \gamma +\alpha  &-1   \end{vmatrix}

Applying R3->R3 – R2

A = (γ – α)(β – α)\begin{vmatrix} \alpha  & \alpha ^{2} & \beta +\gamma  \\ 1   & \beta +\alpha & -1 \\ 0  & \gamma -\beta  &0   \end{vmatrix}

Expanding along R3

A = (γ – α)(β – α)[-(γ – β)(-α – β – γ)]

A = (γ – α)(β – α)(γ – β)(α + β + γ)

A = (β – γ)(γ – α)(α – β)(α + β + γ)

Hence, it is proved.

Question 12. Using the properties of determinants, prove that:

\begin{vmatrix} x & x^{2} & 1+px^{3}\\ y & y^{2} &1+py^{3} \\ z & z^{2} & 1+pz^{3} \end{vmatrix}   =(1 + pxyz)(x – y)(y – z)(z – x)

Solution:

 A = \begin{vmatrix} x & x^{2} & 1+px^{3}\\ y & y^{2} &1+py^{3} \\ z & z^{2} & 1+pz^{3} \end{vmatrix}

Applying R2->R2 – R1 and R3-> R3 – R1

A = \begin{vmatrix} x & x^{2} & 1+px^{3}\\ y-x & y^{2}-x^{2} &p(y^{3}-x^{3}) \\ z-x & z^{2}-x^{2} & p(z^{3}-x^{3}) \end{vmatrix}

A = (y – x)(z – x)\begin{vmatrix} x & x^{2} & 1+px^{3}\\ 1 & y+x &p(y^{2}+x^{2}+xy) \\ 1 & z+x & p(z^{2}+x^{2}+xz) \end{vmatrix}

Applying R3->R3 – R2

A = (y – x)(z – x)\begin{vmatrix} x & x^{2} & 1+px^{3}\\ 1 & y+x &p(y^{2}+x^{2}+xy) \\ 0 & z-y & p(z-y)(x+y+z) \end{vmatrix}

A = (y – x)(z – x)(z – y)\begin{vmatrix} x & x^{2} & 1+px^{3}\\ 1 & y+x &p(y^{2}+x^{2}+xy) \\ 0 & 1 & p(x+y+z) \end{vmatrix}

Expanding along R3

A = (x – y)(y – z)(z – x)[(-1)(p)(xy2 + x3 + x2y) + 1 + px3 + p(x + y + z)(xy)]

= (x – y)(y – z)(z – x)[-pxy2 – px3 – px2y + 1 + px3 + px2y + pxy2 + pxyz]

= (x – y)(y – z)(z – x)(1 + pxyz)

Hence it is proved.

Question 13. using properties of determinants, prove that

\begin{vmatrix} 3a & -a+b & -a+c\\ -b+a & 3b & -b+c\\ -c+a & -c+b & 3c \end{vmatrix}   = 3(a + b + c)(ab + bc + ca)

Solution:

A = \begin{vmatrix} 3a & -a+b & -a+c\\ -b+a & 3b & -b+c\\ -c+a & -c+b & 3c \end{vmatrix}

Applying C1->C1 + C2 + C3

A = \begin{vmatrix} a+b+c & -a+b & -a+c\\ a+b+c & 3b & -b+c\\ a+b+c & -c+b & 3c \end{vmatrix}

A = (a + b + c)\begin{vmatrix} 1 & -a+b & -a+c\\ 1 & 3b & -b+c\\ 1 & -c+b & 3c \end{vmatrix}

Applying R2->R2 – R1 and R3 ->R3 – R1

A = (a + b + c)\begin{vmatrix} 1 & -a+b & -a+c\\ 0 & 2b+a & a-b\\ 0 & a-c & 2c+a \end{vmatrix}

Expanding along C1

A = (a + b + c)[(2b + a)(2c + a) – (a – b)(a – c)]

= (a + b + c)[4bc + 2ab + 2ac + a2 – a2 + ac + ba – bc]

=(a + b + c)(3ab + 3bc + 3ac)

A = 3(a + b + c)(ab + bc + ca)

Hence, it is proved.

Question 14. Using properties of determinants, prove that:

\begin{vmatrix} 1 & 1+p & 1+p+q\\ 2 & 3+2p & 4+3p+2q\\ 3 & 6+3p & 10+6p+3q \end{vmatrix}   = 1

Solution:

A = \begin{vmatrix} 1 & 1+p & 1+p+q\\ 2 & 3+2p & 4+3p+2q\\ 3 & 6+3p & 10+6p+3q \end{vmatrix}

Applying R2->R2 – 2R1 and R3->R3 – 3R1

A = \begin{vmatrix} 1 & 1+p & 1+p+q\\ 0& 1 & 2+p\\ 0 & 3 & 7+3p \end{vmatrix}

Applying R3->R3 – 3R2

\begin{vmatrix} 1 & 1+p & 1+p+q\\ 0& 1 & 2+p\\ 0 & 0 & 1 \end{vmatrix}

Expanding along C1

A = 1(1 – 0)

A = 1

Hence, it is proved.

Question 15. Using properties of determinants, prove that

\begin{vmatrix} sin\alpha  & cos\alpha  &cos(\alpha +\delta ) \\ sin\beta  & cos\beta  & cos(\beta +\delta )\\ sin\gamma  & cos\gamma  & cos(\gamma +\delta ) \end{vmatrix}   = 0

Solution:

A = \begin{vmatrix} sin\alpha  & cos\alpha  &cos(\alpha +\delta ) \\ sin\beta  & cos\beta  & cos(\beta +\delta )\\ sin\gamma  & cos\gamma  & cos(\gamma +\delta ) \end{vmatrix}

A = \frac{1}{sin\delta  cos\delta }\begin{vmatrix} sin\alpha sin\delta   & cos\alpha cos\delta  &cos\alpha cos\delta -sin\alpha sin\delta  \\ sin\beta sin\delta   & cos\beta cos\delta & cos\beta cos\delta-sin\beta sin\delta \\ sin\gamma sin\delta  & cos\gamma cos\delta & cos\gamma cos\delta -sin\gamma sin\delta   \end{vmatrix}

Applying C1->C1 + C3

A = \frac{1}{sin\delta  cos\delta }\begin{vmatrix} cos\alpha cos\delta   & cos\alpha cos\delta  &cos\alpha cos\delta -sin\alpha sin\delta  \\ cos\beta cos\delta   & cos\beta cos\delta & cos\beta cos\delta-sin\beta sin\delta \\ cos\gamma cos\delta  & cos\gamma cos\delta & cos\gamma cos\delta -sin\gamma sin\delta   \end{vmatrix}

from above, you can see that two columns C1 and C2 are identical.

Hence A = 0

Hence, it is proved.

Question 16. Solve the system of the following questions:

2/x + 3/y + 10/z = 4

4/x – 6/y + 5/z = 1

6/x + 9/y – 20/z = 2

Solution:

Assume 1/x = p ; 1/y = q; 1/z = r

then. the above equations will be like

2p + 3Q + 10r = 4

4p – 6q + 5r = 1

6p + 9q – 20r = 2

This can be written in the form of AX=B

where,

A = \begin{bmatrix} 2 & 3 & 10\\ 4 & -6 & 5\\ 6 & 9 & -20 \end{bmatrix}

X = \begin{bmatrix} p\\ q\\ r \end{bmatrix}

B = \begin{bmatrix} 4\\ 1\\ 2 \end{bmatrix}

We have,

|A| = 2(120 – 45) – 3(-80 – 30) + 10(36 + 36)

|A| = 150 + 330 + 720

|A| = 1200 ≠ 0

Hence A is invertible matrix.

A11 = 75

A12 = 110

A13 = 72

A21 = 150

A22 = -100

A23 = 0

A31 = 75

A32 = 30

A33 = -24

A-1 = (adj A)/|A|

A-1\frac{1}{1200}\begin{bmatrix} 75 & 150 & 75\\ 110 & -100 & 30\\ 72 & 0 & -24 \end{bmatrix}

Now,

X = A-1B

\begin{bmatrix} p\\ q\\ r \end{bmatrix}   \frac{1}{1200}\begin{bmatrix} 75 & 150 & 75\\ 110 & -100 & 30\\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4\\ 1\\ 2 \end{bmatrix}

\begin{bmatrix} p\\ q\\ r \end{bmatrix}   \frac{1}{1200}\begin{bmatrix} 300+150+150\\ 440-100+60\\ 288+0-48 \end{bmatrix}

\frac{1}{1200}\begin{bmatrix} 600\\ 400\\ 240 \end{bmatrix}

\begin{bmatrix} \frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5} \end{bmatrix}

From above p = 1/2; q = 1/3 ; r = 1/5

So, x = 2; y = 3; z = 5

Question 17. Choose the correct answer.

If a, b, c are in A.P. then the determinant

\begin{vmatrix} x+2 & x+3 &x+2a  \\ x+3 & x+4 & x+2b\\ x+4 & x+5 & x+2c \end{vmatrix}

(A) 0                                    (B) 1

(C) x                                     (D) 2x

Solution:

A = \begin{vmatrix} x+2 & x+3 &x+2a  \\ x+3 & x+4 & x+2b\\ x+4 & x+5 & x+2c \end{vmatrix}

a, b and c are in A.P So, 2b = a + c

A = \begin{vmatrix} x+2 & x+3 &x+2a  \\ x+3 & x+4 & x+a+c\\ x+4 & x+5 & x+2c \end{vmatrix}

Applying R1->R1 – R2 and R3->R3 – R2

A = \begin{vmatrix} -1 & -1 &a-c  \\ x+3 & x+4 & x+a+c\\ 1 & 1 & c-a \end{vmatrix}

Applying R1->R1 + R3

A = \begin{vmatrix} 0 & 0 &0  \\ x+3 & x+4 & x+a+c\\ 1 & 1 & c-a \end{vmatrix}

All the elements in the first row are 0.

Hence A = 0

So, the correct answer is A.

Question 18. Choose the correct answer.

If x, y, z are non-zero real numbers, then the inverse of matrix A = \begin{bmatrix} x & 0 & 0\\ 0 & y &0 \\ 0 & 0 & z \end{bmatrix}    is 

(A) \begin{bmatrix} x^{-1} & 0 & 0\\ 0 & y^{-1} &0 \\ 0 & 0 & z^{-1} \end{bmatrix}

(B) xyz\begin{bmatrix} x^{-1} & 0 & 0\\ 0 & y^{-1} &0 \\ 0 & 0 & z^{-1} \end{bmatrix}

(C) \frac{1}{xyz}\begin{bmatrix} x & 0 & 0\\ 0 & y &0 \\ 0 & 0 & z \end{bmatrix}

(D) \frac{1}{xyz}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}

Solution:

A = \begin{bmatrix} x & 0 & 0\\ 0 & y &0 \\ 0 & 0 & z \end{bmatrix}

|A| = x(yz – 0) = xyz ≠ 0

Hence, the matrix is invertible

Now,

A11 = yz

A12 = 0

A13 = 0

A21 = 0

A22 = xz

A23 = 0

A31 = 0

A32 = 0

A33 = xy

adj A = \begin{bmatrix} yz & 0 & 0\\ 0 & xz &0 \\ 0 & 0 & xy \end{bmatrix}

A-1 = (adj A)/|A|

A-1\frac{1}{xyz}\begin{bmatrix} yz & 0 & 0\\ 0 & xz &0 \\ 0 & 0 & xy \end{bmatrix}

A-1 \begin{bmatrix} \frac{yz}{xyz} & 0 & 0\\ 0 & \frac{xz}{xyz} &0 \\ 0 & 0 & \frac{xy}{xyz} \end{bmatrix}

A-1\begin{bmatrix} \frac{1}{x} & 0 & 0\\ 0 & \frac{1}{y} &0 \\ 0 & 0 & \frac{1}{z} \end{bmatrix}

A-1 \begin{bmatrix} x^{-1} & 0 & 0\\ 0 & y^{-1} &0 \\ 0 & 0 & z^{-1} \end{bmatrix}

Hence, the correct answer is A.

Question 19. Choose the correct answer

Let A = \begin{bmatrix} 1 & sin\theta  & 1\\ -sin\theta & 1 & sin\theta\\ -1 & -sin\theta & 1 \end{bmatrix}   , where 0 ≤ θ ≤ 2π, then

(A) Det(A) = 0                                      (B) Det(A) ∈ (2, ∞)

(C) Det(A) ∈ (2, 4)                              (D) Det(A) ∈ [2, 4]

Solution:

A = \begin{bmatrix} 1 & sin\theta  & 1\\ -sin\theta & 1 & sin\theta\\ -1 & -sin\theta & 1 \end{bmatrix}

|A| = 1(1 + sin2θ) – sinθ(-sinθ + sinθ) + 1(sin2θ + 1)

|A| = 1 + sin2θ + sin2θ + 1

= 2 + 2 sin2θ

= 2(1 + sin2θ)

Now 0 ≤ θ ≤ 2π

So, 0 ≤ sinθ ≤ 1

0 ≤ sin2θ ≤ 1

0 + 1 ≤ 1 + sin2θ ≤ 1 + 1

2 ≤ 2(1 + sin2θ) ≤ 4

Det(A) ∈ [2, 4]

Hence, the correct answer is D.

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