NCERT Solutions Class 12 Maths (Determinants) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-4 (Determinants) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Question 1. Prove that the determinant $Miscellaneous Exercise on Chapter-4$ is

is independent of θ.

Solution:

A = $Miscellaneous Exercise on Chapter-4$

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 $Miscellaneous Exercise on Chapter-4$ = $Miscellaneous Exercise on Chapter-4$

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 $Miscellaneous Exercise on Chapter-4$

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 = $Miscellaneous Exercise on Chapter-4$ = 0 = 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 $Miscellaneous Exercise on Chapter-4$ == 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 $Miscellaneous Exercise on Chapter-4$ =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 = 2ab2c $\begin{vmatrix} a & c & a+c\\ a+b & b &a\\ 1 & 1 & 0 \end{vmatrix}$

Applying C2 ⇢ C2 – C1

A = 2ab2c $\begin{vmatrix} a & c-a & a+c\\ a+b & -a &a\\ 1 & 0 & 0 \end{vmatrix}$

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 = $Miscellaneous Exercise on Chapter-4$ and B = $Miscellaneous Exercise on Chapter-4$

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 = = $Miscellaneous Exercise on Chapter-4$ 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-1 = $1/13\begin{bmatrix} -14 & -11 & 5\\ -11 & -4 & 3\\ 5 & 3 & 1 \end{bmatrix}$

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

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

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

|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 $Miscellaneous Exercise on Chapter-4$

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 $Miscellaneous Exercise on Chapter-4$

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: $Miscellaneous Exercise on Chapter-4$ =

(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:

$Miscellaneous Exercise on Chapter-4$ =(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

$Miscellaneous Exercise on Chapter-4$ =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:

$Miscellaneous Exercise on Chapter-4$ = 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

$Miscellaneous Exercise on Chapter-4$ = 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

$Miscellaneous Exercise on Chapter-4$

(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 = $Miscellaneous Exercise on Chapter-4$ is
(A) $Miscellaneous Exercise on Chapter-4$
(B) xyz $Miscellaneous Exercise on Chapter-4$
(C) $Miscellaneous Exercise on Chapter-4$
(D) $Miscellaneous Exercise on Chapter-4$