RD Sharma Solutions are primarily designed for CBSE students and are based on the latest syllabus prescribed as per the CCE guidelines by the CBSE Board. The practice is an essential task to learn and score well in Mathematics. Hence the solutions are designed by BYJU’S experts to boost confidence among students in understanding the concepts covered in this chapter. The PDF of RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants is provided here. The students are required to go through RD Sharma Solutions thoroughly before the final exams to score well and intensify their problem-solving abilities. This chapter consists of five exercises and explains the concepts of determinants and its properties.

Let us have a look at some of the important concepts that are discussed in this chapter.

- Definition of determinants
- The determinant of a square matrix of order 1, 2 and 3
- The determinant of a square matrix of order 3 by using Sarrus diagram
- Definition and meaning of singular matrix
- Minors and cofactors of determinants
- Properties of determinants
- Evaluation of determinants
- Evaluation of determinants by using the properties of determinants and proving identities
- The solution of determinant equations
- Addition of determinants
- Evaluation of determinants by using factor theorem
- Applications of determinants to coordinate geometry
- Area of triangle
- Conditions of collinearity of three points
- Equation of a line passing through two given points
- Applications of determinants in solving a system of linear equations
- The solution of a non-homogeneous system of linear equations
- Condition for consistency
- Solutions of a homogeneous system of linear equations

**RD Sharma Solutions For Class 12 Maths Chapter 6 Determinants:-**Download PDF Here

### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 6 – Determinants

Exercise 6.1 Page No: 6.10

**1. Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:**

**Solution:**

(i) Let M_{ij}Â and C_{ij}Â represents the minor and coâ€“factor of an element, where i and j represent the row and column.The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_{ij}Â = (â€“1)^{i+j}Â Ã— M_{ij}

Given,

From the given matrix we have,

M_{11}Â = â€“1

M_{21}Â = 20

C_{11}Â = (â€“1)^{1+1}Â Ã— M_{11}

= 1 Ã— â€“1

= â€“1

C_{21}Â = (â€“1)^{2+1}Â Ã— M_{21}

= 20 Ã— â€“1

= â€“20

Now expanding along the first column we get

|A| = a_{11}Â Ã— C_{11}Â + a_{21}Ã— C_{21}

= 5Ã— (â€“1) + 0 Ã— (â€“20)

= â€“5

(ii) Let M_{ij}Â and C_{ij}Â represents the minor and coâ€“factor 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. Then finding the absolute value of the matrix newly formed.

Also, C_{ij}Â = (â€“1)^{i+j}Â Ã— M_{ij}

Given

From the above matrix we have

M_{11}Â = 3

M_{21}Â = 4

C_{11}Â = (â€“1)^{1+1}Â Ã— M_{11}

= 1 Ã— 3

= 3

C_{21}Â = (â€“1)^{2+1}Â Ã— 4

= â€“1 Ã— 4

= â€“4

Now expanding along the first column we get

|A| = a_{11}Â Ã— C_{11}Â + a_{21}Ã— C_{21}

= â€“1Ã— 3 + 2 Ã— (â€“4)

= â€“11

(iii) Let M_{ij}Â and C_{ij}Â represents the minor and coâ€“factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_{ij}Â = (â€“1)^{i+j}Â Ã— M_{ij}

Given,

M_{31}Â = â€“3 Ã— 2 â€“ (â€“1) Ã— 2

M_{31}Â = â€“4

C_{11}Â = (â€“1)^{1+1}Â Ã— M_{11}

= 1 Ã— â€“12

= â€“12

C_{21}Â = (â€“1)^{2+1}Â Ã— M_{21}

= â€“1 Ã— â€“16

= 16

C_{31}Â = (â€“1)^{3+1}Â Ã— M_{31}

= 1 Ã— â€“4

= â€“4

Now expanding along the first column we get

|A| = a_{11}Â Ã— C_{11}Â + a_{21}Ã— C_{21}+ a_{31}Ã— C_{31}

= 1Ã— (â€“12) + 4 Ã— 16 + 3Ã— (â€“4)

= â€“12 + 64 â€“12

= 40

(iv) Let M_{ij}Â and C_{ij}Â represents the minor and coâ€“factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_{ij}Â = (â€“1)^{i+j}Â Ã— M_{ij}

Given,

M_{31}Â = a Ã— c a â€“ b Ã— bc

M_{31}Â = a^{2}c â€“ b^{2}c

C_{11}Â = (â€“1)^{1+1}Â Ã— M_{11}

= 1 Ã— (ab^{2}Â â€“ ac^{2})

= ab^{2}Â â€“ ac^{2}

C_{21}Â = (â€“1)^{2+1}Â Ã— M_{21}

= â€“1 Ã— (a^{2}b â€“ c^{2}b)

= c^{2}b â€“ a^{2}b

C_{31}Â = (â€“1)^{3+1}Â Ã— M_{31}

= 1 Ã— (a^{2}c â€“ b^{2}c)

= a^{2}c â€“ b^{2}c

Now expanding along the first column we get

|A| = a_{11}Â Ã— C_{11}Â + a_{21}Ã— C_{21}+ a_{31}Ã— C_{31}

= 1Ã— (ab^{2}Â â€“ ac^{2}) + 1 Ã— (c^{2}b â€“ a^{2}b) + 1Ã— (a^{2}c â€“ b^{2}c)

= ab^{2}Â â€“ ac^{2}Â + c^{2}b â€“ a^{2}b + a^{2}c â€“ b^{2}c

(v) Let M_{ij}Â and C_{ij}Â represents the minor and coâ€“factor 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. Then finding the absolute value of the matrix newly formed.

Also, C_{ij}Â = (â€“1)^{i+j}Â Ã— M_{ij}

Given,

M_{31}Â = 2Ã—0 â€“ 5Ã—6

M_{31}Â = â€“30

C_{11}Â = (â€“1)^{1+1}Â Ã— M_{11}

= 1 Ã— 5

= 5

C_{21}Â = (â€“1)^{2+1}Â Ã— M_{21}

= â€“1 Ã— â€“40

= 40

C_{31}Â = (â€“1)^{3+1}Â Ã— M_{31}

= 1 Ã— â€“30

= â€“30

Now expanding along the first column we get

|A| = a_{11}Â Ã— C_{11}Â + a_{21}Ã— C_{21}+ a_{31}Ã— C_{31}

= 0Ã— 5 + 1 Ã— 40 + 3Ã— (â€“30)

= 0 + 40 â€“ 90

= 50

(vi) Let M_{ij}Â and C_{ij}Â represents the minor and coâ€“factor 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. Then finding the absolute value of the matrix newly formed.

Also, C_{ij}Â = (â€“1)^{i+j}Â Ã— M_{ij}

Given,

M_{31}Â = h Ã— f â€“ b Ã— g

M_{31}Â = hf â€“ bg

C_{11}Â = (â€“1)^{1+1}Â Ã— M_{11}

= 1 Ã— (bcâ€“ f^{2})

= bcâ€“ f^{2}

C_{21}Â = (â€“1)^{2+1}Â Ã— M_{21}

= â€“1 Ã— (hc â€“ fg)

= fg â€“ hc

C_{31}Â = (â€“1)^{3+1}Â Ã— M_{31}

= 1 Ã— (hf â€“ bg)

= hf â€“ bg

Now expanding along the first column we get

|A| = a_{11}Â Ã— C_{11}Â + a_{21}Ã— C_{21}+ a_{31}Ã— C_{31}

= aÃ— (bcâ€“ f^{2}) + hÃ— (fg â€“ hc) + gÃ— (hf â€“ bg)

= abcâ€“ af^{2}Â + hgf â€“ h^{2}c +ghf â€“ bg^{2}

(vii) Let M_{ij}Â and C_{ij}Â represents the minor and coâ€“factor 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. Then finding the absolute value of the matrix newly formed.

Also, C_{ij}Â = (â€“1)^{i+j}Â Ã— M_{ij}

Given,

M_{31}Â = â€“1(1 Ã— 0 â€“ 5 Ã— (â€“2)) â€“ 0(0 Ã— 0 â€“ (â€“1) Ã— (â€“2)) + 1(0 Ã— 5 â€“ (â€“1) Ã— 1)

M_{31}Â = â€“9

M_{41}Â = â€“1(1Ã—1 â€“ (â€“1) Ã— (â€“2)) â€“ 0(0 Ã— 1 â€“ 1 Ã— (â€“2)) + 1(0 Ã— (â€“1) â€“ 1 Ã— 1)

M_{41}Â = 0

C_{11}Â = (â€“1)^{1+1}Â Ã— M_{11}

= 1 Ã— (â€“9)

= â€“9

C_{21}Â = (â€“1)^{2+1}Â Ã— M_{21}

= â€“1 Ã— 9

= â€“9

C_{31}Â = (â€“1)^{3+1}Â Ã— M_{31}

= 1 Ã— â€“9

= â€“9

C_{41}Â = (â€“1)^{4+1}Â Ã— M_{41}

= â€“1 Ã— 0

= 0

Now expanding along the first column we get

|A| = a_{11}Â Ã— C_{11}Â + a_{21}Ã— C_{21}+ a_{31}Ã— C_{31}Â + a_{41}Ã— C_{41}

= 2 Ã— (â€“9) + (â€“3) Ã— â€“9 + 1 Ã— (â€“9) + 2 Ã— 0

= â€“ 18 + 27 â€“9

= 0

**2. Evaluate the following determinants:**

**Solution:**

(i) Given

â‡’Â |A| = x (5x + 1) â€“ (â€“7) x

|A| = 5x^{2}Â + 8x

(ii) Given

â‡’Â |A| = cos Î¸ Ã— cos Î¸ â€“ (â€“sin Î¸) x sin Î¸

|A| = cos^{2}Î¸ + sin^{2}Î¸

We know that cos^{2}Î¸ + sin^{2}Î¸ = 1

|A| = 1

(iii) Given

â‡’Â |A| = cos15Â° Ã— cos75Â° + sin15Â° x sin75Â°

We know that cos (A â€“ B) = cos A cos B + Sin A sin B

By substituting this we get, |A| = cos (75 â€“ 15)Â°

|A| = cos60Â°

|A| = 0.5

(iv) Given

â‡’Â |A| = (a + ib) (a â€“ ib) â€“ (c + id) (â€“c + id)

= (a + ib) (a â€“ ib) + (c + id) (c â€“ id)

= a^{2}Â â€“ i^{2}Â b^{2}Â + c^{2}Â â€“ i^{2}Â d^{2}

We know that i^{2} = -1

= a^{2}Â â€“ (â€“1) b^{2}Â + c^{2}Â â€“ (â€“1) d^{2}

= a^{2}Â + b^{2}Â + c^{2}Â + d^{2}

**3. Evaluate:**

**Solution:**

Since |AB|= |A||B|

= 2(17 Ã— 12 â€“ 5 Ã— 20) â€“ 3(13 Ã— 12 â€“ 5 Ã— 15) + 7(13 Ã— 20 â€“ 15 Ã— 17)

= 2 (204 â€“ 100) â€“ 3 (156 â€“ 75) + 7 (260 â€“ 255)

= 2Ã—104 â€“ 3Ã—81 + 7Ã—5

= 208 â€“ 243 +35

= 0

Now |A|^{2}Â = |A|Ã—|A|

|A|^{2}= 0

**4. Show that **

**Solution:**

Given

Let the given determinant as A

Using sin (A+B) = sin A Ã— cos B + cos A Ã— sin B

â‡’Â |A| = sin 10Â° Ã— cos 80Â° + cos 10Â° x sin 80Â°

|A| = sin (10 + 80)Â°

|A| = sin90Â°

|A| = 1

Hence Proved

**Solution:**

Given,

= 2(1 Ã— 1 â€“ 4 Ã— (â€“2)) â€“ 3(7 Ã— 1 â€“ (â€“2) Ã— (â€“3)) â€“ 5(7 Ã— 4 â€“ 1 Ã— (â€“3))

= 2(1 + 8) â€“ 3(7 â€“ 6) â€“ 5(28 + 3)

= 2 Ã— 9 â€“ 3 Ã— 1 â€“ 5 Ã— 31

= 18 â€“ 3 â€“ 155

= â€“140

Now by expanding along the second column

= 2(1 Ã— 1 â€“ 4 Ã— (â€“2)) â€“ 7(3 Ã— 1 â€“ 4 Ã— (â€“5)) â€“ 3(3 Ã— (â€“2) â€“ 1 Ã— (â€“5))

= 2 (1 + 8) â€“ 7 (3 + 20) â€“ 3 (â€“6 + 5)

= 2 Ã— 9 â€“ 7 Ã— 23 â€“ 3 Ã— (â€“1)

= 18 â€“ 161 +3

= â€“140

**Solution:**

Given

â‡’Â |A| = 0 (0 â€“ sinÎ² (â€“sinÎ²)) â€“sinÎ± (â€“sinÎ± Ã— 0 â€“ sinÎ² cosÎ±) â€“ cosÎ± ((â€“sinÎ±) (â€“sinÎ²) â€“ 0 Ã— cosÎ±)

|A| = 0 + sinÎ± sinÎ² cosÎ± â€“ cosÎ± sinÎ± sinÎ²

|A| = 0

Exercise 6.2 Page No: 6.57

**1. Evaluate the following determinant:**

**Solution:**

(i) Given

(ii) Given

= 1[(109) (12) â€“ (119) (11)]

= 1308 â€“ 1309

= â€“ 1

So, Î” = â€“ 1

(iii) Given,

= a (bc â€“ f^{2}) â€“ h (hc â€“ fg) + g (hf â€“ bg)

= abc â€“ af^{2}Â â€“ ch^{2}Â + fgh + fgh â€“ bg^{2}

= abc + 2fgh â€“ af^{2}Â â€“ bg^{2}Â â€“ ch^{2}

So, Î” = abc + 2fgh â€“ af^{2}Â â€“ bg^{2}Â â€“ ch^{2}

(iv) Given

= 2[1(24 â€“ 4)] = 40

So, Î” = 40

(v) Given

= 1[(â€“ 7) (â€“ 36) â€“ (â€“ 20) (â€“ 13)]

= 252 â€“ 260

= â€“ 8

So, Î” = â€“ 8

(vi) Given,

(vii) Given

(viii) Given,

**2. Without expanding, show that the value of each of the following determinants is zero:**

**Solution:**

(i) Given,

(ii) Given,

(iii) Given,

(iv) Given,

(v) Given,

(vi) Given,

(vii) Given,

(viii) Given,

(ix) Given,

As, C_{1}Â = C_{2}, hence determinant is zero

(x) Given,

(xi) Given,

(xii) Given,

(xiii) Given,

(xiv) Given,

(xv) Given,

(xvi) Given,

(xvii) Given,

Hence proved.

**Evaluate the following (3 â€“ 9):**

**Solution:**

Given,

= (a + b + c) (b â€“ a) (c â€“ a) (b â€“ c)

So, Î” = (a + b + c) (b â€“ a) (c â€“ a) (b â€“ c)

**Solution:**

Given,

**Solution:**

Given,

**Solution:**

Given,

**Solution:**

Given,

**Solution:**

Given,

**Solution:**

Given,

= a [a (a + x + y) + az] + 0 + 0

= a^{2 }(a + x + y + z)

So, Î” = a^{2 }(a + x + y + z)

**Solution:**

**Prove the following identities (11 â€“ 45):**

**Solution:**

Given,

**Solution:**

Consider,

= â€“ (a + b + c) [(b â€“ c) (a + b â€“ 2c) â€“ (c â€“ a) (c + a â€“ 2b)]

= 3abc â€“ a^{3}Â â€“ b^{3}Â â€“ c^{3}

Therefore, L.H.S = R.H.S,

Hence the proof.

**Solution:**

Given,

**Solution:**

Consider,

,

**Solution:**

Consider,

L.H.S =Â

Now by applying, R_{1}â†’R_{1}Â + R_{2}Â + R_{3}, we get,

**Solution:**

Consider,

**Solution:**

Consider,

**Solution:**

Consider,

Hence, the proof.

**Solution:**

Given,

= â€“ xyz(x â€“ y) (z â€“ y) [z^{2}Â + y^{2}Â + zy â€“ x^{2}Â â€“ y^{2}Â â€“ xy]

= â€“ xyz(x â€“ y) (z â€“ y) [(z â€“ x) (z + x0 + y (z â€“ x)]

= â€“ xyz(x â€“ y) (z â€“ y) (z â€“ x) (x + y + z)

= R.H.S

Hence, the proof.

**Solution:**

Consider,

= (a^{2}Â + b^{2}Â + c^{2}) (b â€“ a) (c â€“ a) [(b + a) (â€“ b) â€“ (â€“ c) (c + a)]

= (a^{2}Â + b^{2}Â + c^{2}) (a â€“ b) (c â€“ a) (b â€“ c) (a + b + c)

= R.H.S

Hence, the proof.

**Solution:**

Consider,

= [(2a + 4) (1) â€“ (1) (2a + 6)]

= â€“ 2

= R.H.S

Hence, the proof.

**Solution:**

Consider,

= â€“ (a^{2}Â + b^{2}Â + c^{2}) (a â€“ b) (c â€“ a) [(â€“ (b + a)) (â€“ b) â€“ (c) (c + a)]

= (a â€“ b) (b â€“ c) (c â€“ a) (a + b + c) (a^{2}Â + b^{2}Â + c^{2})

= R.H.S

Hence, the proof.

**Solution:**

Consider,

= R.H.S

Hence, the proof.

**Solution:**

Consider,

**Solution:**

Consider,

**Solution:**

Expanding the determinant along R_{1}, we have

Î” = 1[(1) (7) â€“ (3) (2)] â€“ 0 + 0

âˆ´Â Î” = 7 â€“ 6 = 1

Thus,Â

Hence the proof.

Exercise 6.3 Page No: 6.71

**1. Find the area of the triangle with vertices at the points:**

**(i) (3, 8), (-4, 2) and (5, -1)**

**(ii) (2, 7), (1, 1) and (10, 8)**

**(iii) (-1, -8), (-2, -3) and (3, 2)**

**(iv) (0, 0), (6, 0) and (4, 3)**

**Solution: **

(i) Given (3, 8), (-4, 2) and (5, -1) are the vertices of the triangle.

We know that, if vertices of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by:

(ii) Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle.

We know that if vertices of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by:

(iii) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.

We know that if vertices of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by:

As we know area cannot be negative. Therefore, 15 square unit is the area

Thus area of triangle is 15 square units

(iv) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.

We know that if vertices of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by:

**2. Using the determinants show that the following points are collinear:**

**(i) (5, 5), (-5, 1) and (10, 7)**

**(ii) (1, -1), (2, 1) and (10, 8)**

**(iii) (3, -2), (8, 8) and (5, 2)**

**(iv) (2, 3), (-1, -2) and (5, 8)**

**Solution:**

(i) Given (5, 5), (-5, 1) and (10, 7)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by

(ii) Given (1, -1), (2, 1) and (10, 8)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by,

(iii) Given (3, -2), (8, 8) and (5, 2)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by,

Now, by substituting given value in above formula

Since, Area of triangle is zero

Hence, points are collinear.

(iv) Given (2, 3), (-1, -2) and (5, 8)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by,

**3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab**

**Solution:**

Given (a, 0), (0, b) and (1, 1) are collinear

_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by,

â‡’Â

â‡’Â a + b = ab

Hence Proved

**4. Using the determinants prove that the points (a, b), (a’, b’) and (a – a’, b – b) are collinear if a b’ = a’ b.**

**Solution:**

Given (a, b), (a’, b’) and (a – a’, b – b) are collinear

_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by,

â‡’Â a b’ = a’ b

Hence, the proof.

**5. Find the value of Î» so that the points (1, -5), (-4, 5) and (Î», 7) are collinear.**

**Solution:**

Given (1, -5), (-4, 5) and (Î», 7) are collinear

_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by,

â‡’Â – 50 â€“ 10Î» = 0

â‡’Â Î» = â€“ 5

**6. Find the value of x if the area of âˆ† is 35 square cms with vertices (x, 4), (2, -6) and (5, 4).**

**Solution:**

Given (x, 4), (2, -6) and (5, 4) are the vertices of a triangle.

_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by,

â‡’Â [x (â€“ 10) â€“ 4(â€“ 3) + 1(8 â€“ 30)] = Â± 70

â‡’Â [â€“ 10x + 12 + 38] = Â± 70

â‡’Â Â±70 = â€“ 10x + 50

Taking positive sign, we get

â‡’Â + 70 = â€“ 10x + 50

â‡’Â 10x = â€“ 20

â‡’Â x = â€“ 2

Taking â€“negative sign, we get

â‡’Â â€“ 70 = â€“ 10x + 50

â‡’Â 10x = 120

â‡’Â x = 12

Thus x = â€“ 2, 12

Exercise 6.4 Page No: 6.84

**Solve the following system of linear equations by Cramerâ€™s rule:**

**1. x â€“ 2y = 4**

**-3x + 5y = -7**

**Solution:**

Given x â€“ 2y = 4

-3x + 5y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^{st}Â row

â‡’Â D = 5(1) â€“ (â€“ 3) (â€“ 2)

â‡’Â D = 5 â€“ 6

â‡’Â D = â€“ 1

Again,

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 5(4) â€“ (â€“ 7) (â€“ 2)

â‡’Â D_{1}Â = 20 â€“ 14

â‡’Â D_{1}Â = 6

And

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 1(â€“ 7) â€“ (â€“ 3) (4)

â‡’Â D_{2}Â = â€“ 7 + 12

â‡’Â D_{2}Â = 5

Thus by Cramerâ€™s Rule, we have

**2. 2x â€“ y = 1**

**7x â€“ 2y = -7**

**Solution:**

Given 2x â€“ y = 1 and

7x â€“ 2y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 1(â€“ 2) â€“ (â€“ 7) (â€“ 1)

â‡’Â D_{1}Â = â€“ 2 â€“ 7

â‡’Â D_{1}Â = â€“ 9

And

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 2(â€“ 7) â€“ (7) (1)

â‡’Â D_{2}Â = â€“ 14 â€“ 7

â‡’Â D_{2}Â = â€“ 21

Thus by Cramerâ€™s Rule, we have

**3. 2x â€“ y = 17**

**3x + 5y = 6**

**Solution:**

Given 2x â€“ y = 17 and

3x + 5y = 6

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 17(5) â€“ (6) (â€“ 1)

â‡’Â D_{1}Â = 85 + 6

â‡’Â D_{1}Â = 91

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 2(6) â€“ (17) (3)

â‡’Â D_{2}Â = 12 â€“ 51

â‡’Â D_{2}Â = â€“ 39

Thus by Cramerâ€™s Rule, we have

**4. 3x + y = 19**

**3x â€“ y = 23**

**Solution:**

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^{st}Â row

â‡’Â D = 3(â€“ 1) â€“ (3) (1)

â‡’Â D = â€“ 3 â€“ 3

â‡’Â D = â€“ 6

Again,

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 19(â€“ 1) â€“ (23) (1)

â‡’Â D_{1}Â = â€“ 19 â€“ 23

â‡’Â D_{1}Â = â€“ 42

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 3(23) â€“ (19) (3)

â‡’Â D_{2}Â = 69 â€“ 57

â‡’Â D_{2}Â = 12

Thus by Cramerâ€™s Rule, we have

**5. 2x â€“ y = -2**

**3x + 4y = 3**

**Solution:**

Given 2x â€“ y = -2 and

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 3(2) â€“ (â€“ 2) (3)

â‡’Â D_{2}Â = 6 + 6

â‡’Â D_{2}Â = 12

Thus by Cramerâ€™s Rule, we have

**6. 3x + ay = 4**

**2x + ay = 2, a â‰ 0**

**Solution:**

Given 3x + ay = 4 and

2x + ay = 2, a â‰ 0

Let there be a system of n simultaneous linear equations and with n unknown given by

3x + ay = 4

2x + ay = 2, aâ‰ 0

So by comparing with the theorem, let’s find D, D_{1}Â and D_{2}

Solving determinant, expanding along 1^{st}Â row

â‡’Â D = 3(a) â€“ (2) (a)

â‡’Â D = 3a â€“ 2a

â‡’Â D = a

Again,

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 4(a) â€“ (2) (a)

â‡’Â D = 4a â€“ 2a

â‡’Â D = 2a

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 3(2) â€“ (2) (4)

â‡’Â D = 6 â€“ 8

â‡’Â D = â€“ 2

Thus by Cramerâ€™s Rule, we have

**7. 2x + 3y = 10**

**x + 6y = 4**

**Solution:**

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^{st}Â row

â‡’Â D = 2 (6) â€“ (3) (1)

â‡’Â D = 12 â€“ 3

â‡’Â D = 9

Again,

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 10 (6) â€“ (3) (4)

â‡’Â D = 60 â€“ 12

â‡’Â D = 48

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 2 (4) â€“ (10) (1)

â‡’Â D_{2}Â = 8 â€“ 10

â‡’Â D_{2}Â = â€“ 2

Thus by Cramerâ€™s Rule, we have

**8. 5x + 7y = -2**

**4x + 6y = -3**

**Solution:**

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

5x + 7y = â€“ 2

4x + 6y = â€“ 3

So by comparing with the theorem, let’s find D, D_{1}Â and D_{2}

Solving determinant, expanding along 1^{st}Â row

â‡’Â D = 5(6) â€“ (7) (4)

â‡’Â D = 30 â€“ 28

â‡’Â D = 2

Again,

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = â€“ 2(6) â€“ (7) (â€“ 3)

â‡’Â D_{1}Â = â€“ 12 + 21

â‡’Â D_{1}Â = 9

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = â€“ 3(5) â€“ (â€“ 2) (4)

â‡’Â D_{2}Â = â€“ 15 + 8

â‡’Â D_{2}Â = â€“ 7

Thus by Cramerâ€™s Rule, we have

**9. 9x + 5y = 10**

**3y â€“ 2x = 8**

**Solution:**

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^{st}Â row

â‡’Â D = 3(9) â€“ (5) (â€“ 2)

â‡’Â D = 27 + 10

â‡’Â D = 37

Again,

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 10(3) â€“ (8) (5)

â‡’Â D_{1}Â = 30 â€“ 40

â‡’Â D_{1}Â = â€“ 10

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 9(8) â€“ (10) (â€“ 2)

â‡’Â D_{2}Â = 72 + 20

â‡’Â D_{2}Â = 92

Thus by Cramerâ€™s Rule, we have

**10. x + 2y = 1**

**3x + y = 4**

**Solution:**

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^{st}Â row

â‡’Â D = 1(1) â€“ (3) (2)

â‡’Â D = 1 â€“ 6

â‡’Â D = â€“ 5

Again,

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 1(1) â€“ (2) (4)

â‡’Â D_{1}Â = 1 â€“ 8

â‡’Â D_{1}Â = â€“ 7

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 1(4) â€“ (1) (3)

â‡’Â D_{2}Â = 4 â€“ 3

â‡’Â D_{2}Â = 1

Thus by Cramerâ€™s Rule, we have

**Solve the following system of linear equations by Cramerâ€™s rule:**

**11. 3x + y + z = 2**

**2x â€“ 4y + 3z = -1**

**4x + y â€“ 3z = -11**

**Solution:**

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

3x + y + z = 2

2x â€“ 4y + 3z = â€“ 1

4x + y â€“ 3z = â€“ 11

So by comparing with the theorem, let’s find D, D_{1}, D_{2}Â and D_{3}

Solving determinant, expanding along 1^{st}Â row

â‡’Â D = 3[(â€“ 4) (â€“ 3) â€“ (3) (1)] â€“ 1[(2) (â€“ 3) â€“ 12] + 1[2 â€“ 4(â€“ 4)]

â‡’Â D = 3[12 â€“ 3] â€“ [â€“ 6 â€“ 12] + [2 + 16]

â‡’Â D = 27 + 18 + 18

â‡’Â D = 63

Again,

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 2[( â€“ 4)( â€“ 3) â€“ (3)(1)] â€“ 1[( â€“ 1)( â€“ 3) â€“ ( â€“ 11)(3)] + 1[( â€“ 1) â€“ ( â€“ 4)( â€“ 11)]

â‡’Â D_{1}Â = 2[12 â€“ 3] â€“ 1[3 + 33] + 1[â€“ 1 â€“ 44]

â‡’Â D_{1}Â = 2[9] â€“ 36 â€“ 45

â‡’Â D_{1}Â = 18 â€“ 36 â€“ 45

â‡’Â D_{1}Â = â€“ 63

Again

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 3[3 + 33] â€“ 2[â€“ 6 â€“ 12] + 1[â€“ 22 + 4]

â‡’Â D_{2}Â = 3[36] â€“ 2(â€“ 18) â€“ 18

â‡’Â D_{2}Â = 126

â‡’Â

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{3}Â = 3[44 + 1] â€“ 1[â€“ 22 + 4] + 2[2 + 16]

â‡’Â D_{3}Â = 3[45] â€“ 1(â€“ 18) + 2(18)

â‡’Â D_{3}Â = 135 + 18 + 36

â‡’Â D_{3}Â = 189

Thus by Cramerâ€™s Rule, we have

**12. x â€“ 4y â€“ z = 11**

**2x â€“ 5y + 2z = 39**

**-3x + 2y + z = 1**

**Solution:**

Given,

x â€“ 4y â€“ z = 11

2x â€“ 5y + 2z = 39

-3x + 2y + z = 1

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

x â€“ 4y â€“ z = 11

2x â€“ 5y + 2z = 39

â€“ 3x + 2y + z = 1

So by comparing with theorem, now we have to find D, D_{1}Â and D_{2}

Solving determinant, expanding along 1^{st}Â row

â‡’Â D = 1[(â€“ 5) (1) â€“ (2) (2)] + 4[(2) (1) + 6] â€“ 1[4 + 5(â€“ 3)]

â‡’Â D = 1[â€“ 5 â€“ 4] + 4[8] â€“ [â€“ 11]

â‡’Â D = â€“ 9 + 32 + 11

â‡’Â D = 34

Again,

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{1}Â = 11[(â€“ 5) (1) â€“ (2) (2)] + 4[(39) (1) â€“ (2) (1)] â€“ 1[2 (39) â€“ (â€“ 5) (1)]

â‡’Â D_{1}Â = 11[â€“ 5 â€“ 4] + 4[39 â€“ 2] â€“ 1[78 + 5]

â‡’Â D_{1}Â = 11[â€“ 9] + 4(37) â€“ 83

â‡’Â D_{1}Â = â€“ 99 â€“ 148 â€“ 45

â‡’Â D_{1}Â = â€“ 34

Again

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{2}Â = 1[39 â€“ 2] â€“ 11[2 + 6] â€“ 1[2 + 117]

â‡’Â D_{2}Â = 1[37] â€“ 11(8) â€“ 119

â‡’Â D_{2}Â = â€“ 170

And,

â‡’Â

Solving determinant, expanding along 1^{st}Â row

â‡’Â D_{3}Â = 1[â€“ 5 â€“ (39) (2)] â€“ (â€“ 4) [2 â€“ (39) (â€“ 3)] + 11[4 â€“ (â€“ 5)(â€“ 3)]

â‡’Â D_{3}Â = 1 [â€“ 5 â€“ 78] + 4 (2 + 117) + 11 (4 â€“ 15)

â‡’Â D_{3}Â = â€“ 83 + 4(119) + 11(â€“ 11)

â‡’Â D_{3}Â = 272

Thus by Cramerâ€™s Rule, we have

**13. 6x + y â€“ 3z = 5**

**x + 3y â€“ 2z = 5**

**2x + y + 4z = 8**

**Solution:**

Given

6x + y â€“ 3z = 5

x + 3y â€“ 2z = 5

2x + y + 4z = 8

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

6x + y â€“ 3z = 5

x + 3y â€“ 2z = 5

2x + y + 4z = 8

So by comparing with theorem, now we have to find D , D_{1}Â and D_{2}

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D = 6[(4) (3) â€“ (1) (â€“ 2)] â€“ 1[(4) (1) + 4] â€“ 3[1 â€“ 3(2)]

â‡’Â D = 6[12 + 2] â€“ [8] â€“ 3[â€“ 5]

â‡’Â D = 84 â€“ 8 + 15

â‡’Â D = 91

Again, Solve D_{1}Â formed by replacing 1^{st}Â column by B matrices

Here

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D_{1}Â = 5[(4) (3) â€“ (â€“ 2) (1)] â€“ 1[(5) (4) â€“ (â€“ 2) (8)] â€“ 3[(5) â€“ (3) (8)]

â‡’Â D_{1}Â = 5[12 + 2] â€“ 1[20 + 16] â€“ 3[5 â€“ 24]

â‡’Â D_{1}Â = 5[14] â€“ 36 â€“ 3(â€“ 19)

â‡’Â D_{1}Â = 70 â€“ 36 + 57

â‡’Â D_{1}Â = 91

Again, Solve D_{2}Â formed by replacing 1^{st}Â column by B matrices

Here

Solving determinant

â‡’Â D_{2}Â = 6[20 + 16] â€“ 5[4 â€“ 2(â€“ 2)] + (â€“ 3)[8 â€“ 10]

â‡’Â D_{2}Â = 6[36] â€“ 5(8) + (â€“ 3) (â€“ 2)

â‡’Â D_{2}Â = 182

And, Solve D_{3}Â formed by replacing 1^{st}Â column by B matrices

Here

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D_{3}Â = 6[24 â€“ 5] â€“ 1[8 â€“ 10] + 5[1 â€“ 6]

â‡’Â D_{3}Â = 6[19] â€“ 1(â€“ 2) + 5(â€“ 5)

â‡’Â D_{3}Â = 114 + 2 â€“ 25

â‡’Â D_{3}Â = 91

Thus by Cramerâ€™s Rule, we have

**14. x + y = 5**

**y + z = 3**

**x + z = 4**

**Solution:**

Given x + y = 5

y + z = 3

x + z = 4

Let there be a system of n simultaneous linear equations and with n unknown given by

Let D_{j}Â be the determinant obtained from D after replacing the j^{th}Â column by

Now, here we have

x + y = 5

y + z = 3

x + z = 4

So by comparing with theorem, now we have to find D, D_{1}Â and D_{2}

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D = 1[1] â€“ 1[â€“ 1] + 0[â€“ 1]

â‡’Â D = 1 + 1 + 0

â‡’Â D = 2

Again, Solve D_{1}Â formed by replacing 1^{st}Â column by B matrices

Here

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D_{1}Â = 5[1] â€“ 1[(3) (1) â€“ (4) (1)] + 0[0 â€“ (4) (1)]

â‡’Â D_{1}Â = 5 â€“ 1[3 â€“ 4] + 0[â€“ 4]

â‡’Â D_{1}Â = 5 â€“ 1[â€“ 1] + 0

â‡’Â D_{1}Â = 5 + 1 + 0

â‡’Â D_{1}Â = 6

Again, Solve D_{2}Â formed by replacing 1^{st}Â column by B matrices

Here

Solving determinant

â‡’Â D_{2}Â = 1[3 â€“ 4] â€“ 5[â€“ 1] + 0[0 â€“ 3]

â‡’Â D_{2}Â = 1[â€“ 1] + 5 + 0

â‡’Â D_{2}Â = 4

And, Solve D_{3}Â formed by replacing 1^{st}Â column by B matrices

Here

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D_{3}Â = 1[4 â€“ 0] â€“ 1[0 â€“ 3] + 5[0 â€“ 1]

â‡’Â D_{3}Â = 1[4] â€“ 1(â€“ 3) + 5(â€“ 1)

â‡’Â D_{3}Â = 4 + 3 â€“ 5

â‡’Â D_{3}Â = 2

Thus by Cramerâ€™s Rule, we have

**15. 2y â€“ 3z = 0**

**x + 3y = -4**

**3x + 4y = 3**

**Solution:**

Given

2y â€“ 3z = 0

x + 3y = -4

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

2y â€“ 3z = 0

x + 3y = â€“ 4

3x + 4y = 3

So by comparing with theorem, now we have to find D, D_{1}Â and D_{2}

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D = 0[0] â€“ 2[(0) (1) â€“ 0] â€“ 3[1 (4) â€“ 3 (3)]

â‡’Â D = 0 â€“ 0 â€“ 3[4 â€“ 9]

â‡’Â D = 0 â€“ 0 + 15

â‡’Â D = 15

Again, Solve D_{1}Â formed by replacing 1^{st}Â column by B matrices

Here

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D_{1}Â = 0[0] â€“ 2[(0) (â€“ 4) â€“ 0] â€“ 3[4 (â€“ 4) â€“ 3(3)]

â‡’Â D_{1}Â = 0 â€“ 0 â€“ 3[â€“ 16 â€“ 9]

â‡’Â D_{1}Â = 0 â€“ 0 â€“ 3(â€“ 25)

â‡’Â D_{1}Â = 0 â€“ 0 + 75

â‡’Â D_{1}Â = 75

Again, Solve D_{2}Â formed by replacing 2^{nd}Â column by B matrices

Here

Solving determinant

â‡’Â D_{2}Â = 0[0] â€“ 0[(0) (1) â€“ 0] â€“ 3[1 (3) â€“ 3(â€“ 4)]

â‡’Â D_{2}Â = 0 â€“ 0 + (â€“ 3) (3 + 12)

â‡’Â D_{2}Â = â€“ 45

And, Solve D_{3}Â formed by replacing 3^{rd}Â column by B matrices

Here

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D_{3}Â = 0[9 â€“ (â€“ 4) 4] â€“ 2[(3) (1) â€“ (â€“ 4) (3)] + 0[1 (4) â€“ 3 (3)]

â‡’Â D_{3}Â = 0[25] â€“ 2(3 + 12) + 0(4 â€“ 9)

â‡’Â D_{3}Â = 0 â€“ 30 + 0

â‡’Â D_{3}Â = â€“ 30

Thus by Cramerâ€™s Rule, we have

**16. 5x â€“ 7y + z = 11**

**6x â€“ 8y â€“ z = 15**

**3x + 2y â€“ 6z = 7**

**Solution:**

Given

5x â€“ 7y + z = 11

6x â€“ 8y â€“ z = 15

3x + 2y â€“ 6z = 7

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

5x â€“ 7y + z = 11

6x â€“ 8y â€“ z = 15

3x + 2y â€“ 6z = 7

So by comparing with theorem, now we have to find D, D_{1}Â and D_{2}

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D = 5[(â€“ 8) (â€“ 6) â€“ (â€“ 1) (2)] â€“ 7[(â€“ 6) (6) â€“ 3(â€“ 1)] + 1[2(6) â€“ 3(â€“ 8)]

â‡’Â D = 5[48 + 2] â€“ 7[â€“ 36 + 3] + 1[12 + 24]

â‡’Â D = 250 â€“ 231 + 36

â‡’Â D = 55

Again, Solve D_{1}Â formed by replacing 1^{st}Â column by B matrices

Here

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D_{1}Â = 11[(â€“ 8) (â€“ 6) â€“ (2) (â€“ 1)] â€“ (â€“ 7) [(15) (â€“ 6) â€“ (â€“ 1) (7)] + 1[(15)2 â€“ (7) (â€“ 8)]

â‡’Â D_{1}Â = 11[48 + 2] + 7[â€“ 90 + 7] + 1[30 + 56]

â‡’Â D_{1}Â = 11[50] + 7[â€“ 83] + 86

â‡’Â D_{1}Â = 550 â€“ 581 + 86

â‡’Â D_{1}Â = 55

Again, Solve D_{2}Â formed by replacing 2^{nd}Â column by B matrices

Here

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D_{2}Â = 5[(15) (â€“ 6) â€“ (7) (â€“ 1)] â€“ 11 [(6) (â€“ 6) â€“ (â€“ 1) (3)] + 1[(6)7 â€“ (15) (3)]

â‡’Â D_{2 }=Â 5[â€“ 90 + 7] â€“ 11[â€“ 36 + 3] + 1[42 â€“ 45]

â‡’Â D_{2}Â = 5[â€“ 83] â€“ 11(â€“ 33) â€“ 3

â‡’Â D_{2}Â = â€“ 415 + 363 â€“ 3

â‡’Â D_{2}Â = â€“ 55

And, Solve D_{3}Â formed by replacing 3^{rd}Â column by B matrices

Here

Solving determinant, expanding along 1^{st}Â Row

â‡’Â D_{3}Â = 5[(â€“ 8) (7) â€“ (15) (2)] â€“ (â€“ 7) [(6) (7) â€“ (15) (3)] + 11[(6)2 â€“ (â€“ 8) (3)]

â‡’Â D_{3}Â = 5[â€“ 56 â€“ 30] â€“ (â€“ 7) [42 â€“ 45] + 11[12 + 24]

â‡’Â D_{3}Â = 5[â€“ 86] + 7[â€“ 3] + 11[36]

â‡’Â D_{3}Â = â€“ 430 â€“ 21 + 396

â‡’Â D_{3}Â = â€“ 55

Thus by Cramerâ€™s Rule, we have

Exercise 6.5 Page No: 6.89

**Solve each of the following system of homogeneous linear equations:**

**1. x + y â€“ 2z = 0**

**2x + y â€“ 3z =0**

**5x + 4y â€“ 9z = 0**

**Solution:**

Given x + y â€“ 2z = 0

2x + y â€“ 3z =0

5x + 4y â€“ 9z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

= 1(1 Ã— (â€“ 9) â€“ 4 Ã— (â€“ 3)) â€“ 1(2 Ã— (â€“ 9) â€“ 5 Ã— (â€“ 3)) â€“ 2(4 Ã— 2 â€“ 5 Ã— 1)

= 1(â€“ 9 + 12) â€“ 1(â€“ 18 + 15) â€“ 2(8 â€“ 5)

= 1 Ã— 3 â€“1 Ã— (â€“ 3) â€“ 2 Ã— 3

= 3 + 3 â€“ 6

= 0

Since D = 0, so the system of equation has infinite solution.

Now let z = k

â‡’Â x + y = 2k

And 2x + y = 3k

Now using the Cramerâ€™s rule

**2. 2x + 3y + 4z = 0**

**x + y + z = 0**

**2x + 5y â€“ 2z = 0**

**Solution:**

Given

2x + 3y + 4z = 0

x + y + z = 0

2x + 5y â€“ 2z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

= 2(1 Ã— (â€“ 2) â€“ 1 Ã— 5) â€“ 3(1 Ã— (â€“ 2) â€“ 2 Ã— 1) + 4(1 Ã— 5 â€“ 2 Ã— 1)

= 2(â€“ 2 â€“ 5) â€“ 3(â€“ 2 â€“ 2) + 4(5 â€“ 2)

= 1 Ã— (â€“ 7) â€“ 3 Ã— (â€“ 4) + 4 Ã— 3

= â€“ 7 + 12 + 12

= 17

Since D â‰ 0, so the system of equation has infinite solution.

Therefore the system of equation has only solution as x = y = z = 0.