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#### Question 1:

Solve the following system of equations by matrix method:
(i) 5x + 7y + 2 = 0
4x + 6y + 3 = 0

(ii) 5x + 2y = 3
3x + 2y = 5

(iii) 3x + 4y − 5 = 0
xy + 3 = 0

(iv) 3x + y = 19
3xy = 23

(v) 3x + 7y = 4
x + 2y = −1

(vi) 3x + y = 23
5x + 3y = 12

#### Question 2:

Solve the following system of equations by matrix method:
(i) x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1

(ii) x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9

(iii) 6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10

(iv) 3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0

(v)
$\frac{2}{x}-\frac{3}{y}+\frac{3}{z}=10\phantom{\rule{0ex}{0ex}}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=10\phantom{\rule{0ex}{0ex}}\frac{3}{x}-\frac{1}{y}+\frac{2}{z}=13$

(vi) 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25

(vii) 3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2

(viii) 2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6

(ix) 2x + 6y = 2
3x − z = −8
2x − y + z = −3

(x) x − y + z = 2
2x − y = 0
2y − z = 1

(xi) 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5

(xii) x + y + z = 6
x + 2z = 7
3x + y + z = 12

(xiii)

(xiv) x − y + 2z = 7
3x + 4− 5= −5
2− y + 3z = 12

(i)

(v)

(vi)

(vii)

#### Question 3:

Show that each of the following systems of linear equations is consistent and also find their solutions:
(i) 6x + 4y = 2
9x + 6y = 3

(ii) 2x + 3y = 5
6x + 9y = 15

(iii) 5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5

(iv) xy + z = 3
2x + y − z = 2
x −2y + 2z = 1

(v) x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30

(vi) 2x + 2y − 2z = 1
4x + 4yz = 2
6x + 6y + 2z = 3

#### Question 4:

Show that each one of the following systems of linear equation is inconsistent:
(i) 2x + 5y = 7
6x + 15y = 13

(ii) 2x + 3y = 5
6x + 9y = 10

(iii) 4x − 2y = 3
6x − 3y = 5

(iv) 4x − 5y − 2z = 2
5x − 4y + 2z = −2
2x + 2y + 8z = −1

(v) 3xy − 2z = 2
2yz = −1
3x − 5y = 3

(vi) x + y − 2z = 5
x − 2y + z = −2
−2x + y + z = 4

#### Question 5:

If are two square matrices, find AB and hence solve the system of linear equations:
xy = 3, 2x + 3y + 4z = 17, y + 2z = 7

Here,

#### Question 6:

If , find A−1 and hence solve the system of linear equations
2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3

#### Question 7:

Find A−1, if . Hence solve the following system of linear equations:
x + 2y + 5z = 10, xyz = −2, 2x + 3yz = −11

#### Question 8:

(i) If , find A−1. Using A−1, solve the system of linear equations
x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7

(ii) , find A−1 and hence solve the following system of equations:
3x − 4y + 2z = −1, 2x + 3y + 5z = 7, x + z = 2

(iii) , find AB. Hence, solve the system of equations:
x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

(iv) If
, find A−1. Using A−1, solve the system of linear equations
x − 2y = 10, 2x − y − z = 8, −2y + z = 7

(v) Given
, find BA and use this to solve the system of equations
y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17

(vi)
If , find A–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2yz = 8.
(vii) Use product $\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]$ to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.

(vi) We have, .

So, A is invertible.
Let Cij be the co-factors of the elements aij in A[aij]. Then,
${C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}2& 2\\ 1& -1\end{array}\right|=-2-2=-4\phantom{\rule{0ex}{0ex}}{C}_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& 2\\ -3& -1\end{array}\right|=-1\left(-1+6\right)=-5\phantom{\rule{0ex}{0ex}}{C}_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}1& 2\\ -3& 1\end{array}\right|=1+6=7$
${C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}3& 1\\ 1& -1\end{array}\right|=3+1=4\phantom{\rule{0ex}{0ex}}{C}_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}2& 1\\ -3& -1\end{array}\right|=-2+3=1\phantom{\rule{0ex}{0ex}}{C}_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}2& 3\\ -3& 1\end{array}\right|=-1\left(2+9\right)=-11$
${C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}3& 1\\ 2& 2\end{array}\right|=6-2=4\phantom{\rule{0ex}{0ex}}{C}_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}2& 1\\ 1& 2\end{array}\right|=-1\left(4-1\right)=-3\phantom{\rule{0ex}{0ex}}{C}_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}2& 3\\ 1& 2\end{array}\right|=4-3=1$

Now, the given system of equations is expressible as
$\left[\begin{array}{ccc}2& 1& -3\\ 3& 2& 1\\ 1& 2& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}13\\ 4\\ 8\end{array}\right]$
Or AT X = B, where
Now, $\left|{A}^{T}\right|=\left|A\right|=-16\ne 0$
So, the given system of equations is consistent with a unique solution given by
$X={\left({A}^{T}\right)}^{-1}B={\left({A}^{-1}\right)}^{T}B$
$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=-\frac{1}{16}{\left[\begin{array}{ccc}-4& 4& 4\\ -5& 1& -3\\ 7& -11& 1\end{array}\right]}^{T}\left[\begin{array}{c}13\\ 4\\ 8\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{ccc}-4& -5& 7\\ 4& 1& -11\\ 4& -3& 1\end{array}\right]\left[\begin{array}{c}13\\ 4\\ 8\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{c}-52-20+56\\ 52+4-88\\ 52-12+8\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{c}-16\\ -32\\ 48\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]\phantom{\rule{0ex}{0ex}}$
Hence, x = 1, y = 2 and z = −3 is the required solution.

(vii) Suppose, A = $\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]$ $B=\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]$

$A×B=\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccc}-2-9+12& 0-2+2& 1+3-4\\ 0+18-18& 0+4-3& 0-6+6\\ -6-18+24& 0-4+4& 3+6-8\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Since, A × B = I,
$\therefore$ B = A−1                                 .....(1)

Now, the given system of equations is

x + 3z = 9
x + 2y − 2z = 4
2x − 3y + 4z = −3

This can also be represented as,

$\left[\begin{array}{ccc}1& 0& 3\\ -1& 2& -2\\ 2& -3& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right]$
Here, we can observe that $\left[\begin{array}{ccc}1& 0& 3\\ -1& 2& -2\\ 2& -3& 4\end{array}\right]={A}^{T}$
So, ${A}^{T}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right]$

Multiply the above expression by ${\left({A}^{T}\right)}^{-1}$.

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]={\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]}^{T}\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccc}-2& 9& 6\\ 0& 2& 1\\ 1& -3& -2\end{array}\right]\left[\begin{array}{c}9\\ 4\\ -3\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}-18+36-18\\ 0+8-3\\ 9-12+6\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}0\\ 5\\ 3\end{array}\right]$

Hence, x = 0, y = 5 and z = 3.

#### Question 9:

The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.

Let the three numbers be x, y and z.

#### Question 10:

An amount of Rs 10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined income is Rs 1310 and the combined income of first and  second investment is Rs 190 short of the income from the third. Find the investment in each using matrix method.

Let x , y and z be the investments at the rates of interest of 10%, 12% and 15% per annum respectively.

Total investment = Rs 10,000
$⇒x+y+z=10,000$

#### Question 11:

A company produces three products every day. Their production on a certain day is 45 tons. It is found that the production of third product exceeds the production of first product by 8 tons while the total production of first and third product is twice the production of second product. Determine the production level of each product using matrix method.

#### Question 12:

The prices of three commodities P, Q and R are Rs x, y and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R. C purchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B and C earn Rs 6000, Rs 5000 and Rs 13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using matrix method.

#### Question 13:

The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.

As per the information given in the question, the following equations hold true:

.

The above three equations can be represented in the form of a matrix as

Or AX = B, where,

Thus, A is non-singular. Therefore, its inverse exists.

Adj A is given by

Therefore, the number of awardees for Honesty, Cooperation and Supervision are 3, 4, and 5 respectively.

One more value which the management of the colony must include for awards may be Sincerity.

#### Question 14:

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

Let the award money given for Honesty, Regularity and Hard work be x, y and z respectively.

Since total cash award is Rs 6,000.

x + y + z = Rs 6,000 ...(1)

Three times the award money for Hard work and Honesty is Rs 11,000.

x + 3 z = Rs 11,000

x + 0.y + 3 z = Rs 11,000 ...(2)

Award money for Honesty and Hard work is double the one given for regularity.

x + z = 2y

x − 2y + z = 0 ...(3)

The above system can be written in matrix form as,

Or AX = B, where

Thus, A is non-singular. Hence, it is invertible.

Thus, award money given for Honesty, Regularity and Hard work are Rs 500, Rs 2000 and Rs 3500 respectively.

School can include sincerity for awards.

#### Question 15:

Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. x, y and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x, y and z. What values are described in this equations?

#### Question 16:

Two factories decided to award their employees for three values of (a) adaptable tonew techniques, (b) careful and alert in difficult situations and (c) keeping clam in tense situations, at the rate of ₹ x, ₹ y and ₹ z per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of ₹ 29000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of ₹ 30500. If the three prizes per person together cost ₹ 9500, then
i) represent the above situation by matrix equation and form linear equation using matrix multiplication.
ii) Solve these equation by matrix method.
iii) Which values are reflected in the questions?

#### Question 17:

Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹1,600School B wants to spend ₹2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.

Let the award money given for sincerity, truthfulness and helpfulness be ₹x, ₹y and ₹respectively.

Since, the total cash award is ₹900.
x + y + z = 900                    ....(1)

Award money given by school A is ₹1,600.
∴ 3x + 2y + z = 1600              ....(2)

Award money given by school B is ₹2,300.
∴ 4x + y + 3z = 2300              ....(3)

The above system of equations can be written in matrix form CX = D as

Hence, the award money for each value of sincerity, truthfulness and helpfulness is ₹200, ₹300 and ₹400.

One more value which should be considered for award hardwork.

#### Question 18:

Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award ₹x each, ₹y each and ₹z each the three respectively values to its 3, 2 and 1 students with a total award money of ₹1,000School Q wants to spend ₹1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is ₹600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.

​Let the award money given for Discipline, Politeness and Punctuality be ₹x, ₹y and ₹respectively.

Since, the total cash award is ₹600.
∴ x + y + z = 600                    ....(1)

Award money given by school P is ₹1,000.
∴ 3x + 2y + z = 1000              ....(2)

Award money given by school Q is ₹1,500.
∴ 4x + y + 3z = 1500              ....(3)

The above system of equations can be written in matrix form AX = B as

Hence, the award money for each value of Discipline, Politeness and Punctuality is ₹100, ₹200 and ₹300.

​One more value which should be considered for award is Honesty.

#### Question 19:

Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹2,200School Q wants to spend ₹3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.

​​Let the award money given for Tolerance, Kindness and Leadership be ₹x, ₹y and ₹respectively.

Since, the total cash award is ₹1,200.
∴ x + y + z = 1200                    ....(1)

Award money given by school P is ₹2,200.
∴ 3x + 2y + z = 2200                ....(2)

Award money given by school Q is ₹3,100.
∴ 4x + y + 3z = 3100                ....(3)

The above system of equations can be written in matrix form AX = B as

Hence, the award money for each value of Tolerance, Kindness and Leadership is ₹300, ₹400 and ₹500.

​One more value which should be considered for award is Honesty.

#### Question 20:

A total amount of ₹7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8% and $8\frac{1}{2}$% respectively. The total annual interest from these three accounts is ₹550. Equal amounts have been deposited in the 5% and 8% saving accounts. Find the amount deposited in each of the three accounts, with the help of matrices.

​​​Let the amount deposited in each of the three accounts be ₹x, ₹x and ₹respectively.

Since, the total amount deposited is ₹7,000.
∴ x + x + y = 7000
⇒ 2x + y = 7000                      ....(1)

Total annual Interest is ₹550.
∴ $\frac{5}{100}x+\frac{8}{100}x+\frac{17}{200}y=550$
$⇒26x+17y=110000$                ....(2)

The above system of equations can be written in matrix form AX = B as

Hence, the amount deposited in each of the three accounts is ₹1125, ₹1125 and ₹4750.

#### Question 21:

A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of 'A' variety 3 pens of 'B' variety and 2 pens of 'C' variety for Rs 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs 70. Using matrix method, find cost of each variety of pen.

2xy + z = 0
3x + 2yz = 0
x + 4y + 3z = 0

#### Question 2:

2xy + 2z = 0
5x + 3yz = 0
x + 5y − 5z = 0

Here,
2xy + 2z = 0                 ...(1)
5x + 3yz = 0                 ...(2)
x + 5y − 5z = 0                 ...(3)

#### Question 3:

3xy + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0

Here,
3xy + 2z = 0                  ...(1)
4x + 3y + 3z = 0                ...(2)
5x + 7y + 4z = 0                ...(3)

#### Question 4:

x + y − 6z = 0
xy + 2z = 0
−3x + y + 2z = 0

Here,
x + y − 6z = 0                  ...(1)
xy + 2z = 0                  ...(2)
−3x + y + 2z = 0              ...(3)

#### Question 5:

x + y + z = 0
xy − 5z = 0
x + 2y + 4z = 0

Here,
x + y + z = 0             ...(1)
xy − 5z = 0           ...(2)
x + 2y + 4z = 0         ...(3)

#### Question 6:

x + yz = 0
x − 2y + z = 0
3x + 6y − 5z = 0

x + yz = 0                ...(1)
x − 2y + z = 0              ...(2)
3x + 6y − 5z = 0          ...(3)

3x + y − 2z = 0
x + y + z = 0
x − 2y + z = 0

2x + 3yz = 0
xy − 2z = 0
3x + y + 3z = 0

#### Question 1:

If , find x, y and z.

#### Question 2:

If , find x, y and z.

#### Question 3:

If , find x, y and z.

#### Question 4:

Solve the following for x and y:

#### Question 5:

If , find x, y, z.

#### Question 6:

If and AX = B, then find n.

#### Question 1:

The system of equation x + y + z = 2, 3xy + 2z = 6 and 3x + y + z = −18 has
(a) a unique solution
(b) no solution
(c) an infinite number of solutions
(d) zero solution as the only solution

(a) a unique solution

#### Question 2:

The number of solutions of the system of equations
2x + yz = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
is
(a) 3
(b) 2
(c) 1
(d) 0

#### Question 3:

Let . If AX = B, then X is equal to
(a) $\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

(b) $\left[\begin{array}{c}-1\\ -2\\ -3\end{array}\right]$

(c) $\left[\begin{array}{c}-1\\ -2\\ -3\end{array}\right]$

(d)

(e) $\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right]$

(a) $\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

#### Question 4:

The number of solutions of the system of equations:
2x + yz = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
(a) 3
(b) 2
(c) 1
(d) 0

#### Question 5:

The system of linear equations:
x + y + z = 2
2x + yz = 3
3x + 2y + kz = 4 has a unique solution if
(a) k ≠ 0
(b) −1 < k < 1
(c) −2 < k < 2
(d) k = 0

(a) k ≠ 0

#### Question 6:

Consider the system of equations:
a1x + b1y + c1z = 0
a2x + b2y + c2z = 0
a3x + b3y + c3z = 0,
if $\left|\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\right|$ = 0, then the system has
(a) more than two solutions
(b) one trivial and one non-trivial solutions
(c) no solution
(d) only trivial solution (0, 0, 0)

#### Question 7:

Let a, b, c be positive real numbers. The following system of equations in x, y and z

(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions

#### Question 8:

For the system of equations:
x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4
(a) there is only one solution
(b) there exists infinitely many solution
(c) there is no solution
(d) none of these

#### Question 9:

The existence of the unique solution of the system of equations:
x + y + z = λ
5xy + µz = 10
2x + 3yz = 6
depends on
(a) µ only
(b) λ only
(c) λ and µ both
(d) neither λ nor µ

(a) µ only

#### Question 10:

The system of equations:
x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
has a unique solution, if
(a) λ = 5, µ = 13
(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13