Maths M.C.Q (1 Marks)

4. k ≠ 0

Solution:

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

The determination of the coefficient matrix $\begin{bmatrix}1&1&1\\2&1&-1\\3&2&\text{k}\end{bmatrix}$ is

= k + 2 -2k - 3 + 1

=-k

To have a unique solution the determinant ≠ 0

⇒ k ≠ 0

4. 0

Solution:

The given system of equations can be written in matrix form as follows:

$\begin{bmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}7\\1\\5\end{bmatrix}$

$\text{AX}=\text{B}$

Here,

$\text{A}=\begin{bmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7\\1\\5\end{bmatrix}$

Now,

$|\text{A}|=2(9-8)-1(-3-2)-1(4+3)$

$=2+5-7$

$=0$

Let c_ij be the co-factors of the elements a_ij in A = [a_ij]. Then,

$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}-3&2\\4&-3\end{vmatrix}=1,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}1&2\\1&-3\end{vmatrix}=5,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}1&-3\\1&4\end{vmatrix}=7$

$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&-1\\4&-3\end{vmatrix}=-1,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}2&-1\\1&-3\end{vmatrix}=-5,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}2&1\\1&4\end{vmatrix}=-7$

$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&-1\\-3&2\end{vmatrix}=5,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}2&-1\\1&2\end{vmatrix}=-5,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}2&1\\1&-3\end{vmatrix}=-7$

$\text{adj }\text{A}=\begin{bmatrix}1&5&7\\-1&-5&-7\\5&-5&-7\end{bmatrix}^\text{T}=\begin{bmatrix}1&-1&5\\5&-5&-5\\7&-7&-7\end{bmatrix}$

$\Rightarrow(\text{adj }\text{A})\text{B}=\begin{bmatrix}1&-1&5\\5&-5&-5\\7&-7&-7\end{bmatrix}\begin{bmatrix}7\\1\\5\end{bmatrix}$

$=\begin{bmatrix}7-1+25\\35-5-25\\49-7-35\end{bmatrix}=\begin{bmatrix}32\\5\\6\end{bmatrix}\neq0$

The given system of equations is inconsistent. Thus, it has no solution.

4. 0

Solution:

From the given equation we get,

$\triangle=\begin{vmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{vmatrix}$

⇒ 2(9 - 8) -1(-3 - 2) - 1(4 + 3)

⇒ 2(1) - 1(-5) - 1(7)

⇒ 2 + 5 - 7

⇒ 2 + 5 -7

⇒ 0 

$\triangle_1=\begin{vmatrix}7&1&-1\\1&-3&2\\5&4&-3\end{vmatrix}$ 

⇒ 7(9 - 8) - 1(-3 - 10) - 1(4 + 15)

⇒ 7(1) - 1(-13) - 1(19)

⇒ 7 + 13 - 19

⇒ 20 - 19

$\Rightarrow1\neq0$

Hence the gvien system no solution.

2. λ ≠ 5

Solution:

x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ

The determinant of the coefficient matrix $\begin{bmatrix}1&1&1\\1&2&3\\1&3&\lambda\end{bmatrix}$ is

= 2λ - 9 - λ + 3 + 1

= λ - 5

For unique solution determinant ≠ 0

⇒ λ ≠ 5

The right hand side is non zero what so ever be the value of µ.

1. $\begin{bmatrix}1\\2\\3\end{bmatrix}$

Solution:

Here,

$\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$

$|\text{A}|=1(0-2)+1(2-3)+2(4-0)$

$=-2-1+8$

$=5$

Let C_ij be the co-factors of the elements a_ij in A = [a_ij]. Then,

$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}0&1\\2&1\end{vmatrix}=-2,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}2&1\\3&1\end{vmatrix}=1,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}2&0\\3&2\end{vmatrix}=4$

$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}-1&2\\2&1\end{vmatrix}=5,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&2\\3&1\end{vmatrix}=-5,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&-1\\3&2\end{vmatrix}=-5$

$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}-1&2\\0&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&2\\2&1\end{vmatrix}=3,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&-1\\2&0\end{vmatrix}=2$

$\text{adj }\text{A}=\begin{bmatrix}-2&5&-1\\5&-5&-5\\-1&2&3\end{bmatrix}^\text{T}$

$=\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$

$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$

$=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$

$\therefore\ \text{X}=\text{A}^{-1}\text{B}$

$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}\begin{bmatrix}3\\1\\4\end{bmatrix}$

$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-6+5-4\\3-5+12\\12-5+8\end{bmatrix}$

$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\frac{1}{5}\begin{bmatrix}-5\\10\\15\end{bmatrix}$

$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\begin{bmatrix}-1\\2\\3\end{bmatrix}$

1. $\mu$ only.

Solution:

For a unique solution, $|\text{A}|\neq0$

$\Rightarrow\begin{vmatrix}1&1&1\\5&-1&\mu\\2&3&-1\end{vmatrix}\neq0$

$\Rightarrow1(1-3\mu)-1(-5-2\mu)+1(15+2)\neq0$

$\Rightarrow1-3\mu+5+2\mu+17\neq0$

$\Rightarrow-\mu+23\neq0$

$\Rightarrow\mu\neq23$

So, existence of a unique solution depends only on $\mu$.

1. There is only one solution.

Solution:

x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4

The determinant of the coefficient matrix $\begin{bmatrix}1&2&3\\2&1&3\\5&5&9\end{bmatrix}$ is

$= -6 - 2(18 - 15) + 3(10 - 5)$

$= -6 - 6 + 15$

$=3\neq0$

The right hand side is also non zero.

The system has a unique solution.

1. A unique solution.

Solution:

The given system of equations can be written in matrix form as follows:

$\begin{bmatrix}1&1&1\\3&-1&2\\3&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2\\6\\-18\end{bmatrix}$

$\text{AX}=\text{B}$

Here,

$\text{A}=\begin{bmatrix}1&1&1\\3&-1&2\\3&1&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\6\\-18\end{bmatrix}$

$|\text{A}|=1(-1-2)-1(3-6)+1(3+3)$

$=-3+3+6$

$=6\neq0$

So, the given system of equations has a unique solution.

2. Unique solution

Solution:

The given system of equations can be written in matrix form as follows:

$\begin{bmatrix}\frac{1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{-1}{\text{c}^2}\\\frac{1}{\text{a}^2}&\frac{-1}{\text{b}^2}&\frac{1}{\text{c}^2}\\\frac{-1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{1}{\text{c}^2}\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}$

Here,

$\text{A}=\begin{bmatrix}\frac{1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{-1}{\text{c}^2}\\\frac{1}{\text{a}^2}&\frac{-1}{\text{b}^2}&\frac{1}{\text{c}^2}\\\frac{-1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{1}{\text{c}^2}\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}1\\1\\1\end{bmatrix}$

Now,

$|\text{A}|=\begin{vmatrix}\frac{1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{-1}{\text{c}^2}\\\frac{1}{\text{a}^2}&\frac{-1}{\text{b}^2}&\frac{1}{\text{c}^2}\\\frac{-1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{1}{\text{c}^2}\end{vmatrix}$

$=\frac{1}{\text{a}^2\text{b}^2\text{c}^2}\begin{vmatrix}1&1&-1\\1&-1&1\\-1&1&1\end{vmatrix}$

$=\frac{1}{\text{a}^2\text{b}^2\text{c}^2}\times1(-1-1)-1(1+1)-1(1-1)$

$=\frac{1}{\text{a}^2\text{b}^2\text{c}^2}\times(-2-2)$

$=\frac{-4}{\text{a}^2\text{b}^2\text{c}^2}$

$\Rightarrow|\text{A}|\neq0$

So, the given system of equations has a unique solution.