MCQ

MCQ 11 Mark

The system of linear equations:
$x + y + z = 2$
$2x + y − z = 3$
$3x + 2y + kz = 4$
has a unique solution if

✓
$k ≠ 0$
B
$−1 < k < 1$
C
$−2 < k < 2$
D
$k = 0$

Answer

Correct option: A.

$k ≠ 0$

$x + y + z = 2$
$2x + y − z = 3$
$3x + 2y + kz = 4$
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 $\neq 0$
$\Rightarrow k \neq 0$

View full question & answer→

MCQ 21 Mark

The number of solutions of the system of equations:
$2x + y − z = 7$
$x − 3y + 2z = 1$
$x + 4y − 3z = 5$

A
$3$
B
$2$
C
$1$
✓
$0$

Answer

Correct option: D.

$0$

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.

View full question & answer→

MCQ 31 Mark

The number of solutions of the system of equations
$2x + y − z = 7$
$x − 3y + 2z = 1$
$x + 4y − 3z = 5$

A
$3$
B
$2$
C
$1$
✓
$0$

Answer

Correct option: D.

$0$

From the given equation we get,
$\triangle=\begin{vmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{vmatrix}$
$\Rightarrow 2(9 - 8) -1(-3 - 2) - 1(4 + 3)$
$\Rightarrow 2(1) - 1(-5) - 1(7)$
$\Rightarrow 2 + 5 - 7$
$\Rightarrow 2 + 5 -7$
$\Rightarrow 0$
$\triangle_1=\begin{vmatrix}7&1&-1\\1&-3&2\\5&4&-3\end{vmatrix}$
$\Rightarrow 7(9 - 8) - 1(-3 - 10) - 1(4 + 15)$
$\Rightarrow 7(1) - 1(-13) - 1(19)$
$\Rightarrow 7 + 13 - 19$
$\Rightarrow 20 - 19$
$\Rightarrow1\neq0$
Hence the gvien system no solution.

View full question & answer→

MCQ 41 Mark

The system of equations:
$x + y + z = 5$
$x + 2y + 3z = 9$
$x + 3y + \lambda z = µ$
has a unique solution, if

A
$\lambda = 5, µ = 13$
✓
$\lambda \neq 5$
C
$\lambda = 5, µ \neq 13$
D
$µ \neq 13$

Answer

Correct option: B.

$\lambda \neq 5$

$x + y + z = 5$
$x + 2y + 3z = 9$
$x + 3y + \lambda z =\mu$
The determinant of the coefficient matrix $\begin{bmatrix}1&1&1\\1&2&3\\1&3&\lambda\end{bmatrix}$ is
$= 2\lambda - 9 - \lambda + 3 + 1$
$= \lambda - 5$
For unique solution determinant $\neq 0$
$\Rightarrow \lambda \neq 5$
The right hand side is non zero what so ever be the value of $\mu.$

View full question & answer→

MCQ 51 Mark

Consider the system of equations:
$a_1x + b_1y + c_1z = 0$
$a_2x + b_2y + c_2z = 0$
$a_3x + b_3y + c_3z = 0,$
if $\begin{vmatrix}\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\\\text{a}_3&\text{b}_3&\text{c}_3\end{vmatrix}=0$, then the system has

✓
More than two solutions.
B
One trivial and one non$-$trivial solutions.
C
No solutions.
D
Only trivial solution $(0, 0, 0)$.

Answer

Correct option: A.

More than two solutions.

Here, $|A| = 0$ and $B = 0 ($Given$)$
If $|A| = 0$ and $(adj A)B = 0$, then the system is consistent and has infinitely many solutions.
Clearly, it has more than two solutions.

View full question & answer→

MCQ 61 Mark

Let $\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix},\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$. If $AX = B,$ then $X$ is equal to:

✓
$\begin{bmatrix}1\\2\\3\end{bmatrix}$
B
$\begin{bmatrix}-1\\-2\\-3\end{bmatrix}$
C
$\begin{bmatrix}-1\\2\\3\end{bmatrix}$
D
$\begin{bmatrix}0\\2\\1\end{bmatrix}$

Answer

Correct option: A.

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

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,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}0&1\\2&1\end{vmatrix}=-2,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}2&1\\3&1\end{vmatrix}=1,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}2&0\\3&2\end{vmatrix}=4$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}-1&2\\2&1\end{vmatrix}=5,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&2\\3&1\end{vmatrix}=-5,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&-1\\3&2\end{vmatrix}=-5$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}-1&2\\0&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&2\\2&1\end{vmatrix}=3,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&-1\\2&0\end{vmatrix}=2$
$\text{adj }\text{A}=\begin{bmatrix}-2&5&-1\\5&-5&-5\\-1&2&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$
$\therefore\ \text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}\begin{bmatrix}3\\1\\4\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-6+5-4\\3-5+12\\12-5+8\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\frac{1}{5}\begin{bmatrix}-5\\10\\15\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\begin{bmatrix}-1\\2\\3\end{bmatrix}$

View full question & answer→

MCQ 71 Mark

The existence of the unique solution of the system of equations:
$x + y + z = \lambda$
$5x − y + \mu z = 10$
$2x + 3y − z = 6$
depends on

✓
$\mu$ only.
B
$\lambda$ only.
C
$\lambda$ and $\mu$ both.
D
neither $\lambda$ nor $\mu$

Answer

Correct option: A.

$\mu$ only.

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 $\mu$.

View full question & answer→

MCQ 81 Mark

For the system of equations:
$x + 2y + 3z = 1$
$2x + y + 3z = 2$
$5x + 5y + 9z = 4$

✓
There is only one solution.
B
There exists infinitely many solution.
C
There is no solution.
D
None of these.

Answer

Correct option: A.

There is only one 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.

View full question & answer→

MCQ 91 Mark

The system of equation $x + y + z = 2, 3x − y + 2z = 6$ and $3x + y + z = −18$ has:

✓
A unique solution.
B
No solution.
C
An infinite number of solutions.
D
Zero solution as the only solution.

Answer

Correct option: A.

A unique 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.

View full question & answer→

MCQ 101 Mark

Let $a, b, c$ be positive real numbers. The following system of equations in $x, y$ and $z \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=1,$ $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1,$ $-\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1$ has:

A
No solution.
✓
Unique solution.
C
Infinitely many solutions.
D
Finitely many solutions.

Answer

Correct option: B.

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

View full question & answer→

Maths MCQ

Take a timed test