M.C.Q (1 Marks) - Page 4 - Maths STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

MCQ 1511 Mark

If $\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix},$ then $\text{A}^{-1}$ exists if:

A
$\lambda=2$
B
$\lambda\neq2$
C
$\lambda\neq-2$
✓
$\text{None of these}$

Answer

Correct option: D.

$\text{None of these}$

$\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}$

The inverse of a matrix exists if its determinant is not equal to 0.

Consider,

$|\text{A}|=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}\neq0$

$\Rightarrow|\text{A}| = 2 (6 – 5) – \lambda (0 – 5) + (-3) (0 – 2)\neq0$

$\Rightarrow2 + 5\lambda + 6 \neq 0$

$\Rightarrow5\lambda + 8 \neq 0$

$\Rightarrow5\lambda \neq -8$

$\Rightarrow\lambda\neq\frac{-8}{5}$

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MCQ 1521 Mark

Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|\text{adj.} \ A|$ is equal to:

A
$|A|$
✓
$|A|^2$
C
$|A|^3$
D
$3|A|$

Answer

Correct option: B.

$|A|^2$

If $A$ is a non $-$ singular matrix of order $\text{n}\times\text{n},$ then $|\text{adj}. A| = |A|^{n-1}$
$\therefore$ Putting $n = 3, |\text{adj.} A| = |A|^2$
Therefore, option $(b) $ is correct.

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MCQ 1531 Mark

If $A$ satisfies the equation $\text{x}^2-5\text{x}^2+4\text{x}+\lambda=0$ then $A^{-1}$ exists if:

A
$\lambda\neq1$
B
$\lambda\neq2$
C
$\lambda\neq-1$
✓
$\lambda\neq0$

Answer

Correct option: D.

$\lambda\neq0$

A satisfies $\text{x}^3-5\text{x}^2+4\text{x}+\lambda=0$
$\Rightarrow\text{A}^3-5\text{A}^2+4\text{A}=-\lambda$
Assuming $A^{-1}$ exists, we get
$\text{A}^{-1}(\text{A}^3-5\text{A}^2+4\text{A})=-\lambda\text{A}^{-1}$
$\Rightarrow\text{A}^2-5\text{A}+4=-\text{A}^{-1}\lambda$
$\Rightarrow\text{A}-1=\frac{-(\text{A}^2-5\text{A}+4)}{\lambda}$
Thus$, A^{-1}$ exists if $\lambda\neq0$.

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MCQ 1541 Mark

What is the value of a + b + c + d ?

A
62
✓
63
C
65
D
68

Answer

Correct option: B.

63

$\text{ax}^3+\text{bx}^2+\text{cx}+\text{d}=\begin{bmatrix}\text{x}+1&\text{amp;}2\text{x}&\text{amp; 3}\text{x}\\2\text{x}+3&\text{amp;}\text{x+1}&\text{amp;}\text{x}\\2-\text{x}&\text{amp;}3\text{x}+4&\text{amp;}5\text{x}-1\end{bmatrix}$ if

$\text{x}=1\text{a}+\text{b}+\text{c}+\text{d}=\begin{bmatrix}2&\text{amp;}2&\text{amp;3}\\5&\text{amp;}2&\text{amp;1}\\1&\text{amp;}7&\text{amp;4} \end{bmatrix}$

$\text{a}+\text{b}+\text{c}+\text{d}=2-38+99=101-38=63$

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MCQ 1551 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,

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

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.

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MCQ 1561 Mark

If $B$ is a non$-$singular matrix and $A$ is a square matrix, then $\text{det} (B^{-1} AB)$ is equal to:

A
$\text{Det} (A^{-1})$
B
$\text{Det} (B^{-1})$
✓
$\text{Det} (A)$
D
$\text{Det} (B)$

Answer

Correct option: C.

$\text{Det} (A)$

$B$ is non$-$singular.
This implies that $|\text{B}|\neq0,$ that $B$ is invertible and that $B^{-1}$ exists.
Here $B$ is invertible.
$\therefore|\text{B}^{-1}|=|\text{B}|^{-1}=\frac{1}{|\text{B}|}$
$\Rightarrow|\text{B}^{-1}\text{AB}|=|\text{B}^{-1}||\text{AB}|$
$\Rightarrow|\text{B}^{-1}\text{AB}|=|\text{B}|^{-1}|\text{A}||\text{B}|$
$\Rightarrow|\text{B}^{-1}\text{AB}|=\frac{1}{|\text{B}|}|\text{A}||\text{B}|$
$\Rightarrow|\text{B}^{-1}\text{AB}|=|\text{A}|$

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MCQ 1571 Mark

There are two value of a which makes the determinant $\triangle=\begin{vmatrix}1&-2&5\\2&\text{a}&-1\\0&4&2\text{a}\end{vmatrix}$ equal to $86$. The sum of these two values is:

A
$4$
B
$5$
✓
$-4$
D
$9$

Answer

Correct option: C.

$-4$

$\triangle=\begin{vmatrix}1&-2&5\\2&\text{a}&-1\\0&4&2\text{a}\end{vmatrix}=86$
$\Rightarrow 1(2a^2 + 4) - 2(-4a - 20) = 86$
$\Rightarrow 2a^2 + 4 + 8a + 40 = 86$
$\Rightarrow 2a^2 + 8a - 42 = 0$
$\Rightarrow a^2 + 4a - 21 = 0$
$\Rightarrow a^2+ 7a - 3a - 21 = 0$
$\Rightarrow a(a + 7) - 3(a + 7) = 0$
$\Rightarrow a = -7, 3$
Sum of the two values of $a = -7 + 3 = -4$
Hence, the correct option is $(c)$

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MCQ 1581 Mark

If $\text{A}=\begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix},$ then $\text{det (adj (adj A))}$ is:

✓
$14^4$
B
$14^3$
C
$14^2$
D
$14$

Answer

Correct option: A.

$14^4$

$\text{A}=\begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix}$
$|\text{A}|=14$
$\text{det(adj(adj A))}=|\text{A}|^{{\text{n}-1}^{2}}$
$\text{det(adj(adj A))}=|14|^{{3-1}^{2}}=14^4$

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MCQ 1591 Mark

If $\text{A}=\begin{vmatrix} 1 &\text{amp; 2} \\ 2 &\text{amp; 1} \end{vmatrix}$ and $\text{f}\text{(x)}=\frac{1+\text{x}}{1-\text{x}},$ then $\text{f}(|\text{A}|)$ is:

✓
$\dfrac{-1}{2}$
B
$\dfrac{1}{2}$
C
$\dfrac{-1}{3}$
D
$\text{None of these}$

Answer

Correct option: A.

$\dfrac{-1}{2}$

Here, $|\text{A}| =1\times 1-2\times 2 = -3$

$\therefore\text{f}(|\text{A}|)=\cfrac{1+(-3)}{1+3}=-\cfrac{1}{2}$

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MCQ 1601 Mark

If $d$ is the determinant of a square matrix $A$ of order $n,$ then the determinant of its adjoint is:

A
$d^n$
✓
$d^{n-1}$
C
$d^{n+1}$
D
$d$

Answer

Correct option: B.

$d^{n-1}$

We know,
$\text{|adj} A| = |A|^{n-1}$
$\Rightarrow \text{|adj} A| = d^{n-1}$

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MCQ 1611 Mark

Evaluate $ |\text{A}|^2-5|\text{A}|+1,$ if $\text{A}=\begin{bmatrix}7&4\\5&5\end{bmatrix}$ is:

A
161
B
251
C
150
✓
151

Answer

Correct option: D.

151

Given that, $\text{A}=\begin{bmatrix}7&4\\5&5\end{bmatrix}$
$|\text{A}|=(7(5)-5(4))=35-20=15$

$|\text{A}|^2-5|\text{A}|+1=(15)^2-5(15)+1=225-75+1=151.$

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MCQ 1621 Mark

If $\left|\begin{array}{lll}<\text{br}> &1 &\text{amp; } 0 &\text{amp; } 0\\ <\text{br}>&2 &\text{amp; } 3 &\text{amp; } 4\\ <\text{br}>&5 &\text{amp; } -6 &\text{amp; x}<\text{br}> \end{array}\right|=45$ then $\text{x}=$

A
4
✓
7
C
-5
D
-7

Answer

Correct option: B.

7

Given, $\left|\begin{array}{lll}<\text{br}> &1 &\text{amp; } 0 &\text{amp; } 0\\ <\text{br}>&2 &\text{amp; } 3 &\text{amp; } 4\\ <\text{br}>&5 &\text{amp; } -6 &\text{amp; x}<\text{br}> \end{array}\right|=45$
By operation of matrix (5),

$1(3\text{x}+24)=45$

$3\text{x}=21$

$\Rightarrow \text{x}=7$

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MCQ 1631 Mark

Choose the correct answer from given four options in each of the Exercise:If $x, y, z$ are all different from zero and $\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0,$ then the value of $x^{-1} + y^{-1} + z^{-1}$ is:

A
$xyz$
B
$x^{-1} + y^{-1} + z^{-1}$
C
$-x - y - z$
✓
$-1$

Answer

Correct option: D.

$-1$

We have, $\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0$
Applying $\text{C}_1\rightarrow\text{C}_1-\text{C}_3$ and $\text{C}_2\rightarrow\text{C}_2-\text{C}_3,$
$\Rightarrow\ \begin{vmatrix}\text{x}&0&1\\0&\text{y}&1\\-\text{z}&-\text{z}&1+\text{z}\end{vmatrix}=0$
Expanding along $R_1$,
$\text{x}\big[\text{y}(1+\text{z})+\text{z}\big]-0+1(\text{yz})=0$
$\Rightarrow\text{x}(\text{y}+\text{yz}+\text{z})+\text{zy}=0$
$\Rightarrow\text{xy}+\text{xyz}+\text{xz}+\text{xz}=0$
$\Rightarrow\frac{\text{xy}}{\text{xyz}}+\frac{\text{xyz}}{\text{xyz}}+\frac{\text{xz}}{\text{xyz}} +\frac{\text{yz}}{\text{xyz}} =0\ [$On dividing by $(xyz)$ from both sides$]$
$\Rightarrow\ \frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+1=0$
$\Rightarrow\ \frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=-1$
$\therefore\text{x}^{-1}+\text{y}^{-1}+\text{z}^{-1}=-1$

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MCQ 1641 Mark

Choose the correct answer from given four options in each of the Exercise:The value of the determinant $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$is:

A
$9x^2(x + y)$
B
$9y^2(x + y)$
C
$3y^2(x + y)$
✓
$7x^2(x + y)$

Answer

Correct option: D.

$7x^2(x + y)$

We have, $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}3(\text{x}+\text{y})&\text{x}+\text{y}&\text{y}\\3(\text{x}+\text{y})&\text{x}&\text{y}\\3(\text{x}+\text{y})&\text{x}+2\text{y}&-2\text{y}\end{vmatrix}\big[\because\ \text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3$ and $C_3\rightarrow\text{C}_3 -\text{C}_2\big]$
$=3(\text{x}+\text{y})\begin{vmatrix}1&(\text{x}+\text{y})&\text{y}\\1&\text{x}&\text{y}\\1&(\text{x}+2\text{y})&-2\text{y}\end{vmatrix} [$Taking $3(x + y)$ common from first column$]$
$=3(\text{x}+\text{y})\begin{vmatrix}0&\text{y}&0\\1&\text{x}&\text{y}\\1&(\text{x}+2\text{y})&-2\text{y}\end{vmatrix}[\because\ \text{R}_1\rightarrow\text{R}_1-\text{R}_2]$
Expanding along $R_1$,
$=3(\text{x}+\text{y})\big[-\text{y}(-2\text{y})-\text{y}\big]$
$=3\text{y}^2.3(\text{x}+\text{y})=9\text{y}^2(\text{x}+\text{y})$

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MCQ 1651 Mark

If $\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$ is prthogonal, than x + y =

A
3
✓
0
C
-3
D
1

Answer

Correct option: B.

0

$\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$
$\text{A}^\text{T}\text{A}=\text{I}$

$\frac{1}{3}\begin{bmatrix} 1 & 2 & \text{x} \\ 1 & 1 & 2 \\ 2 & -2 & \text{y} \end{bmatrix}\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 &\text{y} \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\frac{1}{9}\begin{bmatrix} 1+4+\text{x}^2 & 1+2+2\text{x} & \text{xy}-2 \\ 1+2+2\text{x} & 1+1+4 & 2-2+2\text{y} \\ 2-4+\text{xy} & 2-2+2\text{y} & 4+4+\text{y}^2 \end{bmatrix}$

$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\begin{bmatrix} 5+\text{x}^2 & 3+2\text{x} & \text{xy}-2 \\ 3+2\text{x} & 6 & 2\text{y} \\ -6+\text{xy} & 2\text{y} & 8+\text{y}^2 \end{bmatrix}$

$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Equality of two matrices does not hold Matrix A is not orthogonal.

Hence, the given question is incorrect.

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MCQ 1661 Mark

The matrix $\begin{bmatrix}1 &\text{amp; } 0 &\text{amp; 1} \\ 2 &\text{amp; 1}&\text{amp; 0} \\ 3 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$ is:

A
non-singular
✓
singular
C
skew-symmetric
D
symmetric

Answer

Correct option: B.

singular

Given $\text{A}=\begin{bmatrix}1 &\text{amp; } 0 &\text{amp; 1} \\ 2 &\text{amp; 1}&\text{amp; 0} \\ 3 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$

$\text{|A|}=\begin{bmatrix}1 &\text{amp; } 0 &\text{amp; 1} \\ 2 &\text{amp; 1}&\text{amp; 0} \\ 3 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$

$\Rightarrow∣\text{A}∣=1−1=0$

Hence, A is singular.

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MCQ 1671 Mark

If $\text{A}=\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix},$ find $|\text{A}|$

✓
352
B
356
C
325
D
532

Answer

Correct option: A.

352

$\Rightarrow|\text{A}|=\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix}$Evaluating along the first row, we get

$\triangle=2\begin{bmatrix}1&3\\8&2\end{bmatrix}-5\begin{bmatrix}6&3\\4&2\end{bmatrix}+9\begin{bmatrix}6&1\\4&8\end{bmatrix}$

$\triangle=2(2-24)-5(12-12)+9(48-4)$

$\triangle=2(-22)-0+9(44) $

$\triangle=-44+9(44)=44(-1+9)=352 $

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MCQ 1681 Mark

If $\text{S}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix},$ then adj A is:

A
$\begin{bmatrix} -\text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
✓
$\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
C
$\begin{bmatrix} \text{d} & \text{b} \\ \text{c} & \text{a} \end{bmatrix}$
D
$\begin{bmatrix} \text{d} & \text{c} \\ \text{b} & \text{a} \end{bmatrix}$

Answer

Correct option: B.

$\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$

Adjoint of a square matrix of order 2 is obtained by interchancing the diagoinal elements and changing the signs of off-diagonal elements.

Here,

$\text{A}=\begin{bmatrix}\text{a} & \text{bc} & \text{d} \end{bmatrix}$

$\Rightarrow\text{adj A}=\begin{bmatrix}\text{d} & -\text{b}-\text{c} & \text{a} \end{bmatrix}$

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MCQ 1691 Mark

A set of points which do not lie on the same line are called as

A
collinear
✓
non-collinear
C
concurrent
D
square

Answer

Correct option: B.

non-collinear

A set of points which do not lie on the same line are called as non collinear points
solution

solution

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MCQ 1701 Mark

If $\text{A} = \begin{bmatrix}1&\text{amp; } \log_{\text{b}}\text{a}\\ \log_\text{a}\text{b}&\text{amp; } 1\end{bmatrix}$then $ |\text{A}|$ is equal to:

✓
$0$
B
$\log_\text{a}\text{b}$
C
$-1$
D
$\log_\text{b}\text{a}$

Answer

Correct option: A.

$0$

On solving the given matrix,
$|\text{A}|=1-\log_\text{a}\text{b}.\log_\text{b} \text{a}=1-1=0$

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M.C.Q (1 Marks) - Page 4 - Maths STD 12 Science Questions - Vidyadip