MATHS M.C.Q (1 Marks)

2. 2

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.

3. ±6

2. λ ≠ 5

We know that $\big|\text{A}^{-1}\big|=\frac{1}{|\text{A}|}$

1. -4

1. -20

3. $\pm6$

4. 32

3. $\pm6$

$\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$=\begin{vmatrix}0&0&2(\text{a}+\text{c}-2\text{b})\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$[$Applying $R_1 \rightarrow R_1 + R_3- R_2, R_1 \rightarrow R_1 - R_2]$
$=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$ $[\because a, b, c$ are in $A.P.]$
$=0$

1. 0

$\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$
$=\begin{vmatrix}1&\sin\theta&2\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix} [$Applying $C_3 \rightarrow C_3 + C_1]$
$=2\times\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix} [$Expanding along $C_3]$
$=2(\sin^2\theta+1)$
Given, $0\leq\theta\leq2\pi$
$-1\leq\sin\theta\leq1$
$0\leq\sin^2\theta\leq1$
$|\text{A}|=2(\sin^2\theta+1)$
$|\text{A}|=2\times1=2$ $[\theta=0]$
$|\text{A}|=2\times2=4$ $[\theta=2\pi]$
$\text{Det (A)}\in[2,4]$

2. 0

3. $5\sqrt{3}+4$

1. 0

1. sq.units

$\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}&0&-\text{z}\\0&\text{y}&-\text{z}\\1&1&1+\text{z}\end{vmatrix}=0 [$Applying $R_{2 }\rightarrow R_2 - R_3$ and $R_1 \rightarrow R_1 - R_3]$
$\Rightarrow\text{x}\big[\text{y}(1+\text{z})+\text{z}\big]+1(\text{yz})=0[ $Expanding along first column$]$
$\Rightarrow\text{x}[\text{y}+\text{yz}+\text{z}]+\text{yz}=0$
$\Rightarrow\text{xy}+\text{xyz}+\text{xz}+\text{yz}=0$
$\Rightarrow\text{xy}+\text{yz}+\text{zx}=-\text{xyz}$
$\Rightarrow\frac{\text{xy}}{\text{xyz}}+\frac{\text{yz}}{\text{xyz}}+\frac{\text{zx}}{\text{xyz}}=-\frac{\text{xyz}}{\text{xyz}}$
$\Rightarrow\frac{1}{\text{z}}+\frac{1}{\text{x}}+\frac{1}{\text{y}}=-1$
$\Rightarrow\text{x}^{-1}+\text{y}^{-1}+\text{z}^{-1}=-1$
Hence, the correct option is $(d)$

3. -8

1. A unique solution.

1. n

2. 0

$A$ and $B$ are non-singular matrices of order $n \times n.$
$\therefore|\text{A}|\neq0\text{ and }|\text{B}|\neq0\ .....(\text{i})$
$A$ and $B$ are of the same order, so $AB$ is defined and is of the same order.
Thus,
$|AB| = |A||B|$
$\Rightarrow|\text{AB}|\neq0\ \big[\text{Using (1)}\big]$
Thus, $AB$ is non-singular.

3. 10

2. 9

4. −13

1. 0

2. 63

2. Unique solution

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

1. $\dfrac{-1}{2}$

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