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Maths MCQ

Determinants · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 33 questions.

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33 questions · auto-graded multiple-choice test.

MCQ 11 Mark
The value of $\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^4&5^5&5^6\end{vmatrix}$ is:
  • A
    $5^2$
  • ✓
    $0$
  • C
    $5^{13}$
  • D
    $5^9$
Answer
Correct option: B.
$0$
$\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^4&5^5&5^6\end{vmatrix}$
$=5^2\times5^3\times5^4\begin{vmatrix}1&5&5^2\\1&5&5^2\\1&5&5^2\end{vmatrix} [$Taking out common factors from $R_1, R_2, R_3]$
$=5^2\times5^3\times5^4\times5\begin{vmatrix}1&1&5^2\\1&1&5^2\\1&1&5^2\end{vmatrix}$
$=5^2\times5^3\times5^4\times0$
$=0$
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MCQ 21 Mark
$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times\begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$
  • A
    $7$
  • ✓
    $10$
  • C
    $1$
  • D
    $17$
Answer
Correct option: B.
$10$
$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times\begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$
$=\begin{vmatrix}\log_32^9&\log_{2^{2}}3\\\log_32^3&\log_{2^2}3^3\end{vmatrix}\times\begin{vmatrix}\log_23&\log_{2^{3}}3\\\log_32^3&\log_32^2\end{vmatrix}$
$=\begin{vmatrix}9\log_32&\frac{1}{2}\log_23\\3\log_32&\frac{1}{2}\times2\log_23\end{vmatrix}\times\begin{vmatrix}\log_23&\frac{1}{3}\log_23\\2\log_32&2\log_32\end{vmatrix}$
$=\Big(\big(9\log_32\times\log_23\big)-\big(3\log_32\times\frac{1}{2}\log_23\big)\Big)\times\Big(\big(\log_23\times2\log_32\big)-\Big(\frac{1}{3}\log_23\times2\log_32\Big)\Big)$
$=\Big(9-\frac{3}{2}\Big)\times\Big(2-\frac{2}{3}\Big)$
$=\frac{15}{2}\times\frac{4}{3}$
$=10$
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MCQ 31 Mark
If $\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\2\text{k}&\text{n}^2+\text{n}+2&\text{n}^2+\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix} $ and $\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=48,$ then $n$ equals:
  • ✓
    $4$
  • B
    $6$
  • C
    $8$
  • D
    None of these.
Answer
Correct option: A.
$4$
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\2\text{k}&\text{n}^2+\text{n}+2&\text{n}^2+\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix} $
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\1&\text{n}+2&-2\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_3$
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_1$
Now,
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\3&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+.......+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{n}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+1(-2-\text{n})\Big)+\text{n}\Big(0+2\Big)\Big]\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+3(-2-\text{n})\Big)+\text{n}\Big(0+6\Big)\Big]+......\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+(2\text{n}-1)(-2-\text{n})\Big)+\text{n}\Big(0+2(2\text{n}-1)\Big)\Big]$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(1+3+5+.....+\text{n}\Big)$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(\text{n}^2\Big)$
$\Rightarrow\ 2\text{n}^2+4\text{n}=48$
$\Rightarrow\ (\text{n}+6)(\text{n}-4)=0$
$\Rightarrow\ \text{n}=4$
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MCQ 41 Mark
Let $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}$ be an identity in $x,$ where $a, b, c, d, e$ are independent of $x.$ Then the value of $e$ is:
  • A
    $4$
  • ✓
    $0$
  • C
    $1$
  • D
    None of these.
Answer
Correct option: B.
$0$
Let $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}$
$=(\text{x}^2+3\text{x})\begin{vmatrix}-2\text{x}&\text{x}-4\\\text{x}+4&3\text{x}\end{vmatrix}-(\text{x}-1)\begin{vmatrix}\text{x}+1&\text{x}-4\\\text{x}-3&3\text{x}\end{vmatrix}+(\text{x}+3)\begin{vmatrix}\text{x}+1&-2\text{x}\\\text{x}-3&\text{x}+4\end{vmatrix}$
$= (x^2 + 3x)(-6x - x^2 + 16) - (x - 1)(3x^2 + 3x - x^2 + 7x - 12) + (x + 3)(x^2 + 5x + 4 + 2x^2 - 6x)$
$= -7x^4 + 16x^2 + 48x + 21x^3 + 8x^2 - 22x - 2x^3 - 12 + 8x^2 + x + 3x^3 + 12$
$= -7x^4 + 22x^3 + 32x^2 + 27x + 0$
But $x$ is a root of $ax^4 + bx^3 + cx^2 + dx + e$
$e = 0$
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MCQ 51 Mark
If $\begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}=16,$ then the value of $\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b}+\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+\text{z}&\text{c}+\text{r}\end{vmatrix}$ is:
  • A
    $4$
  • B
    $8$
  • C
    $16$
  • ✓
    $32$
Answer
Correct option: D.
$32$
$\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b}+\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+\text{z}&\text{c}+\text{r}\end{vmatrix}=\begin{vmatrix}\text{p}&\text{a}&\text{a}\\\text{q}&\text{b}&\text{b}\\\text{r}&\text{c}&\text{c}\end{vmatrix}+​​​​\begin{vmatrix}\text{p}&\text{a}&\text{p}\\\text{q}&\text{b}&\text{q}\\\text{r}&\text{c}&\text{r}\end{vmatrix}\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}\\+​​​​\begin{vmatrix}\text{p}&\text{x}&\text{p}\\\text{q}&\text{y}&\text{q}\\\text{r}&\text{z}&\text{r}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{a}&\text{a}\\\text{y}&\text{b}&\text{b}\\\text{r}&\text{c}&\text{c}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{r}&\text{c}&\text{r}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{x}&\text{a}\\\text{y}&\text{y}&\text{b}\\\text{z}&\text{z}&\text{c}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{x}&\text{p}\\\text{y}&\text{y}&\text{q}\\\text{z}&\text{z}&\text{r}\end{vmatrix}$
$=0+0+\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}+0+0+\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{z}&\text{c}&\text{r}\end{vmatrix}+0+0$
$=\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
$=2\begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}$
$=2\times16=32$
Hence, the correct option is $(b)$
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MCQ 61 Mark
If $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:
  • A
    $f(a) = 0$
  • B
    $f(b) = 0$
  • ✓
    $f(0) = 0$
  • D
    $f(1) = 0$
Answer
Correct option: C.
$f(0) = 0$
Let $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix}$
Now, $\text{f(a)}=\begin{vmatrix}0&\text{a}-\text{a}&\text{a}-\text{b}\\\text{a}+\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&0&\text{a}-\text{b}\\2\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$
$=(\text{a}-\text{b})(2\text{a}^2+2\text{ac})\neq0$
$\text{f(b)}=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}-\text{b}\\\text{b}+\text{a}&0&\text{b}-\text{c}\\\text{b}+\text{b}&\text{b}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&\text{b}-\text{a}&0\\\text{b}+\text{a}&0&\text{b}-\text{c}\\2\text{a}&\text{b}+\text{c}&0\end{vmatrix}$
$=(\text{b}-\text{a})(2\text{ab}-2\text{ac})\neq0$
$\text{f(0)}=\begin{vmatrix}0&\text{0}-\text{a}&\text{0}-\text{b}\\\text{0}+\text{a}&0&\text{0}-\text{c}\\\text{0}+\text{b}&\text{0}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&\text{c}\\\text{b}&\text{c}&0\end{vmatrix}$
$=\text{a}(\text{bc})-\text{b}(\text{ac})=0$
Hence, the correct option is $(c)$
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MCQ 71 Mark
If $w$ is a non$-$real cube root of unity and $n$ is not a multiple of $3$, then $\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}$ is equal to:
  • ✓
    $0$
  • B
    $\omega$
  • C
    $\omega^2$
  • D
    $1$
Answer
Correct option: A.
$0$
$\triangle=\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}$
$=\begin{vmatrix}1+\omega^{\text{n}}+\omega^{2\text{n}}&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}+1+\omega^{\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}+\omega^{2\text{n}}+1&\omega^{2\text{n}}&1\end{vmatrix} [$Applying $C_1 \rightarrow C_1+ C_2 + C_3]$
Now, $1+\omega+\omega^2=0$ $[\because$ is a complex cube root of unity$]$
$1+\omega^{\text{n}}+\omega^{2\text{n}}=0$
$[\because n$ is not a multiple of $3]$
$\triangle=\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}=0$
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MCQ 81 Mark
The determinant $\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$ equals:
  • A
    $abc(b - c)(c - a)(a - b)$
  • B
    $(b - c)(c - a)(a - b)$
  • C
    $(a + b + c)(b - c)(c - a)(a - b)$
  • ✓
    None of these
Answer
Correct option: D.
None of these
$\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$
$=\begin{vmatrix}\text{b}(\text{b}-\text{a})&\text{b}-\text{c}&\text{c}(\text{b}-\text{a})\\\text{a}(\text{b}-\text{a})&\text{a}-\text{b}&\text{b}(\text{b}-\text{a})\\\text{c}(\text{b}-\text{a})&\text{c}-\text{a}&\text{a}(\text{b}-\text{a})\end{vmatrix}$
$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}&\text{b}-\text{c}&\text{c}\\\text{a}&\text{a}-\text{b}&\text{b}\\\text{c}&\text{c}-\text{a}&\text{a}\end{vmatrix} [$Taking $(b - a)$ common from $C_1$ and $C_3]$
$=(\text{b}-\text{a})^2\begin{vmatrix}0&\text{b}-\text{c}&\text{c}\\0&\text{a}-\text{b}&\text{b}\\0&\text{c}-\text{a}&\text{a}\end{vmatrix} [$Applying $C_1\rightarrow C_1 - C_2 - C_3]$
$=0$
Hence, the correct option is $(d)$
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MCQ 91 Mark
If $A$ and $B$ are square matrices of order $2$, then det $(A + B) = 0$ is possible only when:
  • A
    Det $(A) = 0$ or det $(B) = 0$
  • B
    Det $(A) +$ det $(B) = 0$
  • C
    Det $(A) = 0$ and det $(B) = 0$
  • ✓
    $A + B = 0$
Answer
Correct option: D.
$A + B = 0$
Let $\text{A}=[\text{a}_{\text{ij}}]$ and $\text{B}=[\text{b}_{\text{ij}}]$ be a square matrix of order $2$
As their orders are same,
$A + B$ is defined as
$\text{A}+\text{B}=[\text{a}_{\text{ij}}+\text{b}_\text{ij}]$
$\Rightarrow|\text{A}+\text{B}|=|\text{a}_{\text{ij}}+\text{b}_\text{ij}|$
Now,
$|\text{A}+\text{B}|=0$
$\Rightarrow|\text{a}_{\text{ij}}+\text{b}_\text{ij}|=0$
$\Rightarrow[\text{a}_{\text{ij}}+\text{b}_\text{ij}]=0$
$[$corrsponding term is $0]$
$\Rightarrow\text{A}+\text{B}=0$
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MCQ 101 Mark
The value of the determinant $\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{c}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$ is:
  • A
    $a^3 + b^3 + c^3$
  • B
    $3bc$
  • ✓
    $a^3 + b^3 + c^3 - 3abc$
  • D
    None of these
Answer
Correct option: C.
$a^3 + b^3 + c^3 - 3abc$
$\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$
$=\begin{vmatrix}-\text{b}&\text{b}+\text{c}+\text{a}&\text{a}\\-\text{c}&\text{c}+\text{a}+\text{b}&\text{b}\\-\text{a}&\text{a}+\text{b}+\text{c}&\text{c}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 - C_3$ and $C_2 \rightarrow C_2 + C_3]$
$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}&1&\text{b}\\\text{a}&1&\text{c}\end{vmatrix} [$Taking $(-1)$ common from $C_1$ and $(a + b + c)$ common from $C_2]$
$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}-\text{b}&0&\text{b}-\text{a}\\\text{a}-\text{b}&0&\text{c}-\text{a}\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_1$ and $R_3\rightarrow R_3 - R_1]$
$= (-1)(a + b + c)[-(c - b)(c - a) + (b - a)(a - b)]$
$= (-1)(a + b + c)[-c^2 + ac + bc - ab + ba - b^2 - a^2 + ab]$
$= (-1)(a + b + c)(-a^2 - b^2 - c^2 + ab + bc + ac)$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$
$= a^3 + ab^2 + ac^2 - a^2b - abc - a^2c + ba^2 + b^3 + bc^2 - ab^2-b^2c - abc + ca^2 + cb^2 + c^3 - acb - bc^2 - ac^2$
$= a^3 + b^3 + c^3- 3abc$
Hence, the correct option is $(c)$
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MCQ 111 Mark
Using the factor theorem it is found that $a + b, b + c$ and $c + a$ are three factors of the determinant $\begin{vmatrix}-2\text{a}&\text{a}+\text{b}&\text{a}+\text{c}\\\text{b}+\text{a}&-2\text{b}&\text{b}+\text{c}\\\text{c}+\text{a}&\text{c}+\text{b}&-2\text{c}\end{vmatrix}$ the other factor in the value of the determinant is:
  • ✓
    $4$
  • B
    $2$
  • C
    $a + b + c$
  • D
    None of these.
Answer
Correct option: A.
$4$
$\triangle=\begin{vmatrix}-2\text{a}&\text{a}+\text{b}&\text{a}+\text{c}\\\text{b}+\text{a}&-2\text{b}&\text{b}+\text{c}\\\text{c}+\text{a}&\text{c}+\text{b}&-2\text{c}\end{vmatrix}$
Let $a + b = 2C, b + c = 2A$ and $c + a = 2B$
$\Rightarrow a + b + b + c + c + a = 2A + 2B + 2C$
$\Rightarrow 2(a + b + c) = (A + B + C)$
Also, $a = (a + b + c) - (b + c) = (A + B + C) - 2A = B + C - A$
Similarly, $b = C + A - B, c = A + B - C$
Hence, $4$ is the order factor of the determinant.
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MCQ 121 Mark
Which of the following is not correct in a given determinant of $A,$ where $A = [a_{ij}]_{3\times 3}$:
  • A
    Order of minor is less than order of the det $(A).$
  • ✓
    Minor of an element can never be equal to cofactor of the same element.
  • C
    Value of determinant is obtained by multiplying elements of a row or column by corresponding cofactors.
  • D
    Order of minors and cofactors of elements of $A$ is same.
Answer
Correct option: B.
Minor of an element can never be equal to cofactor of the same element.
$C_{ij} = (-1)^{i+j}M_{ij}$
So, for even values of $i + j, C_{ij} = M_{ij}$.
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MCQ 131 Mark
If $\triangle_1=\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix},\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix},$ then:
  • ✓
    $\triangle_1+\triangle_2=0$
  • B
    $\triangle_1+2\triangle_2=0$
  • C
    $\triangle_1=\triangle_2$
  • D
    None of these.
Answer
Correct option: A.
$\triangle_1+\triangle_2=0$
$\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}1&\text{abc}&\text{a}^2\\1&\text{bca}&\text{b}^2\\1&\text{cab}&\text{c}^2\end{vmatrix}$ [$R_1, R_2, R_3$ are multiplies by $a, b$ and $c$ respectively, therefore we divide by $abc]$
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}1&1&\text{a}^2\\1&1&\text{b}^2\\1&1&\text{c}^2\end{vmatrix}$
$[$Taking abc common from $C_2]$
$=-\begin{vmatrix}1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$
We know that the value of a determinant remains unchanged if its rows and columns are interchanged.
so, $\triangle_2=-\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix} $
$=-\triangle_1$
$\triangle_1+\triangle_2=0$
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MCQ 141 Mark
If $a > 0$ and discriminant of $ax^2 + 2bx + c$ is negative, then $\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{ax}+\text{b}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$ is:
  • A
    Positive
  • B
    $(ac - b^2)(ax^2 + 2bx + c)$
  • ✓
    Negative
  • D
    $0$
Answer
Correct option: C.
Negative
Discriminant D of $ax^2 + 2bx + c = (2b)^2- 4ac < 0 [$Given$]$
$\Rightarrow 4b^2 - 4ac < 0$
$\Rightarrow b^2 - ac < 0$, where $a > 0 .....(i)$
$\Rightarrow\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{ax}+\text{b}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}\text{ax}&\text{bx}&\text{ax}^2+\text{bx}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix} [$Applying $R_1 \rightarrow xR_1]$
$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}\text{ax}+\text{b}&\text{bx}+\text{c}&\text{ax}^2+\text{bx}+\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix} [$Applying $R_1 \rightarrow R_1 + R_2]$
$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}0&0&\text{ax}^2+2\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_3]$
$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{Bmatrix}\text{ax}^2+2\text{bx}+\text{c}\begin{vmatrix}\text{b}&\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}\end{vmatrix}\end{Bmatrix} [$Expanding along $R_1]$
$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{x}+\text{bc}-\text{acx}-\text{bc})$
$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})\text{ x }(\text{b}^2-\text{ac})$
$\Rightarrow\triangle=(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{ac})<0 [$From eq. $(i)]$
$\Rightarrow\triangle<0$
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MCQ 151 Mark
If $a, b, c$ are distinct, then the value of $x$ satisfying $\begin{vmatrix}0&\text{x}^2-\text{a}&\text{x}^3-\text{b}\\\text{x}^2+\text{a}&0&\text{x}^2+\text{c}\\\text{x}^4+\text{b}&\text{x}-\text{c}&0\end{vmatrix}=0$ is:
  • A
    $c$
  • B
    $a$
  • C
    $b$
  • ✓
    $0$
Answer
Correct option: D.
$0$
When we put $x = 0$ in the given matrix,
then it turns out to be the skew symmetric matrix of order $3$ and the determinant of the skew symmetric matrix of odd order is always $0.$
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MCQ 161 Mark
The number of distinct real roots of $\begin{vmatrix}\text{cosec}&\sec\text{x}&\sec\text{x}\\\sec\text{x}&\text{cosec}\text{x}&\sec\text{x}\\\sec\text{x}&\sec\text{x}&\text{cosecx}\end{vmatrix}=0$ lies in the interval $-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$ is:
  • A
    $1$
  • ✓
    $2$
  • C
    $3$
  • D
    $0$
Answer
Correct option: B.
$2$

Let $\triangle=\begin{vmatrix}\text{cosec x}&\sec\text{x}&\sec\text{x}\\\sec\text{x}&\text{cosec}\text{x}&\sec\text{x}\\\sec\text{x}&\sec\text{x}&\text{cosec x}\end{vmatrix}$
$=(\text{cosec x})^3\begin{vmatrix}1&\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&1&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}&1\end{vmatrix}$
$=(\text{cosec x})^3\begin{vmatrix}1&\tan\text{x}&\tan\text{x}\\\tan\text{x}&1&\tan\text{x}\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$
$=(\text{cosec x})^3\begin{vmatrix}1-\tan\text{x}&\tan\text{x}-1&0\\0&1-\tan\text{x}&\tan\text{x}-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 - R_2, R_2 \rightarrow R_2 - R_3$]
$=(\text{cosec x})^3(1-\tan\text{x})^2\begin{vmatrix}1&-1&0\\0&1&-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$ [Taking out $(1-\tan\text{x})$ common from $R_1$ and $R_2$]
$=(\text{cosec x})^3(1-\tan\text{x})^2\begin{Bmatrix}1\begin{vmatrix}1&-1\\\tan\text{x}&1\end{vmatrix}+\tan\text{x}\begin{vmatrix}-1&0\\1&-1\end{vmatrix}\end{Bmatrix}$ [Expanding along $C_1$]
$=(\text{cosec x})^3(1-\tan\text{x})^2\{1+\tan\text{x}+\tan\text{x}\}$
$=(\text{cosec x})^3(1-\tan\text{x})^2\{1+2\tan\text{x}\}$
$\triangle=0$
$=(\text{cosec x})^3(1-\tan\text{x})^2(1+2\tan\text{x})=0$
$(1-\tan\text{x})=0,(\text{coses x})^3=0$ and $(1+2\tan\text{x})=0$
Or $\tan\text{x}=1,\text{cosec x}=0$ and $\tan\text{x}=\frac{-1}{2}$
$\Rightarrow-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$ $\Big[\tan\text{x}=1,\text{x}=\frac{-1}{2}$ are $2$ real roots as $\text{cosec x}=0$ has no solution$\Big]$
Thus, these are $2$ solutions.

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MCQ 171 Mark
If $\text{A}_{\text{r}}=\begin{vmatrix}1&\text{r}&2^{\text{r}}\\2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix},$ then the value of $\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}$ is:
  • A
    $n$
  • B
    $2n$
  • ✓
    $-2n^3$
  • D
    $n^2$
Answer
Correct option: C.
$-2n^3$
$\text{A}_\text{r}=\begin{vmatrix}1&\text{r}&2^{\text{r}}\\2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\sum\limits_{\text{r}=1}^\text{n}1&\sum\limits_{\text{r}=1}^\text{n}\text{r}&\sum\limits_{\text{r}=1}^\text{n}2\text{r}\\\sum\limits_{\text{r}=1}^\text{n}2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
As $\sum\limits_{\text{r}=1}^\text{r}1=1+1+1\ ......+1(\text{n times})=\text{n}$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{r}\text{r}=1+2+3+\ .....+\text{n}=\frac{\text{n}(\text{n}+1)}{2}$
Let $\text{S}=\sum\limits_{\text{r}=1}^\text{r}2^\text{r}=2+2^2+2^3=\ .....+2^{\text{n}}$
$\Rightarrow2\text{S}=2^2+3^2=\ ....+2^{\text{n}}+2^{\text{n}+1}$
$\Rightarrow2\text{S}-\text{S}$
$\Rightarrow\text{S}=\sum\limits_{\text{r}=1}^\text{n}2^{\text{r}}=2^{\text{n+1}}-2$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1}-2\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$[$Applying $R_1 \rightarrow R_1 - R_2]$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\text{n}-\text{n}&\frac{\text{n}(\text{n}+1)}{2}-\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1}-2-2^{\text{n}+1}\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$=\begin{vmatrix}0&0&-2\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$=-2\times\begin{vmatrix}2\text{n}&\text{n}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}\end{vmatrix}$
$=-2\big[\text{n}^{3}+\text{n}^2-\text{n}^2\big]$
$=-2\text{n}^3$
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MCQ 181 Mark
The value of $\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$ is:
  • A
    $2$
  • B
    $4$
  • ✓
    $8$
  • D
    $n^2$
Answer
Correct option: C.
$8$
$\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\\text{n}&\text{n}+2&\text{n}+3\\\frac{\text{n}(\text{n}-1)}{2}&\frac{(\text{n}+2)(\text{n}+1)}{2}&\frac{(\text{n}+4)(\text{n}+3)}{2}\end{vmatrix}$
$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&\frac{4\text{n}+2}{2}&\frac{8\text{n}+12}{2}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&(2\text{n}+1)&(4\text{n}+6)\end{vmatrix}$
$=8\text{n}+12-8\text{n}-4$
$=8$
Hence, the correct option is $(c)$
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MCQ 191 Mark
If $\text{x},\text{ y}\in\text{R},$ then the determinant $\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$ lies in the interval:
  • ✓
    $\Big[-\sqrt{2},\sqrt{2}\Big]$
  • B
    $[-1,1]$
  • C
    $\Big[-\sqrt{2},1\Big]$
  • D
    $\Big[-1,-\sqrt{2}\Big]$
Answer
Correct option: A.
$\Big[-\sqrt{2},\sqrt{2}\Big]$
$\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$
$=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\0&0&\sin\text{y}-\cos\text{y}\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - \cos y R_1 + \sin y\ R_2]$
$=(\sin\text{y}-\cos\text{y})(\cos^2\text{x}+\sin^2\text{x})$
$=\sin\text{y}-\cos\text{y}$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)$
$=\sqrt{2}\Big(\cos\frac{\pi}{4}\sin\text{y}-\sin\frac{\pi}{4}\cos\text{y}\Big)$
$=\sqrt{2}\sin\Big(\text{y}-\frac{\pi}{4}\Big)$
Therefore, $-\sqrt{2}\leq\triangle\leq\sqrt{2}$
Hence, the correct option is $(a)$
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MCQ 201 Mark
If $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix},$ then $x =$
  • A
    $3$
  • B
    $\pm3$
  • ✓
    $\pm6$
  • D
    $6$
Answer
Correct option: C.
$\pm6$
$\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}$
$\Rightarrow2\text{x}^2-40=18+14$
$\Rightarrow2\text{x}^2-40=32$
$\Rightarrow2\text{x}^2=72$
$\Rightarrow\text{x}^2=36$
$\Rightarrow\text{x}=\pm6$
Hence, the correct option is $(C)$
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MCQ 211 Mark
If $a, b, c$ are in $A.P.$, then the determinant $\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
  • ✓
    $0$
  • B
    $1$
  • C
    $x$
  • D
    $2x$
Answer
Correct option: A.
$0$
$\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$=\begin{vmatrix}0&0&2(\text{a}+\text{c}-2\text{b})\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$[$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 $A.P.]$
$=0$
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MCQ 221 Mark
Let $\text{A}=\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix},$ where $0\leq\theta\leq2\pi.$ Then:
  • A
    $\text{Det (A)}=0$
  • B
    $\text{Det (A)}\in(2,\infty)$
  • C
    $\text{Det (A)}\in(2,4)$
  • ✓
    $\text{Det (A)}\in[2,4]$
Answer
Correct option: D.
$\text{Det (A)}\in[2,4]$
$\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$
$=\begin{vmatrix}1&\sin\theta&2\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix} [$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, $0\leq\theta\leq2\pi$
$-1\leq\sin\theta\leq1$
$0\leq\sin^2\theta\leq1$
$|\text{A}|=2(\sin^2\theta+1)$
$|\text{A}|=2\times1=2$ $[\theta=0]$
$|\text{A}|=2\times2=4$ $[\theta=2\pi]$
$\text{Det (A)}\in[2,4]$
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MCQ 231 Mark
If the determinant $\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}=0,$ then:
  • A
    $a, b, c$ are in $\text{H.P.}$
  • ✓
    $\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$ or $a, b, c$ are in $\text{G.P.}$
  • C
    $a, b, c$ are in $\text{G.P.}$ only.
  • D
    $a, b, c$ are in $\text{A.P.}$
Answer
Correct option: B.
$\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$ or $a, b, c$ are in $\text{G.P.}$
Let $\triangle=\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}-2\text{b}\alpha-3&2\text{b}\alpha+3\text{c}&0\end{vmatrix} [$Applying $C_1 \rightarrow C_1- C_2]$
$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2(\text{a}-\text{b})\alpha+3(\text{b}-\text{c})&2\text{b}\alpha+3\text{b}&0\end{vmatrix}$
$=2\alpha(2\text{a}\alpha+3\text{b})-3(2\text{b}\alpha+3\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}\\\text{b}-\text{c}&\text{c}\end{vmatrix} [$Expanding along $R_3]$
$=-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)$
But $\triangle=0 [$Given$]$
$\Rightarrow-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)=0$
$\Rightarrow(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})=0$
Or $(\text{ac}-\text{b}^2)=0$
$\Rightarrow\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$
Or $ac = b^2$,
​​​​​​​i.e. $a, b, c$ are in $\text{G.P.}$
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MCQ 241 Mark
If $\text{A}+\text{B}+\text{C}=\pi,$ then the value of $\begin{vmatrix}\sin(\text{A}+\text{B}+\text{C})&\sin(\text{A}+\text{C})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\text{A}+\text{B})&\tan(\text{B}+\text{C})&0\end{vmatrix}$ is equal to:
  • ✓
    $0$
  • B
    $1$
  • C
    $2\sin\text{B}\tan\text{A}\cos\text{C}$
  • D
    None of these.
Answer
Correct option: A.
$0$
$\text{A}+\text{B}+\text{C}=\pi$
$\text{A}+\text{C}=\pi-\text{B},\text{A}+\text{B}=\pi-\text{C}$ and $\text{B}+\text{C}=\pi-\text{A}$
Thus the determinant becomes
$\begin{vmatrix}\sin\pi&\sin(\pi-\text{B})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\pi-\text{C})&\tan(\pi-\text{A})&0\end{vmatrix}$
$=\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$
$[\sin\pi=0,\sin(\pi-\text{B}),\cos(\pi-\text{C})=-\cos\text{C},\tan(\pi-\text{A})=-\tan\text{A}]$
It is a skew symmetric matrix of the odd order $3.$
Thus by property of determinants, we get
$|\triangle|=0$
$\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$
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MCQ 251 Mark
Let $\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix},$ then $\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}$ is equal to:
  • ✓
    $0$
  • B
    $-1$
  • C
    $2$
  • D
    $3$
Answer
Correct option: A.
$0$
$\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix}$ $[$Applying $R_2 \rightarrow R_2 - R_3]$
$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}-\cos\text{x}&0&\text{x}-1\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - R_1]$
$=-\text{x}[\text{x}\sin\text{x}-\sin\text{x}-\text{x}\sin\text{x}+\text{x}\cos\text{x}]$
$=-\text{x}(\text{x}\cos\text{x}-\sin\text{x})$
$\therefore\ \lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{x}(\sin\text{x}-\text{x}\cos\text{x})}{\text{x}^2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}^2}-\lim_\limits{\text{x}\rightarrow0}\cos\text{x}$
$=1-1=0$
Hence, the correct option is $(a)$
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MCQ 261 Mark
Which of the following is not correct?
  • A
    $|\text{A}|=|\text{A}^{\text{T}}|,$ where $\text{A}=[\text{a}_{\text{ij}}]_{3\times3}$
  • B
    $|\text{kA}|=|\text{k}^3|,$ where $\text{A}=[\text{a}_{\text{ij}}]_{3\times3}$
  • C
    If a is a skew$-$symmetric of odd order, then $|A| = 0$
  • ✓
    $\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$
Answer
Correct option: D.
$\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$
$\begin{vmatrix}\text{a}+\text{b}&\text{c}+\text{d}\\\text{e}+\text{f}&\text{g}+\text{h} \end{vmatrix}=\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{e}+\text{f}&\text{h} \end{vmatrix}+\begin{vmatrix}\text{a}+\text{b}&\text{d}\\\text{e}+\text{f}&\text{h}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$
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MCQ 271 Mark
If $x, y, z$ are 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 $x^{-1} + y^{-1} + z^{-1}$ is:
  • A
    $xyz$
  • B
    $x^{-1} + y^{-1} + z^{-1}$
  • C
    $-x - y - z$
  • ✓
    $-1$
Answer
Correct option: D.
$-1$
$\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}&0&-\text{z}\\0&\text{y}&-\text{z}\\1&1&1+\text{z}\end{vmatrix}=0$ $[$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 $(d)$
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MCQ 281 Mark
The maximum value of $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1+\cos\theta&1&1\end{vmatrix}$ is $(\theta$ is real$):$
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{\sqrt{3}}{2}$
  • C
    $\sqrt{2}$
  • D
    $-\frac{\sqrt{3}}{2}$
Answer
Correct option: A.
$\frac{1}{2}$
$\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1+\cos\theta&1&1\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\\cos\theta&0&0\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=-\sin\theta\cos\theta$
$=-\frac{\sin2\theta}{2}$
Now, maximum and minimum value of $\sin\theta$ is $1$ and $-1.$
So, the maximum value of $-\sin\theta$ is $1.$
So, the maximum value of $-\sin2\theta$ is $1.$
Therefore, the maximum value of $-\frac{\sin2\theta}{2}$ is $\frac{1}{2}$
Hence, the correct option is $(a)$
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MCQ 291 Mark
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
    $9x^2(x + y)$
  • ✓
    $9y^2(x + y)$
  • C
    $3y^2(x + y)$
  • D
    $7x^2(x + y)$
Answer
Correct option: B.
$9y^2(x + y)$
$\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}-2\text{y}&\text{y}&\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-\text{y}&2\text{y}&-\text{y}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_2$ and $R_3\rightarrow R_3 - R_2]$
$=\text{y}^2\begin{vmatrix}-2&1&1\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-1&2&-1\end{vmatrix} [$Taking $(y)$ common from $R_1$ and from $R_3]$
$=\text{y}^2\begin{vmatrix}-2&-3&3\\\text{x}+2\text{y}&3\text{x}+4\text{y}&-\text{y}\\-1&0&0\end{vmatrix}$ $[$Applying $C_2 \rightarrow C_2 + 2C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\text{y}^2\big[-1(3\text{y}-9\text{x}-12\text{y})\big]$
$=\text{y}^2[9\text{y}+9\text{x}]$
$=9\text{y}^2(\text{y}+\text{x})$
Hence, the correct option is $(b)$
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MCQ 301 Mark
The value of the determinant $\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{nx}&\cos(\text{n}+1)\text{x}&\cos(\text{n}+2)\text{x}\\\sin\text{nx}&\sin(\text{n}+1)\text{x}&\sin(\text{n}+2)\text{x}\end{vmatrix}$ is independent of:
  • ✓
    $n$
  • B
    $a$
  • C
    $x$
  • D
    None of these.
Answer
Correct option: A.
$n$
Let $A = nx, B = (n - 1)x, C = (n + 2)x$
$\Rightarrow C - B = x, B - A = x, C - A = 2x$
Thus, the given determinant is
$\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{A}&\cos\text{B}&\cos\text{C}\\\sin\text{A}&\sin\text{B}&\sin\text{C}\end{vmatrix}$
$=\text{a}^2(\cos\text{B}\sin\text{C}-\cos\text{C}\sin\text{B})-\text{a}\times(\cos\text{A}\sin\text{C}-\cos\text{C}\sin\text{A})+1\times(\cos\text{A}\sin\text{B}-\sin\text{A}\cos\text{B})$
$=\text{a}^2\sin(\text{C}-\text{B})-\text{a}\sin(\text{C}-\text{A})+\sin(\text{B}-\text{A})$
$=\text{a}^2\sin{\text{x}}-\text{a}\sin2\text{x}+\sin\text{x} [$Independent of $n]$
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MCQ 311 Mark
Let $\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}.$ Then, the value of $5a + 4b + 3c + 2d + e$ is equal to:
  • A
    $0$
  • B
    $-16$
  • C
    $16$
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
$\triangle=\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}$
$=\text{x}\begin{vmatrix}\text{x}&6\\\text{x}&6\end{vmatrix}-\text{x}^2\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}+\text{x}\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}$
$= 0 - x^2(12 - x^2) + x(12 - x^2)$
$= x^4 - 12x^2 + 12x - x^3$
$= ax^4 + bx^3 + cx^2 + dx + e$
$\Rightarrow x^4 - 12x^2 + 12x - x^3 = ax^4 + bx^3 + cx^2 + dx + e$
$\Rightarrow a = 1, b = -1, c = -12, d = 12, e = 0$
Thus,
$5a + 4b + 3c + 2d + e $
$= 5 - 4 - 36 + 24 + 0$
$ = -11$
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MCQ 321 Mark
If $\text{A}=\begin{vmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}\end{vmatrix}$ and $C_{ij}$ is cofactor of $a_{ij}$ in a, then value of $|A|$ is given by:
  • A
    $a_{11}C_{31} + a_{12}C_{32} + a_{13}C_{33}$
  • B
    $a_{11}C_{11} + a_{12}C_{21} + a_{13}C_{31}$
  • C
    $a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13}$
  • ✓
    $a_{11}C_{11} + a_{21}C_{21} + a_{13}C_{31}$
Answer
Correct option: D.
$a_{11}C_{11} + a_{21}C_{21} + a_{13}C_{31}$
Properties of determinants state that if a is a square matrix of the order $n,$ then Det $(A)$ is the sum of products of elements of a row $($or a column$)$ with the corresponding cofactor of that element.
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MCQ 331 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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