1 Marks Question - MATHS STD 12 Science Questions - Vidyadip

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19 questions · timed · auto-graded

Question 11 Mark

Let $A = \left[\begin{array}{ccc} {1} & {\sin \theta} & {1} \\ {-\sin \theta} & {1} & {\sin \theta} \\ {-1} & {-\sin \theta} & {1} \end{array}\right]$ where $0 \leq \theta \leq 2 \pi$. Then

Answer

$A=\left[\begin{array}{ccc} {1} & {\sin \theta} & {1} \\ {-\sin \theta} & {1} & {\sin \theta} \\ {-1} & {-\sin \theta} & {1} \end{array}\right]$
$|A| = 1 (1 \times 1 – \sin \theta\times (-\sin \theta)) – \sin \theta (-\sin \theta+\sin\theta) + 1 [(- \sin \theta)~\times (-\sin \theta )-(-1)\times1]$
$|A| = 1+\sin^2\theta+\sin^2\theta +1$
$|A| = 2 + 2~\sin^2\theta$
$|A| = 2(1 + \sin2\theta)$
Now, 0 $\leq \theta \leq$ 2$\pi$
$\Rightarrow \sin 0 \leq \sin \theta \leq \sin 2 \pi$
$\Rightarrow 0 \leq \sin^2 \theta \leq 1$
$\Rightarrow 1 + 0 \leq 1 + \sin^2 \theta \leq 1 + 1$
$\Rightarrow 2 \leq 2(1 + \sin^2 \theta ) \leq 4$
$\therefore$ Det $(A) \in [2, 4]$

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Question 21 Mark

If x, y, z are non-zero real numbers, then the inverse of matrix $A=\left[\begin{array}{lll} {x} & {0} & {0} \\ {0} & {y} & {0} \\ {0} & {0} & {z} \end{array}\right]$ is

Answer

Here, $A=\left[\begin{array}{lll} {x} & {0} & {0} \\ {0} & {y} & {0} \\ {0} & {0} & {z} \end{array}\right]$
Clearly, we can see that
$adjA$$=\left[\begin{array}{lll} {yz} & {0} & {0} \\ {0} & {xz} & {0} \\ {0} & {0} & {xy} \end{array}\right]~~and ~~~|A|=xyz$
$\therefore ~A^{-1}=\frac {adjA}{|A|}=\frac1{xyz}\left[\begin{array}{lll} {{yz}} & {0} & {0} \\ {0} & {{xz}} & {0} \\ {0} & {0} & {xy} \end{array}\right]$
$=\left[\begin{array}{lll} {x^{-1}} & {0} & {0} \\ {0} & {y^{-1}} & {0} \\ {0} & {0} & {z^{-1}} \end{array}\right]$

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Question 31 Mark

If $A$ is an invertible matrix of order $2,$ then det $(A^{–1})$ is equal to

Answer

We know that, $A^{-1} = \frac{1}{|\mathrm{A}|} \ \operatorname{Adj}(A)$
So, $\left|A^{-1}\right|$ = $\left|\frac{1}{|A|} \operatorname{Adj}(A)\right|$
$= \frac{1}{|A|^{n}}|\operatorname{Adj}(A)|$
$= \frac{1}{|A|^{n}}|A|^{n-1}$
$=\frac{1}{|A|^{1}}$

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Question 41 Mark

Let A be a non-singular square matrix of order 3 $\times$ 3. Then |adj A| is equal to

Answer

For a square matrix of order n$\times$n,
We know that $A.adjA=|A|I$
Here, n=3
$\therefore$ $|A.adjA|=|A|^n$
$|adjA|=|A|^{n-1}$
So, $|AdjA| = |A|^{3-1} = |A|^2$

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Question 51 Mark

Find adjoint of the matrix $\left|\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right|$

Answer

Here $A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}} \\ {{a_{21}}}&{{a_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right]$$\Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right|$
$\therefore A_{11} =$ Cofactor of ${a_{11}} = {\left( { - 1} \right)^2}\left( 4 \right) = 4$
$A_{12} =$ Cofactor of ${a_{12}} = {\left( { - 1} \right)^3}\left( 3 \right) = - 3$
$A_{21} =$ Cofactor of ${a_{21}} = {\left( { - 1} \right)^3}\left( 2 \right) = - 2$
$A_{22} =$ Cofactor of ${a_{22}} = {\left( { - 1} \right)^4}\left( 1 \right) = 1$
$\therefore adj_A = \left| {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}} \\ {{A_{21}}}&{{A_{22}}} \end{array}} \right|$
$= \left| {\begin{array}{*{20}{c}} 4&{ - 3} \\ { - 2}&1 \end{array}} \right|$

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Question 61 Mark

If $\Delta=\left|\begin{array}{lll} {a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}} \end{array}\right|$ and $A_{ij}$ is Cofactors of $a_{ij},$ then value of $\Delta$ is given by

Answer

$\Delta=\left|\begin{array}{lll} {a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}} \end{array}\right|$
Expanding along Column $1$
$\Delta=(-1)^{1+1} \times a_{11} \times\left|\begin{array}{ll} {a_{22}} & {a_{23}} \\ {a_{32}} & {a_{33}} \end{array}\right| + (-1)^{2+1} \times a_{21} \times\left|\begin{array}{ll} {a_{12}} & {a_{13}} \\ {a_{32}} & {a_{33}} \end{array}\right| + (-1)^{3+1} \times a_{31} \times\left|\begin{array}{ll} {a_{12}} & {a_{13}} \\ {a_{22}} & {a_{23}} \end{array}\right|$
$\Delta = a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$

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Question 71 Mark

If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is

Answer

$\frac{1}{2}\left|\begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right|=35$
$\left|\begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right|=70$
$|2(4-4)+6(5-k)+1(20-4 k)|$ = 70
$|50-10 k|=70$
50 - 10k = $\pm$70
k = -2, 12

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Question 81 Mark

If $\left|\begin{array}{cc} {x} & {2} \\ {18} & {x} \end{array}\right|=\left|\begin{array}{cc} {6} & {2} \\ {18} & {6} \end{array}\right|$ , then $x$ is equal to

Answer

We have $\left|\begin{array}{ll} {x} & {2} \\ {18} & {x} \end{array}\right|=\left|\begin{array}{ll} {6} & {2} \\ {18} & {6} \end{array}\right|$
We know that determinant of $A$ is calculated as $|A|=\left|\begin{array}{ll} {a} & {b} \\ {c} & {d} \end{array}\right| = ad - bc$
$\Rightarrow x(x) – 2(18) = 6(6) – 2(18)$
$\Rightarrow x^2 - 36 = 36 – 36$
$\Rightarrow x^2 =36 – 36 + 36$
$\Rightarrow x^2 = 36$
$\Rightarrow x = \pm6$

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Question 91 Mark

Find values of x, if $\left|\begin{array}{ll} {2} & {3} \\ {4} & {5} \end{array}\right|=\left|\begin{array}{cc} {x} & {3} \\ {2 x} & {5} \end{array}\right|$

Answer

We know that determinant of A is calculated as $|A|=\left|\begin{array}{ll} {a} & {b} \\ {c} & {d} \end{array}\right|$ = ad - bc
We have $\left|\begin{array}{cc} {2} & {3} \\ {4} & {5} \end{array}\right|=\left|\begin{array}{cc} {x} & {3} \\ {2 x} & {5} \end{array}\right|$
$\Rightarrow$ 2(5) – 3(4) = x(5) – 3(2x)
$\Rightarrow$ 10 – 12 = 5x – 6x
$\Rightarrow$ -2 = -x
$\Rightarrow$ x = 2
The value of x is 2.

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Question 101 Mark

Find value of $x$, if $\left|\begin{array}{cc}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$

Answer

$(2 – 20) = (2x^2– 24)$
$-18 = 2x^2– 24$
$-2x^2 = -24 + 18$
$-2x^2 = -6$
$2x^2 = 6$
$x^2 = 3$
$x = \pm \sqrt 3 $

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Question 111 Mark

If $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right]$ find $| A |$.

Answer

Given: $A = \left[ {\begin{array}{*{20}{c}} 1&1&{ - 2} \\ 2&1&{ - 3} \\ 5&4&9 \end{array}} \right]$
$ \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&1&{ - 2} \\ 2&1&{ - 3} \\ 5&4&9 \end{array}} \right|$
Expanding along first row, $1\left| {\begin{array}{*{20}{c}} 1&{ - 3} \\ 4&{ - 9} \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} 2&{ - 3} \\ 5&{ - 9} \end{array}} \right| + \left( { - 2} \right)\left| {\begin{array}{*{20}{c}} 2&1 \\ 5&4 \end{array}} \right|$
= {-9(-12)} - {-18 - (-15)} - 2(8 - 5)
= -9 + 12 - (-18 + 15) -2(3)
= 3(-3) - 6
= 3 + 3 - 6 = 0

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Question 121 Mark

Evaluate the determinant $\left|\begin{array}{rrr} {2} & {-1} & {-2} \\ {0} & {2} & {-1} \\ {3} & {-5} & {0} \end{array}\right|$

Answer

ATQ,
We know that a determinant of a 3 $\times$ 3 matrix is calculated as
$|A|=\left|\begin{array}{ccc} {i} & {j} & {k} \\ {a} & {b} & {c} \\ {d} & {e} & {f} \end{array}\right|=i\left|\begin{array}{cc} {b} & {c} \\ {e} & {f} \end{array}\right|-j\left|\begin{array}{cc} {a} & {c} \\ {d} & {f} \end{array}\right|+k\left|\begin{array}{cc} {a} & {b} \\ {d} & {e} \end{array}\right|$
= $2\left|\begin{array}{cc} {2} & {-1} \\ {-5} & {0} \end{array}\right|-(-1)\left|\begin{array}{cc} {0} & {-1} \\ {3} & {0} \end{array}\right|+(-2)\left|\begin{array}{cc} {0} & {2} \\ {3} & {-5} \end{array}\right|$
= 2[0 – (-1)(-5)] + 1[0 – (-1)(3)] – 2[0 – 3(2)]
= 2[0 – 5] + 1[0 + 3] – 2[-6]
= 2[-5] + 1[3] -2[-6]
= -10 + 3 + 12
= 5
The determinant of the above matrix is 5.

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Question 131 Mark

Evaluate the determinant $\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$

Answer

Given: $\left| {\begin{array}{*{20}{c}} 0&1&2 \\ { - 1}&0&{ - 3} \\ { - 2}&3&0 \end{array}} \right|$
Expanding along first row, $0\left| {\begin{array}{*{20}{c}} 0&{ - 3} \\ 3&0 \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} { - 1}&{ - 3} \\ { - 2}&0 \end{array}} \right| + 2\left| {\begin{array}{*{20}{c}} { - 1}&0 \\ { - 2}&3 \end{array}} \right|$
= 0(0 + 9) - (0 - 6) + 2( - 3 - 0)
= 0 + 6 - 6 = 0

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Question 141 Mark

Evaluate the determinant $\left|\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right|$

Answer

Given: $\left| {\begin{array}{*{20}{c}} 3&{ - 4}&5 \\ 1&1&{ - 2} \\ 2&3&1 \end{array}} \right|$
Expanding along first row, $3\left| {\begin{array}{*{20}{c}} 1&{ - 2} \\ 3&1 \end{array}} \right| - \left( { - 4} \right)\left| {\begin{array}{*{20}{c}} 1&{ - 2} \\ 2&1 \end{array}} \right| + 5\left| {\begin{array}{*{20}{c}} 1&1 \\ 2&3 \end{array}} \right|$
= 3(1 + 6) + 4{1 - ( - 4)} + 5(3 - 2)
$ = 3 \times 7 + 4 \times 5 + 5 \times 1$
= 21 + 20 + 5 = 46

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Question 151 Mark

Evaluate the determinant $\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$

Answer

Given: $\left| {\begin{array}{*{20}{c}} 3&{ - 1}&{ - 2} \\ 0&0&{ - 1} \\ 3&{ - 5}&0 \end{array}} \right|$
Expanding along first row, $3\left| {\begin{array}{*{20}{c}} 0&{ - 1} \\ { - 5}&0 \end{array}} \right| - \left( { - 1} \right)\left| {\begin{array}{*{20}{c}} 0&{ - 1} \\ 3&0 \end{array}} \right| + \left( { - 2} \right)\left| {\begin{array}{*{20}{c}} 0&0 \\ 3&{ - 5} \end{array}} \right|$
= 3(0 - 5) + 1(0 - (-3)) - 2(0 - 0)
= -15 + 3 - 0 = -12

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Question 161 Mark

Evaluate the determinant $\left|\begin{array}{cc}x^2-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$

Answer

Applying $C_1\rightarrow C_1-C_2,$
we get, .$\left| {\begin{array}{*{20}{c}} {{x^2} - x + 1}&{x - 1} \\ {x + 1}&{x + 1} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{x^2} - 2x + 2}&{x - 1} \\ 0&{x + 1} \end{array}} \right|$
$= \left( {{x^2} - 2x + 2} \right).\left( {x + 1} \right) - \left( {x - 1} \right).0$
$= x^3 - 2x^2 + 2x + x^2 - 2x + 2$
$= x^3 - x^2 + 2$

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Question 171 Mark

Evaluate the determinant $\left|\begin{array}{cc} {\cos \theta} & {-\sin \theta} \\ {\sin \theta} & {\cos \theta} \end{array}\right|$

Answer

We know that determinant of $A$ is calculated as $|A|=\left|\begin{array}{ll} {a} & {b} \\ {c} & {d} \end{array}\right| = ad - bc$
Now, $\left|\begin{array}{cc} {\cos \theta} & {-\sin \theta} \\ {\sin \theta} & {\cos \theta} \end{array}\right|$
$= \cos \theta(\cos \theta) - (-\sin \theta)(\sin \theta)$
$= \cos^2\theta + \sin^2\theta$
$= 1... [\because \cos^2\theta + \sin^2\theta = 1]$
$\therefore$ The determinant of the above matrix is $1.$

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Question 181 Mark

Evaluate $\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|$

Answer

$\left| {\begin{array}{*{20}{c}} 2&4 \\ { - 5}&{ - 1} \end{array}} \right| $ = 2(-1) - 4(-5) = -2 + 20 = 18

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Question 191 Mark

If matrix $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right]$ then find adjoint matrix $(\operatorname{adj} A)$.

Answer

Adjoint of $A:$ $\left[\begin{array}{cc}4 & -3 \\ -1 & 2\end{array}\right]$

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1 Marks Question - MATHS STD 12 Science Questions - Vidyadip