1 Marks - Maths STD 12 Science Questions - Vidyadip

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18 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 $\leq$ sin²$\theta$ $\leq$ 1
$\Rightarrow$ 1 + 0 $\leq$ 1 + sin²$\theta$ $\leq$ 1 + 1
$\Rightarrow$ 2 $\leq$ 2(1 + sin²$\theta$) $\leq$ 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 = $\frac{1}{|\mathrm{A}|}$ Adj (A)
So, $\left|A^{-1}\right|$ = $\left|\frac{1}{|A|} \operatorname{Adj}(A)\right|$
  = $\frac{1}{|A|^{n}}|\operatorname{Adj}(A)|$
  = $\frac{1}{|A|^{n}}|A|^{n-1}=\frac{1}{|A|^{1}}$
  $=\frac{1}{|A|^{1}}$
{since adj(A) is of order n and |Adj(A)| = |A|^n-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 $A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}} \\ {{a_{21}}}&{{a_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right]$

$\Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right|$

$\therefore$ A₁₁ = Cofactor of ${a_{11}} = {\left( { - 1} \right)^2}\left( 4 \right) = 4$

A₁₂ = Cofactor of ${a_{12}} = {\left( { - 1} \right)^3}\left( 3 \right) = - 3$

A₂₁ = Cofactor of ${a_{21}} = {\left( { - 1} \right)^3}\left( 2 \right) = - 2$

A₂₂ = Cofactor of ${a_{22}} = {\left( { - 1} \right)^4}\left( 1 \right) = 1$

$\therefore adjA = \left| {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}} \\ {{A_{21}}}&{{A_{22}}} \end{array}} \right|$

$= \left| {\begin{array}{*{20}{c}} 4&{ - 3} \\ { - 2}&1 \end{array}} \right|$

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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
$\Delta=(-1)^{1+1} \times a_{11} \times\left|\begin{array}{ll} {a_{22}} & {a_{23}} \\ {a_{32}} & {a_{33}} \end{array}\right|$ + $(-1)^{2+1} \times a_{21} \times\left|\begin{array}{ll} {a_{12}} & {a_{13}} \\ {a_{32}} & {a_{33}} \end{array}\right|$ + $(-1)^{3+1} \times a_{31} \times\left|\begin{array}{ll} {a_{12}} & {a_{13}} \\ {a_{22}} & {a_{23}} \end{array}\right|$
$\Delta$ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

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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)
$\Rightarrow$ x² - 36 = 36 – 36
$\Rightarrow$ x² =36 – 36 + 36
$\Rightarrow$ x² = 36
$\Rightarrow$ x = $\pm$6.

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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) = (2x²– 24)
-18 = 2x$^2$– 24
-2x² = -24 + 18
-2x² = -6
2x² = 6
x² = 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|$$$

$= \left( {{x^2} - 2x + 2} \right).\left( {x + 1} \right) - \left( {x - 1} \right).0$

= x³ - 2x² + 2x + x² - 2x + 2

= x³ - x² + 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 $\theta$)
= cos²$\theta$ + sin²$\theta$
= 1 ... [$\because$ cos²$\theta$ + sin²$\theta$ = 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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1 Marks - Maths STD 12 Science Questions - Vidyadip