Maths Case study (4 Marks)

1. (b) $\text{A}^\text{n}=\begin{bmatrix}1+2\text{n}&-4\text{n}\\\text{n}&1-2\text{n}\end{bmatrix}$

Solution:

We have, $\text{A}=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$

$\therefore\text{A}^2=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}\begin{bmatrix}3&-4\\1&-1\end{bmatrix}=\begin{bmatrix}5&-8\\2&-3\end{bmatrix},$ which can be obtained from $\text{A}^\text{n}=\begin{bmatrix}1+2\text{n}&-4\text{n}\\\text{n}&1-2\text{n}\end{bmatrix}$ for n = 2.

2. (d) 1

Solution:

We have, $\text{A}=\begin{bmatrix}1&2\\0&1\end{bmatrix}$

$\therefore|\text{A}|=\begin{vmatrix}1&2\\0&1\end{vmatrix}=1-0=1$

Also, |Aⁿ| = |A· A ...... A(n times)| = |A|ⁿ = 1ⁿ = 1

3. (a) Aⁿ = nA - (n - 1)I

Solution:

For n = 1, all options are true.

$\text{A}^2=\text{A}\cdot\text{A}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&0\\2&1\end{bmatrix}$

and $\text{A}^3=\text{A}^2\cdot\text{A}=\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&0\\3&1\end{bmatrix}$

Putting n = 3, in (a), we get A³ = 3A - 2I

$=3\begin{bmatrix}1&0\\1&1\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}$

$=\begin{bmatrix}3&0\\3&3\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}=\begin{bmatrix}1&0\\3&1\end{bmatrix},$ which is true.

All other options are different from A³ = 3A -2I for n = 3.

4. (d) 0

Solution:

We have, $\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}$

$\therefore\text{A}^2=\text{A}\cdot\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}$

$=\begin{bmatrix}\text{a}^2&0&0\\0&\text{a}^2&0\\0&0&\text{a}^2\end{bmatrix}$

Similarly, $\text{A}^\text{n}=\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$

Now, cofactor of $\text{a}_{13}=(-1)^{1+3}\begin{vmatrix}0&\text{a}^\text{n}\\0&0\end{vmatrix}=0$

5. (c) 2ⁿ

Solution:

We have, |A| = 2 and |Aⁿ| = |A·A ...... A(n - times)|

= |A| |A| ...... |A|(n - times) = |A|ⁿ = 2ⁿ

1. (d) 15.5 sq. units

Solution:

Let $\Delta$ be the area of the triangle then,

$\Delta=\frac{1}{2}\begin{vmatrix}\begin{vmatrix}-2&6&1\\3&-6&1\\1&5&1\end{vmatrix}\end{vmatrix}$

$=\frac{1}{2}|-2(-6-5)-6(3-1)+1(15+6)|$

$\Rightarrow\Delta=\frac{1}{2}|43-12|=15.5\text{ sq. units}$

2. (d) $\frac{40}{7}$

Solution:

The given points are collinear.

$\therefore\frac{1}{2}\begin{vmatrix}2&-3&1\\\text{k}&-1&1\\0&4&1\end{vmatrix}=0$

Expanding along R₁ we get

2(-1 - 4) + 3(k) + 1(4k) = 0

$\Rightarrow7\text{k}-10=0\Rightarrow\text{k}=\frac{10}{7}\Rightarrow4\text{k}=\frac{40}{7}$

3. (a) 2

Solution:

Area of $\Delta\text{ABC}=3 \text{ sq.units}$ [Given]

$\Rightarrow\frac{1}{2}\begin{vmatrix}1&3&1\\0&0&1\\\text{k}&0&1\end{vmatrix}=\pm3\Rightarrow\begin{vmatrix}1&3&1\\0&0&1\\\text{k}&0&1\end{vmatrix}=\pm6$

$\Rightarrow1(0-0)-3(0-\text{k})+1(0-0)=\pm6$

$\Rightarrow3\text{k}=\pm6\Rightarrow\text{k}=\pm2$

4. (a) y = 2x

Solution:

Let Q(x, y) be any point on the line joining A(1, 2) and B(3, 6). Then, area of $\Delta\text{ABQ}=0$

$\Rightarrow\frac{1}{2}\begin{vmatrix}1&2&1\\3&6&1\\\text{x}&\text{y}&1\end{vmatrix}=0$

⇒ 1(6 - y) - 2(3 - x) + 1 (3y - 6x) = 0

⇒ 6 - y - 6 + 2x + 3y - 6x = 0

⇒ -4x = -2y ⇒ 2x = y.

5. (c) A, B and C are collinear.

Solution:

Area of $\Delta\text{ABC}$ is given by

$\frac{1}{2}\begin{vmatrix}11&7&1\\5&5&1\\-1&3&1\end{vmatrix}=\frac{1}{2}\big[11(5-3)-7(5+1)+1(15+5)\big]$

$=\frac{1}{2}[22-42+20]=0$

$\therefore$ Points are collinear.

1. (c) $\begin{bmatrix}3\\-3\end{bmatrix}$

Solution:

We have, $\text{A}=\begin{bmatrix}1&0\\2&1\end{bmatrix}$

Let $\text{U}_1=\begin{bmatrix}\text{a}\\\text{b}\end{bmatrix}$ and $\text{AU}_1=\begin{bmatrix}1\\0\end{bmatrix}$

$\Rightarrow\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}\text{a}\\\text{b}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}\Rightarrow\begin{bmatrix}\text{a}\\2\text{a}+\text{b}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$

⇒ a = 1 and 2a + b = 0 ⇒ a = 1 and b = -2.

Let $\text{U}_2=\begin{bmatrix}\text{c}\\\text{d}\end{bmatrix}$ then  $\text{AU}_2=\begin{bmatrix}2\\3\end{bmatrix}$

$\Rightarrow\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}\text{c}\\\text{d}\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}\Rightarrow\begin{bmatrix}\text{c}\\2\text{c}+\text{d}\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}$

⇒ c = 2 and 2c + d = 3

⇒ c = 2 and d = 3 - 4= -1

Thus, $\text{U}_1+\text{U}_2=\begin{bmatrix}1\\-2\end{bmatrix}+\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}3\\-3\end{bmatrix}$

2. (c) 3

Solution:

Clearly, $\text{U}=\begin{bmatrix}1&2\\-2&-1\end{bmatrix}$

$\therefore|\text{U}|=\begin{vmatrix}1&2\\-2&-1\end{vmatrix}=-1+4=3$

3. (d) 5

Solution:

We have, $\text{X}=\begin{bmatrix}3&2\end{bmatrix}\begin{bmatrix}1&2\\-2&-1\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}$

$=\begin{bmatrix}3&2\end{bmatrix}\begin{bmatrix}7\\-8\end{bmatrix}=[21-16]=[5]$

$\therefore|\text{X}|=5$

4. (a) 1

Solution:

a₂₂ in U is -1 and its minor is 1.

5. (d) 3

Solution:

Since, the sum of products of elements of any row (or column) with their corresponding cofactors is equal to the value of determinant.

$\therefore$ a₁₁A₁₁ + a₁₂A₁₂ = |U| = 3

1. (c) $3\sqrt{3}\text{ sq}.\text{units}$

Solution:

Area of triangle is given by $\begin{vmatrix}\frac{1}{2}\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_2&\text{y}_2&1\\\text{x}_3&\text{y}_3&1\end{vmatrix}\end{vmatrix}.$

$\therefore$ Required area $=\begin{vmatrix}\frac{1}{2}\begin{vmatrix}0&0&1\\3&\sqrt{3}&1\\3&-\sqrt{3}&1\end{vmatrix}\end{vmatrix}$

$=\begin{vmatrix}\frac{1}{2}[1(-3\sqrt{3}-3\sqrt{3})]\end{vmatrix}$ [Expanding along R₁]

$=3\sqrt{3}\text{ sq}.\text{units}$

2. (c) $75\sqrt{3}\text{ sq}.\text{units}$

Solution:

Since a face of the Pyramid consist of 25 smaller equilateral triangles.

$\therefore$ Required area $=25\times3\sqrt{3}=75\sqrt{3}\text{ sq}.\text{units}$

3. (b) 3 units

Solution:

Area of equilateral triangle $=\frac{\sqrt{3}}{4}\text{a}^2$

$\therefore3\sqrt{3}=\frac{\sqrt{3}}{4}\text{a}^2\Rightarrow\text{a}^2=12\Rightarrow\text{a}=2\sqrt{3}$

Let h be the length of altitude of a smaller equilateral triangle. Then,

$\frac{1}{2}\times\text{base}\times\text{height}=3\sqrt{3}$

$\Rightarrow\frac{1}{2}\times2\sqrt{3}\times\text{h}=3\sqrt{3}\Rightarrow\text{h}=3\text{ units}$

4. (c) $\text{x}=2+\sqrt3$

Solution:

Let the third vertex be (x, y), then we get $\frac{1}{2}\begin{vmatrix}2&4&1\\2&6&1\\\text{x}&\text{y}&1\end{vmatrix}=\pm3\sqrt3$

$\Rightarrow\frac{1}{2}[2(6-\text{y})-4(2-\text{x})+1(2\text{y}-6\text{x})]=\pm3\sqrt3$

$\Rightarrow12-2\text{y}-8+4\text{x}+2\text{y}-6\text{x}=\pm6\sqrt3$

$\Rightarrow4-2\text{x}=\pm6\sqrt3$

$\Rightarrow2-\text{x}=\pm3\sqrt3\Rightarrow\text{x}=2\pm3\sqrt3$

5. (c) Collinear.

Solution:

Area of $\triangle\text{ABC}=\frac{1}{2}\begin{vmatrix}\text{a}&0&1\\0&\text{b}&1\\1&1&1\end{vmatrix}$

$=\frac{1}{2}[\text{a}(\text{b}-1)-0+1(0-\text{b})]$

$=\frac{1}{2}(\text{ab}-\text{a}-\text{b})=0$

$\Big[\because\frac{1}{\text{a}}+\frac{1}{\text{b}}=1\Rightarrow\text{b}+\text{a}=\text{ab}\Big]$

$\therefore$ Points A, Band Care collinear.

1. (a) 2

Solution:

Let $\Delta=\begin{vmatrix}1&-2\\4&3\end{vmatrix}$

Cofactor of 1 = 3, cofactor of -2 = -4

Cofactor of 4 = 2, cofactor of 3 = 1

$\therefore$ Required sum = 3 - 4 + 2 + 1 = 2

2. (b) -3

Solution:

Let $\Delta=\begin{vmatrix}5&6&-3\\-4&3&2\\-4&-7&3\end{vmatrix}$

Minor of $\text{a}_{21}=\begin{vmatrix}6&-3\\-7&3\end{vmatrix}=18-21=-3$

3. (c) 110

Solution:

Let $\Delta=\begin{vmatrix}2&-3&5\\6&0&4\\1&5&-7\end{vmatrix}$

Clearly, a₃₂ = 5

and A₃₂ = cofactor of a₃₂ in $\Delta=(-1)^{3+2}\begin{vmatrix}2&5\\6&4\end{vmatrix}$

= (-1) (8 - 30) = 22

$\therefore$ a₃₂.A₃₂ = 5 × 22 = 110

4. (d) 7

Solution:

Here, $\Delta=\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}$

$\therefore$ Minor of $\text{a}_{23}=\begin{vmatrix}5&3\\1&2\end{vmatrix}=10-3=7$

5. (b) 28

Solution:

Here, $\Delta=\begin{vmatrix}2&-3&5\\6&0&4\\1&5&-7\end{vmatrix}$

$\text{A}_{11}=(-1)^{1+1}\begin{vmatrix}0&4\\5&-7\end{vmatrix}=1(0-20)=-20,$

$\text{A}_{12}=(-1)^{1+2}\begin{vmatrix}6&4\\1&-7\end{vmatrix}=-1(-42-4)=46,$

$\text{A}_{13}=(-1)^{1+3}\begin{vmatrix}6&0\\1&5\end{vmatrix}=1(30-0)=30$

$\therefore\Delta=\text{a}_{11}\text{A}_{11}+\text{a}_{12}\text{a}_{12}+\text{a}_{13}\text{A}_{13}$

$= 2(-20) -3(46) + 5(30) = -28$

$\Rightarrow|\Delta|=28$

1. (d) All of these.

Solution:

According to given condition, we have the following system of linear equations.

x + y + z = 45

x + 8 = z or x + O.y - z = -8

and x + z = 2y or x - 2y + z = 0

2. (c) $\begin{pmatrix}\frac{1}{3}&\frac{1}{2}&\frac{1}{6}\\\frac{1}{3}&0&\frac{-1}{3}\\\frac{1}{3}&\frac{-1}{2}&\frac{1}{6}\end{pmatrix}$

Solution:

Let A $=\begin{pmatrix}1&1&1\\1&0&-2\\1&-1&1\end{pmatrix}$ then we have,

$\text{A}^{-1}=\frac{1}{6}\begin{pmatrix}2&2&2\\3&0&-3\\1&-2&1\end{pmatrix}$

Now, (A')^-1 = (A^-1)'

$=\frac{1}{6}\begin{pmatrix}2&3&1\\2&0&-2\\2&-3&1\end{pmatrix}=\begin{pmatrix}\frac{1}{3}&\frac{1}{2}&\frac{1}{6}\\\frac{1}{3}&0&\frac{-1}{3}\\\frac{1}{3}&\frac{-1}{2}&\frac{1}{6}\end{pmatrix}$

3. (b) 11 : 15 : 19

Solution:

The above system of equations can be written in matrix form as

A'X =B, where,

$\text{A}'=\begin{pmatrix}1&1&1\\1&0&-1\\1&-2&1\end{pmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ and $\text{B}=\begin{bmatrix}45\\-8\\0\end{bmatrix}$

$\Rightarrow\text{X}=(\text{A}')^{-1}\text{B}=\frac{1}{6}\begin{bmatrix}2&3&1\\2&0&-2\\2&-3&1\end{bmatrix}\begin{bmatrix}45\\-8\\0\end{bmatrix}$

$=\frac{1}{6}\begin{bmatrix}90-24\\90\\90+24\end{bmatrix}=\frac{1}{6}\begin{bmatrix}66\\90\\114\end{bmatrix}=\begin{bmatrix}11\\15\\19\end{bmatrix}$

Thus, x : y : z = 11 : 15 : 19

4. (d) |AB| = |A| + |B|

Solution:

Clearly, |AB| = |A|.|B|

5. (b) Minor of an element can never be equal to cofactor of the same element.

1. (c) ₹ 1

Solution:

Cost of one pen is ₹ 1.​​​​​​

2. (a) ₹ 3

Solution:

Cost of one pen and one bag= ₹(1 + 2) = ₹ 3

3. (b) ₹ 6

Solution:

Cost of one pen and one instrument box = ₹(1 + 5) = ₹ 6

4. (c) Determinant is a number associated to a square matrix.

Solution:

According to the definition of determinant, determinant is a number associated to a square matrix.

5. (b) A is non-singular.

Solution:

Given matrix equation is AB = AC

Pre-multiplying by A^-1 on both sides, we get

A^-1 AB = A^-1 AC ⇒ (A^-1A) B = (A^-1A)C

⇒ IB = IC ($\because$ AA^-1 = A^-1A = I)

⇒ B = C

Since A^-1 exists only if A is non-singular.

$\therefore$ For B = C, A should be non-singular.

1. (b) 21

2. (c) $\frac{4}{3}$

3. (b) $\frac{1}{3}\text{t}^2+20\text{t}+1$

Solution:

$\text{v}(\text{t})=\frac{1}{3}\text{t}^2+20\text{t}+1$

4. (d) 376 miles/ sec

Solution:

Clearly, required speed = v(15)

$=\frac{1}{3}\times225+20\times15+1$

= 75 + 300 + 1 = 376 miles per second.

5. (d) 27 seconds

Solution:

Consider, v(t) = 784

$\Rightarrow\frac{1}{3}\text{t}^2+20\text{t}+1=784$

⇒ t² + 60t = 2349

⇒ t² + (87 - 27)t - 2349 = 0

⇒ t(t + 87) -27(t + 87) = 0

⇒ (t - 27)(t + 87) = 0

⇒ t = 27 seconds [$\because$ Time can't be negative)]

1. (b) ₹ 300

2. (d) ₹ 500

3. (b) ₹ 400

4. (c) 0

Solution:

If a matrix P is both symmetric and skew-symmetric then it will be a zero matrix. So, |P| = 0.

5. (a) |Q|

Solution:

We have, Q² = QQ = Q(PQ)

= (QP)Q = PQ = Q

$\therefore$ |Q²| = |Q|

1. (a) ₹ 1

2. (d) ₹ 5

3. (b) ₹ 2

4. (b) Vijay

Solution:

Vijay investing most of the money on hand -rnade bags.

5. (a) Salim

Solution:

Salim investing less amount of money on polythene bags.