### Mathematics Assignments | PDF | 8 to 12

PDF Files of Mathematics Assignments From VIII Standard to XII Standard PDF of mathematics Assignments for the students from VIII standard to XII standard.These assignments are strictly according to the CBSE and DAV Board Final question Papers

## Math Assignment  Class XII  Ch - 04DETERMINANTS

Extra questions of chapter 04 Determinants class XII  with answers and  hints to the difficult questions, strictly according to the CBSE syllabus. Important and useful math. assignment for the students of class XII

## Strictly according to the CBSE Board

Question 1

Find the value of k for which matrix A is a singular matrix

$\inline \bg{black}{A=\left [ \begin{matrix}k & 8 \\4 & 2k \\\end{matrix} \right ]}$

Ans: åœŸ 4
Question 2

If A is a square matrix of order 3 and |A| = - 4, then find the value of |adj A|

Ans: 16

Question 3

$\inline \bg{black}If\: A=\left [ \begin{matrix}\alpha & 2 \\2 & \alpha \\\end{matrix} \right ]$ ,

and |A3| = 27, then find the value of Î±

Ans: åœŸ√7
Question 4

$\large \bg{black}\mathbf{If\: \: \left|\begin{matrix}5 & 3 & -1 \\-7 & x & -3 \\9 & 6 & -2 \\\end{matrix} \right|=0}$ , then find the value of x

Ans : x = 9
Question 5

$\large \bg{black}\mathbf{If\: A=\left [ \begin{matrix}2 & -3 \\5 & -4 \\\end{matrix} \right ],}$ $\large \bg{black}\mathbf{\: B=\left [ \begin{matrix}1 & -2 \\3 & 5 \\\end{matrix} \right ]}$
then verify that |AB| = |A||B|

Solution Hint: |AB| = 77 and |A||B| = 7 x 11 = 77

Question 6
Evaluate:  $\large \bg{black}\mathbf{ \left | \begin{matrix}a+ib & c+id \\-c+id & a-ib \\\end{matrix} \right |}$
Ans: a2 + b2 + c2 + d2
Question 7

$\inline \bg{black}Evaluate\: \left | \begin{matrix}cos15^{o} & sin15^{o} \\sin75^{o} & cos75^{o} \\\end{matrix} \right |$

Ans: 0
Question 8

For what value of x, the matrix A is singular

$\inline \bg{black}A= \left [ \begin{matrix}5-x & x+1 \\2 & 4 \\\end{matrix} \right ]$
Solution Hint:
A matrix is said to be singular if |A| = 0
Question 9
Find the area of triangle with vertices A(5, 4), B(-2, 4), C(2, -6)
Ans: 35 sq unit

Question 10

Using determinants show that the points (2, 3), (-1, -2) and (5, 8) are collinear

Solution Hint:
Find area of triangle by taking above given points as vertices.
If area of triangle = 0 then points are collinear.

Question 11

Using determinants find the value of k so that the points (k, 2 - 2k), (- k + 1, 2k), and (- 4 - k, 6 - 2k) may be collinear

Ans: k = -1, 1/2

Question 12
Using determinants, find the equation of line joining the points (3,1), and (9,3)

Ans: x - 3y = 0

Question 13

Find the value of k, if area of triangle is 4 square units whose vertices are (-2,0), (0,4), and (0, k)

Ans: K = 0, 8

Question 14

Find the value of |AB| if matrices A and B are given below

$\bg{black}A=\left [ \begin{matrix}1 & 2 \\3 & -1 \\\end{matrix} \right ]$ and $\bg{black}B=\left [ \begin{matrix}1 & 0 \\-1 & 0 \\\end{matrix} \right ]$

Ans: 0

Question 15

Find the value of x if matrix A is a singular matrix
$A=\left [ \begin{matrix}5x & 2 \\-10 & 1 \\\end{matrix} \right ]$

Ans: x = - 4

Question 16

Find the product : $\left [ \begin{matrix}1 & -1 & 0 \\2 & 3 & 4 \\0 & 1 & 2 \\\end{matrix} \right ]\left [ \begin{matrix}2 & 2 & -4 \\-4 & 2 & -4 \\2 & -1 & 5 \\\end{matrix} \right ]$

Hence solve the following system of equations
x - y = 3, 2x + 3y + 4z = 17, y + 2z = 7

Ans: x = 2, y = -1, z = 4

Solution Hint
Let given matrices are A and C

$AC=\left [ \begin{matrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{matrix} \right ]=6I$
Now find the product AC we get

$A\times \frac{1}{6}C=I\: \: \Rightarrow \: \: A^{-1}=\frac{1}{6}C$
Now given system of equations can be written as

$A=\left [ \begin{matrix}1 & -1 & 0 \\2 & 3 & 4 \\0 & 1 & 2 \\\end{matrix} \right ], X=\left [ \begin{matrix}x \\y \\z\end{matrix} \right ], B=\left [ \begin{matrix} 3\\17 \\7\end{matrix} \right ]$

AX = B ⇒ X =
A-1B
$\Rightarrow X=\frac{1}{6}CB$
⇒ x = 2, y = -1, z = 4 is the required solution

Question 17

$If \: \: A =\left [ \begin{matrix}3 & -4 & 2 \\2 & 3 & 5 \\1 & 0 & 1 \\\end{matrix} \right ],$
Find A-1  and hence solve following system of equations

3x - 4y + 2z = -1, 2x + 3y + 5z = 7, x + z = 2

Answer: x = 3, y = 2, z = -1

Solution Hint:

Find |A| we find |A| = -9 ⇒ A is invertible

Find cofactors of A and then find Adj. A we get
$Adj. A =\left [ \begin{matrix}3 & 4 & -26 \\3 & 1 & -11 \\-3 & -4 & 17 \\\end{matrix} \right ],$
Find A-1  we get

$A^{-1} =\frac{1}{9}\left [ \begin{matrix}-3 & -4 & 26 \\-3 & -1 & 11 \\3 & 4 & -17 \\\end{matrix} \right ],$

Given system of equations can be written as AX = B
⇒ X= A-1B

$\left [ \begin{matrix}x \\y \\z\end{matrix} \right ] =\frac{1}{9}\left [ \begin{matrix}-3 & -4 & 26 \\-3 & -1 & 11 \\3 & 4 & -17 \\\end{matrix} \right ]\left [ \begin{matrix} -1\\7 \\2\end{matrix} \right ]=\left [ \begin{matrix} 3\\2 \\-1\end{matrix} \right ]$

⇒ x = 3, y = 2, z = -1

Question 18

$\inline \bg{black} \mathbf{\; If\; \; A=\left [ \begin{matrix}2 & -3 & 5 \\3 & 2 & -4 \\1 & 1 & -2 \\\end{matrix} \right ]}$

Find A-1. Use A-1 to solve the system of equations.

2x – 3y + 5z = 11, 3x + 2y – 4z = - 5, x + y - 2z = - 3

Ans: x = 1, y = 2, z = 3

Solution Hint

Find the |A| we get |A| = -1

$\inline \bg{black}Adjoint \: A=\left[ \begin{matrix}0 & -1 & 2 \\2 & -9 & 23 \\1 & -5 & 13 \\\end{matrix}\right]$

Solve: X = A-1B we get x = 1, y = 2, z = 3

Question 19

Show that the matrix A satisfies the equation A - 4A - 5I = O and hence find A-1
$\bg{black}A=\left [ \begin{matrix}1 & 2 & 2 \\2 & 1 & 2 \\2 & 2 & 1 \\\end{matrix} \right ]$

Ans:
$\bg{black}A^{-1}=\frac{1}{5}\left [ \begin{matrix}-3 & 2 & 2 \\2 & -3 & 2 \\2 & 2 & -3 \\\end{matrix} \right ]$

Question 20
Find the matrix X for which

$\bg{black}\left [ \begin{matrix}1 & -4 \\3 & -2 \\\end{matrix} \right ]X=\left [ \begin{matrix}-16 & -6 \\ 7&2 \\\end{matrix} \right ]$

$\bg{black}X=\left [ \begin{matrix}6 & 2 \\11/2 & 2 \\\end{matrix} \right ]$

Question 21

$\inline If\: \: A^{-1}=\left [ \begin{matrix}3 & -1 & 1 \\-15 & 6 & -5 \\5 & -2 & 2 \\\end{matrix} \right ]$  $\inline and\: \: B=\left [ \begin{matrix}1 & 2 & -2 \\-1 & 3 & 0 \\0 & -2 & 2 \\\end{matrix} \right ]$

Find  (AB)-1

Ans:  Find B-1

$B^{-1}=\frac{1}{6}\left[\begin{matrix}6&0&6\\2&2&2\\2&2&5\\\end{matrix}\right]$

$(AB)^{-1}=B^{-1}A^{-1}$

$(AB)^{-1}=\frac{1}{6}\left [ \begin{matrix}48 & -18 & 18 \\-14 & 6 & -4 \\1 & 0 & 2 \\\end{matrix} \right ]$
Question 22
Solve the following system of equations
x + y + z = 3, 2x - y + z = -1, 2x + y - 3z = -9
Ans:  x = -8/7,  y=10/7,  z = 19/7
Question 23
Solve the following system of equations

$\inline \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4,$
$\inline \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1,$
$\inline \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$  where x, y, z ≠ 0
Ans: x = 2, y = 3, z = 5

Question 24

If A =$\begin{bmatrix}1 & 2 \\3 & -5 \\\end{bmatrix}$,  B = $\begin{bmatrix}1 & 0 \\0 & 2 \\\end{bmatrix}$  and X be a matrix such that  A = BX, then find X

Answer:  $\frac{1}{2}\begin{bmatrix}2 & 4 \\3 & -5 \\\end{bmatrix}$

Question 25

$\large \bg{black}\mathbf{If\;\;A=\left[\begin{matrix}1&-2&0\\2&-1&-1\\0&-2&1\\\end{matrix}\right]}$ ,

Use it to solve the following system of equations

x - 2y = 10, 2x - y - z = 8,  - 2y + z = 7

Solution Hint

$\large \bg{black}\mathbf{A^{-1}=\left[\begin{matrix}-3&2&2\\-2&1&1\\-4&2&3\\\end{matrix}\right]}$

x = 0,   y = -5,   z = -3

Question 26

A scholarship is a sum of money provided to a student to help him or her pay for education. Some students are granted scholarships based on their academic achievements, while others are rewarded based on their financial needs. Every year a school offers scholarships to girl children and meritorious achievers based on certain criteria. In the session 2023 – 24, the school offered monthly scholarship of ₹ 3,000 each to some girl students and ₹ 4,000 each to meritorious achievers in academics as well as sports. In all, 50 students were given the scholarships and monthly expenditure incurred by the school on scholarships was ₹ 1,80,000.

Based on the above information, answer the following questions :
i) Express the given information algebraically using matrices.
ii) Check whether the system of matrix equations so obtained is consistent or not.
iii) (a) Find the number of scholarships of each kind given by the school, using matrices.
OR
iii) (b) Had the amount of scholarship given to each girl child and meritorious student been interchanged, what would be the monthly expenditure incurred by the school ?

Let No. of girl child scholarships = x
No. of meritorious achievers = y
x + y = 50
3000x + 4000y = 180000 ⇒ 3x + 4y = 180

$\large \bg{black}\mathbf{\left[\begin{matrix}1&1\\3&4\\\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right]=\left[\begin{matrix}50\\180\end{matrix}\right]}$

$\large \bg{black}\mathbf{\left|\begin{matrix}1&1\\3&4\\\end{matrix}\right|=1\neq 0}$

Therefore the system of equations are consistent

X  = A-1

$\large \bg{black}\mathbf{\left[\begin{matrix}x\\y\end{matrix}\right]=\left[\begin{matrix}4&-1\\-3&1\\\end{matrix}\right]\left[\begin{matrix}50\\180\end{matrix}\right]=\left[\begin{matrix}20\\30\end{matrix}\right]}$

⇒ x = 20, y = 30

Required expenditure = ₹ [30(3000) + 20(4000)] = ₹ 1,70,000