11th Business Maths Monthly Test - 1 ( Matrices and Determinants )

MABS Institution

11th Business Maths Monthly Test - 1 ( Matrices and Determinants )-Aug 2020

Time : 2.30 Hours

Part - A

Multiple Choice Question - Answer All Question

The co-factor of -7 in the determinant \(\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}\) is

-18
18
-7
7

If A \(=\begin{pmatrix} -1 & 2 \\ 1 & -4 \end{pmatrix}\) then A (adj A) is

\(\begin{pmatrix} -4 & -2 \\ -1 & -1 \end{pmatrix}\)
\(\begin{pmatrix} 4 & -2 \\ -1 & 1 \end{pmatrix}\)
\(\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\)
\(\begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}\)

The value of \(\begin{vmatrix} x & x^2 & -yz & 1 \\ y & y^2 & -zx & 1 \\ z & z^2 & -xy &1 \end{vmatrix}\) is

1
0
-1
-xyz

If A and B are non-singular matrices then, which of the following is incorrect?

A² = Iimplies A^-1= A
I^-1 = I
If AX = B, then X = B^-1 A
If A is square matrix of order 3 then |adj A|= |A|²

If \(\begin{vmatrix} x & 2 \\ 8 &5 \end{vmatrix}=0\) then the value of x is

\({{-5}\over{6}}\)
\({{5}\over{6}}\)
\({{-16}\over{5}}\)
\({{16}\over{5}}\)

If any three-rows or columns of a determinant are identical, then the value of the determinant is

If A = \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\)such that ad - bc \(\neq\) 0 then A^-1 is

\({{1}\over{ad-bc}}\begin{pmatrix} d & b \\-c & a\end{pmatrix}\)
\({{1}\over{ad-bc}}\begin{pmatrix} d & b \\c & a\end{pmatrix}\)
\({{1}\over{ad-bc}}\begin{pmatrix} d & -b \\-c & a\end{pmatrix}\)
\({{1}\over{ad-bc}}\begin{pmatrix} d & -b \\c & a\end{pmatrix}\)

If A is 3 x 3 matrix and |A|= 4, then |A^-1| is equal to

\({{1}\over{4}}\)
\({{1}\over{16}}\)
2
4

The inverse matrix of \(\begin{pmatrix} \frac { 1 }{ 5 } & \frac { 5 }{ 25 } \\ \frac { 2 }{ 5 } & \frac { 1 }{ 2 } \end{pmatrix}\) is

\({{7}\over{30}}\begin{pmatrix} \frac { 1 }{ 2 } & \frac { 5 }{ 12 } \\ \frac { 2 }{ 5 } & \frac { 4 }{ 5 } \end{pmatrix}\)
\({{7}\over{30}}\begin{pmatrix} \frac { 1 }{ 2 } & \frac { -5 }{ 12 } \\ \frac { -2 }{ 5 } & \frac { 1 }{ 5 } \end{pmatrix}\)
\({{30}\over{7}}\begin{pmatrix} \frac { 1 }{ 2 } & \frac { 5 }{ 12 } \\ \frac { 2 }{ 5 } & \frac { 4 }{ 5 } \end{pmatrix}\)
\({{30}\over{7}}\begin{pmatrix} \frac { 1 }{ 2 } & \frac { -5 }{ 12 } \\ \frac { -2 }{ 5 } & \frac { 4 }{ 5 } \end{pmatrix}\)

If \(\begin{vmatrix} 4 & 3 \\ 3 & 1 \end{vmatrix}=-5\) then value of \(\begin{vmatrix} 20 & 15 \\ 15 & 5 \end{vmatrix}\) is

-5
-125
-25
0

Part - B

Generic 2 - Answer All Question

Find the minors and cofactors of all the elements of the following determinants.
\(\begin{bmatrix} 1&-3&2\\4&-1&2\\3&5&2 \end{bmatrix}\)

If \(A=\left| \begin{matrix} -2 & 6 \\ 3 & -9 \end{matrix} \right| \)then, find A^-1

Evaluate \(\begin{vmatrix} 2 &-1 &-2 \\0 & 2 & -1\\3 & -5& 0 \end{vmatrix}.\)
Find |AB| if \(A=\begin{bmatrix} 3&-1\\2&1 \end{bmatrix}and \begin{bmatrix} 3&0\\1&-2 \end{bmatrix}\)

If \(A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\) then show that |2A| = 4 |A|.

The technology matrix of an economic system of two industries is\(\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}\) .Test whether the system is viable as per Hawkins-Simon conditions.

Evaluate:\(\left| \begin{matrix} x & x+1 \\ x-1 & x \end{matrix} \right| \)

Find the values of x if \(\begin{vmatrix} 2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4\\6 & x \end{vmatrix}.\)

If A \(=\begin{bmatrix} 1 \\ -4\\3 \end{bmatrix}\) and B = [-1 2 1], verify that (AB)^T = B^T. A^T.

Suppose the inter-industry flow of the product of two sectors X and Yare given as under.

Production Sector Consumption Sector Domestic demand Gross output

X Y

X 15 10 10 35

Y 20 30 15 65

Find the gross output when the domestic demand changes to 12 for X and 18 for Y.

Production Sector	Consumption Sector	Domestic demand	Gross output
	X	Y
X	15	10	10	35
Y	20	30	15	65

Part - C

Generic 3 - Answer All Question

Show that \(\begin{vmatrix}0 &ab^2 &ac^2 \\a^2b & 0 & bc^2\\a^2c&b^2c&0\end{vmatrix}=2a^3b^3c^3.\)

If A=\(\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\) and B=\(\begin{bmatrix} 4 & 6 \\ 3 & 2 \end{bmatrix}\), verify that (AB)^-1=B^-1A^-1

Find the inverse of each of the following matrices.
\(\begin{bmatrix} 1&-1\\2&3 \end{bmatrix}\)

Find the adjoint of the matrix \(A=\begin{bmatrix}2&3\\1&4 \end{bmatrix}\)

The data below are about an economy of two industries P and Q. The values are in lakhs of rupees.

Producer User Final Demand Total output

P Q

P 16 12 12 40

Q 12 8 4 24

Find the technology matrix and check whether the system is viable as per Hawkins-Simon conditions.

Producer	User	Final Demand	Total output
P	16	12	12	40
Q	12	8	4	24

Using the properties of determinants, show that \(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| \)=0
If A=\(\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\)and B=\(\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}\)then, verify that (AB)^-1=B^-1A^-1

Verify that A(adj A) = (adj A) A = IAI·I for the matrix A = \(\begin{bmatrix}2 & 3 \\-1 & 4\end{bmatrix}\)

Find the adjoint of the matrix \(\left[ \begin{matrix} 2 & -1 & 3 \\ 0 & 5 & 1 \\ 3 & 6 & 8 \end{matrix} \right] \)

Find the minor and cofactor of each element of the determinant\(\left| \begin{matrix} 1 & -2 \\ 4 & 3 \end{matrix} \right| \)

Part - D

Generic 5 - Answer All Question

The following inter – industry transactions table was constructed for an economy of the year 2016

Industry 1 2 Final consumption Total output

1 500 1,600 400 2,500

2 1,750 1,600 4,650 8,000

Labours 250 4,800 - -

Construct technology co-efficient matrix showing direct requirements. Does a solution exist for this system.

An economy produces only coal and steel. These two commodities serve as intermediate inputs in each other’s production. 0.4 tonne of steel and 0.7 tonne of coal are needed to produce a tonne of steel. Similarly 0.1 tonne of steel and 0.6 tonne of coal are required to produce a tonne of coal. No capital inputs are needed. Do you think that the system is viable? 2 and 5 labour days are required to produce a tonne s of coal and steel respectively. If economy needs 100 tonnes of coal and 50 tonnes of steel, calculate the gross output of the two commodities and the total labour days required.

Weekly expenditure in an office for three weeks is given as follows. Assuming that the salary in all the three weeks of different categories of staff did not vary, calculate the salary for each type of staff, using matrix inversion method.

Week Number of employees Total weekly
Salary (in rupees)

A B C

1^st week 4 2 3 4900

2^nd week 3 3 2 4500

3^rd week 4 3 4 5800

Solve by matrix inversion method: 2x + 3y - 5 = 0, x - 2y + 1 = 0.

You are given the following transaction matrix for a two sector economy.

Sector Sales Final demand Gross output

1 2

1 4 3 13 20

2 5 4 3 12

i) Write the technology matrix.
ii) Determine the output when the final demand for the output sector 1 alone increases to 23 units.
If \(A=\left[ \begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix} \right] \)satisfies the equation A² kA +I₂ - + 2 = then, find k and also A^-1.

Sector	Sales	Final demand	Gross output
1	4	3	13	20
2	5	4	3	12

Find the minor and cofactor of each element of the determinant.\(\left| \begin{matrix} 3 & 1 & 2 \\ 2 & 2 & 5 \\ 4 & 1 & 0 \end{matrix} \right| \)

Solve by using matrix inversion method:
3x-2y+3z=8;2x+y-z=1;4x-3y+2z =4

Two commodities A and B are produced such that 0.4 tonne of A and 0.7 tonne of B are required to produce a tonne of A. Similarly 0.1 tonne of A and 0.7 tonne of B are needed to produce a tonne of B. Write down the technology matrix. If 6.8 tonnes of A and 10.2 tonnes of B are required, find the gross production of both of them.

If A = \(\begin{bmatrix}1 & 1 & 1 \\ 3 & 4 & 7\\1 & -1 & 1 \end{bmatrix}\) verify that A ( adj A ) = ( adj A ) A = |A| I₃.

Comment

Add Comment

Industry	1	2	Final consumption	Total output
1	500	1,600	400	2,500
2	1,750	1,600	4,650	8,000
Labours	250	4,800	-	-

Week	Number of employees			Total weekly Salary (in rupees)
Week	A	B	C	Total weekly Salary (in rupees)
1^st week	4	2	3	4900
2^nd week	3	3	2	4500
3^rd week	4	3	4	5800

InstaQB

Institutions

12th Standard English Medium

12th Standard Tamil Medium

11th Standard English Medium

11th Standard Tamil Medium

10th Standard English Medium

9th Standard English Medium

8th Standard English Medium

7th Standard English Medium

Class XII

Class XI

12Th Standard

11Th Standard

10Th Standard

9Th Standard

11th Business Maths Monthly Test - 1 ( Matrices and Determinants )

MABS Institution

11th Business Maths Monthly Test - 1 ( Matrices and Determinants )-Aug 2020

Part - A

Multiple Choice Question - Answer All Question

The co-factor of -7 in the determinant \(\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}\) is

If A \(=\begin{pmatrix} -1 & 2 \\ 1 & -4 \end{pmatrix}\) then A (adj A) is

The value of \(\begin{vmatrix} x & x^2 & -yz & 1 \\ y & y^2 & -zx & 1 \\ z & z^2 & -xy &1 \end{vmatrix}\) is

If A and B are non-singular matrices then, which of the following is incorrect?

If \(\begin{vmatrix} x & 2 \\ 8 &5 \end{vmatrix}=0\) then the value of x is

If any three-rows or columns of a determinant are identical, then the value of the determinant is

If A = \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\)such that ad - bc \(\neq\) 0 then A-1 is

If A is 3 x 3 matrix and |A|= 4, then |A-1| is equal to

The inverse matrix of \(\begin{pmatrix} \frac { 1 }{ 5 } & \frac { 5 }{ 25 } \\ \frac { 2 }{ 5 } & \frac { 1 }{ 2 } \end{pmatrix}\) is

If \(\begin{vmatrix} 4 & 3 \\ 3 & 1 \end{vmatrix}=-5\) then value of \(\begin{vmatrix} 20 & 15 \\ 15 & 5 \end{vmatrix}\) is

Part - B

Generic 2 - Answer All Question

Find the minors and cofactors of all the elements of the following determinants. \(\begin{bmatrix} 1&-3&2\\4&-1&2\\3&5&2 \end{bmatrix}\)

If \(A=\left| \begin{matrix} -2 & 6 \\ 3 & -9 \end{matrix} \right| \)then, find A-1

Evaluate \(\begin{vmatrix} 2 &-1 &-2 \\0 & 2 & -1\\3 & -5& 0 \end{vmatrix}.\)

Find |AB| if \(A=\begin{bmatrix} 3&-1\\2&1 \end{bmatrix}and \begin{bmatrix} 3&0\\1&-2 \end{bmatrix}\)

If \(A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\) then show that |2A| = 4 |A|.

The technology matrix of an economic system of two industries is\(\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}\) .Test whether the system is viable as per Hawkins-Simon conditions.

Evaluate:\(\left| \begin{matrix} x & x+1 \\ x-1 & x \end{matrix} \right| \)

Find the values of x if \(\begin{vmatrix} 2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4\\6 & x \end{vmatrix}.\)

If A \(=\begin{bmatrix} 1 \\ -4\\3 \end{bmatrix}\) and B = [-1 2 1], verify that (AB)T = BT. AT.

Suppose the inter-industry flow of the product of two sectors X and Yare given as under. Production Sector Consumption Sector Domestic demand Gross output X Y X 15 10 10 35 Y 20 30 15 65 Find the gross output when the domestic demand changes to 12 for X and 18 for Y.

Part - C

Generic 3 - Answer All Question

Show that \(\begin{vmatrix}0 &ab^2 &ac^2 \\a^2b & 0 & bc^2\\a^2c&b^2c&0\end{vmatrix}=2a^3b^3c^3.\)

If A=\(\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\) and B=\(\begin{bmatrix} 4 & 6 \\ 3 & 2 \end{bmatrix}\), verify that (AB)-1=B-1A-1

Find the inverse of each of the following matrices. \(\begin{bmatrix} 1&-1\\2&3 \end{bmatrix}\)

Find the adjoint of the matrix \(A=\begin{bmatrix}2&3\\1&4 \end{bmatrix}\)

The data below are about an economy of two industries P and Q. The values are in lakhs of rupees. Producer User Final Demand Total output P Q P 16 12 12 40 Q 12 8 4 24 Find the technology matrix and check whether the system is viable as per Hawkins-Simon conditions.

Using the properties of determinants, show that \(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| \)=0

If A=\(\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\)and B=\(\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}\)then, verify that (AB)-1=B-1A-1

Verify that A(adj A) = (adj A) A = IAI·I for the matrix A = \(\begin{bmatrix}2 & 3 \\-1 & 4\end{bmatrix}\)

Find the adjoint of the matrix \(\left[ \begin{matrix} 2 & -1 & 3 \\ 0 & 5 & 1 \\ 3 & 6 & 8 \end{matrix} \right] \)

Find the minor and cofactor of each element of the determinant\(\left| \begin{matrix} 1 & -2 \\ 4 & 3 \end{matrix} \right| \)

Part - D

Generic 5 - Answer All Question

Solve by matrix inversion method: 2x + 3y - 5 = 0, x - 2y + 1 = 0.

You are given the following transaction matrix for a two sector economy. Sector Sales Final demand Gross output 1 2 1 4 3 13 20 2 5 4 3 12 i) Write the technology matrix. ii) Determine the output when the final demand for the output sector 1 alone increases to 23 units.

If \(A=\left[ \begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix} \right] \)satisfies the equation A2 kA +I2 - + 2 = then, find k and also A-1.

Find the minor and cofactor of each element of the determinant.\(\left| \begin{matrix} 3 & 1 & 2 \\ 2 & 2 & 5 \\ 4 & 1 & 0 \end{matrix} \right| \)

Solve by using matrix inversion method: 3x-2y+3z=8;2x+y-z=1;4x-3y+2z =4

If A = \(\begin{bmatrix}1 & 1 & 1 \\ 3 & 4 & 7\\1 & -1 & 1 \end{bmatrix}\) verify that A ( adj A ) = ( adj A ) A = |A| I3.

Comment

Other Study material

11th Business Maths Monthly Test - 3 ( Differentia

11th Business Maths Monthly Test - 4 (Operations R

11th Business Maths Weekly Test - 2 ( Applications

11th Business Maths Weekly Test - 1 ( Applications

11th Business Maths Monthly Test - 4 ( Differentia

11th Business Maths Weekly Test - 4 ( Differential

11th Business Maths Weekly Test - 3 ( Differentia

11th Business Maths Weekly Test -2 ( Differential

11th Business Maths Weekly Test - 1 ( Differential

11th Business Maths Monthly Test - 4 ( Trigonometr

11th Business Maths Monthly Test - 3 ( Trigonometr

11th Business Maths Monthly Test - 2 ( Trigonometr

11th Business Maths Monthly Test - 1 ( Trigonometr

11th Business Maths Weekly Test - 4 ( Trigonometry

11th Business Maths Monthly Test - 1 ( Differentia

If A = \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\)such that ad - bc \(\neq\) 0 then A^-1 is

If A is 3 x 3 matrix and |A|= 4, then |A^-1| is equal to

Find the minors and cofactors of all the elements of the following determinants.
\(\begin{bmatrix} 1&-3&2\\4&-1&2\\3&5&2 \end{bmatrix}\)

If \(A=\left| \begin{matrix} -2 & 6 \\ 3 & -9 \end{matrix} \right| \)then, find A^-1

If A \(=\begin{bmatrix} 1 \\ -4\\3 \end{bmatrix}\) and B = [-1 2 1], verify that (AB)^T = B^T. A^T.

Suppose the inter-industry flow of the product of two sectors X and Yare given as under.

Production Sector Consumption Sector Domestic demand Gross output

X Y

X 15 10 10 35

Y 20 30 15 65

Find the gross output when the domestic demand changes to 12 for X and 18 for Y.

If A=\(\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\) and B=\(\begin{bmatrix} 4 & 6 \\ 3 & 2 \end{bmatrix}\), verify that (AB)^-1=B^-1A^-1

Find the inverse of each of the following matrices.
\(\begin{bmatrix} 1&-1\\2&3 \end{bmatrix}\)

The data below are about an economy of two industries P and Q. The values are in lakhs of rupees.

Producer User Final Demand Total output

P Q

P 16 12 12 40

Q 12 8 4 24

Find the technology matrix and check whether the system is viable as per Hawkins-Simon conditions.

If A=\(\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\)and B=\(\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}\)then, verify that (AB)^-1=B^-1A^-1

You are given the following transaction matrix for a two sector economy.

Sector Sales Final demand Gross output

1 2

1 4 3 13 20

2 5 4 3 12

i) Write the technology matrix.
ii) Determine the output when the final demand for the output sector 1 alone increases to 23 units.

If \(A=\left[ \begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix} \right] \)satisfies the equation A² kA +I₂ - + 2 = then, find k and also A^-1.

Solve by using matrix inversion method:
3x-2y+3z=8;2x+y-z=1;4x-3y+2z =4

If A = \(\begin{bmatrix}1 & 1 & 1 \\ 3 & 4 & 7\\1 & -1 & 1 \end{bmatrix}\) verify that A ( adj A ) = ( adj A ) A = |A| I₃.