3

3

Q1. For what real value of y will matrix A be equal to matrix B, where

1) 1, 3
2) 1/3, 1/2
3) No real value
4) 2 and 3

Solution

Q2. Find the values of a,b,c,d from the following matrix equation.

Solution

By space the space equality space of space matrices comma space the space corresponding space elements space are space equal. table row cell 2 straight a plus 5 equals 1 end cell cell space space space space space space space straight b plus 7 equals 2 end cell cell space space space space space space space 3 straight c minus 8 equals 1 end cell cell space space space space space space space space 3 straight d plus 9 equals 3 end cell end table therefore space 2 straight a equals negative 4 space space space space space space space straight b equals negative 5 space space space space space space space space space space space space 3 straight c equals 9 space space space space space space space space space space space space space space space space 3 straight d equals negative 6 therefore space straight a equals negative 2 space space space space space space space space space straight b equals negative 5 space space space space space space space space space space space space space straight c equals 3 space space space space space space space space space space space space space space space space space space space space space straight d equals negative 2

Q3. To construct a 2 x 3 matrix [ a_ij], such that a_ij = -

The values that i and j can take are …….

1) i = 1, 2 ; j = 1, 2, 3
2) i = 1, 2, 3 ; j = 1, 2, 3
3) i = 1, 2, 3 ; j = 1, 2
4) i = 1, 2 ; j = 1, 2

Solution

2 ´3 matrix [ a _ij] will have 2 rows and 3 columns, so i = 1, 2 ; j = 1, 2, 3

Q4.

begin mathsize 12px style If space straight A equals open square brackets table row cosθ cell negative sinθ end cell row sinθ cosθ end table close square brackets comma space then space for space what space value space of space straight theta comma space is space straight A space an space identity space matrix ? end style

Solution

begin mathsize 12px style straight A equals open square brackets table row cosθ cell negative sinθ end cell row sinθ cosθ end table close square brackets space will space be space an space identity space matrix space straight I equals open square brackets table row 1 0 row 0 1 end table close square brackets space if space straight theta equals 0 degree. Thus comma space we space have straight A equals open square brackets table row cell cos space 0 degree end cell cell negative sin space 0 degree end cell row cell sin space 0 degree end cell cell cos space 0 degree end cell end table close square brackets equals open square brackets table row 1 0 row 0 1 end table close square brackets space space space space space space space space space space space space space open parentheses because cos space 0 degree equals 1 space and space sin space 0 degree equals 0 close parentheses end style

Q5.

1) $open square brackets table row 4 row 0 row cell 2 straight x end cell end table close square brackets$
2) None
3) $open square brackets table row 4 row straight x row cell 2 straight x end cell end table close square brackets$
4) $open square brackets table row 4 row straight x row straight x end table close square brackets$

Solution

open square brackets table row 1 2 0 row 2 0 1 row 1 0 2 end table close square brackets cross times open square brackets table row 0 row 2 row straight x end table close square brackets equals open square brackets table row cell 1 cross times 0 plus 2 cross times 2 plus 0 cross times straight x end cell row cell 2 cross times 0 plus 0 cross times 2 plus 1 cross times straight x end cell row cell 1 cross times 0 plus 0 cross times 2 plus 2 cross times straight x end cell end table close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open square brackets table row cell 0 plus 4 plus 0 end cell row cell 0 plus 0 plus straight x end cell row cell 0 plus 0 plus 2 straight x end cell end table close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open square brackets table row 4 row straight x row cell 2 straight x end cell end table close square brackets

Q6.

Solution

Q7. If A = [aij] be a matrix of order m

n, then write A in the expanded form if m = 3 and n = 1.

Solution

If m

n = 3

1, the matrix has 3 rows and 1 column. It is a column martix and can be written in the expanded form as

Q8. If

find the values of x and y.

Solution

By space the space equality space of space matrices comma space the space corresponding space elements space are space equal comma space if space two space matrices space are space equal. Thus comma space we space have straight x plus 3 straight y equals 4 straight y equals negative 1 7 minus straight x equals 0 space rightwards double arrow straight x equals 7 Hence comma space straight x equals 7 space and space straight y equals negative 1

Q9.

Solution

Q10. Calculate the inverse of matrix .

Solution

In order to use elementary row operations, write A = IA.

Q11. Calculate the inverse of matrix

Solution

In order to use elementary row operations, write A = IA.

Q12.

If space straight A space equals space open square brackets table row cosθ cell negative sinθ end cell row sinθ cosθ end table close square brackets space and space straight A plus straight A to the power of straight T equals straight I space then space find space straight theta.

Solution

straight A plus straight A to the power of straight T equals straight I therefore space open square brackets table row cosθ cell negative sinθ end cell row sinθ cosθ end table close square brackets plus open square brackets table row cosθ sinθ row cell negative sinθ end cell cosθ end table close square brackets equals open square brackets table row 1 0 row 0 1 end table close square brackets therefore space open square brackets table row cell 2 cosθ end cell 0 row 0 cell 2 cosθ end cell end table close square brackets equals open square brackets table row 1 0 row 0 1 end table close square brackets By space equality space of space matrices comma space we space have 2 cosθ equals 1 therefore space cosθ equals 1 half therefore space straight theta equals 60 degree

Q13. Find , if it exist, given

Solution

In order to use elementary row operations, write A = IA.

Q14. Write a 2

3 matrix A = [aij] whose elements are given by aij = (i - 2j)².

Solution

A 2

3 matrix can be represented as

. The element aij = (i - 2j)², therefore can be written asa₁₁ = (1 - 2)² = (-1)² = 1a₁₂ = (1 - 4)² = (-3)² = 9a₁₃ = (1 - 6)² = (-5)² = 25a₂₁ = (2 - 2)² = 0² = 0a₂₂ = (2 - 4)² = (-2)² = 4a₂₃ = (2 - 6)² = (-4)² = 16

Q15. Calculate the inverse of matrix

Solution

Q16.

begin mathsize 12px style For space the space matrices space straight A space equals space open square brackets table row 2 7 cell negative 1 end cell row 3 cell negative 1 end cell 4 end table close square brackets space and space straight B space equals space open square brackets table row 1 cell negative 2 end cell row cell negative 4 end cell 0 row 3 1 end table close square brackets space verify space that space left parenthesis AB right parenthesis to the power of straight T space equals space straight B to the power of straight T straight A to the power of straight T end style

Solution

Q17. Find X and Y if .

Solution

Q18.

begin mathsize 12px style If space straight A to the power of straight T space equals space open square brackets table row 1 cell negative 1 end cell 2 row 4 3 cell negative 2 end cell end table close square brackets space space and space space straight B space equals space open square brackets table row 2 cell negative 3 end cell row 1 cell negative 4 end cell row 2 cell negative 2 end cell end table close square brackets space then space show space that space open parentheses straight A space plus space straight B close parentheses to the power of straight T space equals space straight A to the power of straight T space end exponent plus space straight B to the power of straight T end style

Solution

Therefore, and Thus, .

Q19.

begin mathsize 12px style Given space straight A space equals space open square brackets table row 1 2 row 3 4 end table close square brackets space comma space space straight B space equals space open square brackets table row 3 1 row 4 2 end table close square brackets space and space straight C space equals open square brackets table row 5 1 row 7 4 end table close square brackets space show space that space left parenthesis straight A space plus space straight B right parenthesis space straight C space equals space AC space plus space BC. end style

Solution

A equals open square brackets table row 1 2 row 3 4 end table close square brackets comma space B equals open square brackets table row 3 1 row 4 2 end table close square brackets space a n d space C equals open square brackets table row 5 1 row 7 4 end table close square brackets

Therefore,

Q20. Compute where _{$begin mathsize 12px style straight A space equals space open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets end style$}and I is the unit matrix.

Solution

Q21. Express the matrix as the sum of a symmetric and a skew symmetric matrix.

Solution

Given:

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