Question:

Construct a 2×2 matrix, A=$\begin{array}{}{\mathrm{[a}}_{\mathrm{ij}}]\end{array}$, whose elements are given by: (i) $\begin{array}{}{a}_{\mathrm{ij}}\end{array}$=$\begin{array}{}{\mathrm{(i+j)}}^2\end{array}$/2 (ii) $\begin{array}{}{a}_{\mathrm{ij}}\end{array}$=i​/j (iii) $\begin{array}{}{a}_{\mathrm{ij}}\end{array}$=$\begin{array}{}{\mathrm{(i+2j)}}^2\end{array}$/2

Answer:

Answer:

(i) A = $\begin{array}{cc}1& 3\\ 2& 4\end{array}$

(ii) A = $\begin{array}{cc}1& \mathrm{1/2}\\ 2& 1\end{array}$

(iii) A = $\begin{array}{cc}1& \mathrm{7/4}\\ 2& 4\end{array}$

Question:

In the matrix, write : A=$\left[\begin{array}{cccc}2& 5& 19& -7\\ 35& -2& \mathrm{5/2}& 12\\ \mathrm{\surd 3}& 1& -5& 17\end{array}\right]$(i) The order of the matrix (ii) The number of elements (iii) Write the elements a13​,a21​,a33​,a24​,a23

Answer:

(i) The order of the matrix is 3x4

(ii) The number of elements is 12

(iii) a13 = √3, a21 = 35, a33 = 17, a24 = 12, a23 = 5/2

Question:

If a matrix has 18 elements , what are the possible orders it can have? what, if it has 5 elements?

Answer:

If a matrix has 18 elements, the possible orders it can have are 2x9, 3x6, 6x3, 9x2.

If a matrix has 5 elements, the possible orders it can have are 1x5, 5x1.

Question:

Which of the given values of x and y make the following pair of matrices equal.$\left[\begin{array}{cc}\mathrm{3x+7}& 5\\ \mathrm{y+1}& \mathrm{2-3x}\end{array}\right]$=$\left[\begin{array}{cc}0& \mathrm{y-2}\\ 8& 4\end{array}\right]$ A x=−1​/3,y=7 B Not possible to find C y=7,x=−2​/3 D x=−1​/3,y=−2/3

Answer:

The answer is C. y=7, x=-2/3

Question:

The number of all possible matrices of order 3×3 with each entry 0 or 1 is: A 27 B 18 C 81 D 512

Answer:

Answer: C 81

Question:

Find the value of x,y and z from the following equation: (i) $\left[\begin{array}{cc}4& 3\\ \mathrm{x}& 5\end{array}\right]$=$\left[\begin{array}{cc}\mathrm{y}& \mathrm{z}\\ 1& 5\end{array}\right]$ (ii) $\left[\begin{array}{cc}\mathrm{x+y}& 2\\ \mathrm{5+z}& \mathrm{xy}\end{array}\right]$=$\left[\begin{array}{cc}6& 2\\ 5& 8\end{array}\right]$ (iii) $\left[\begin{array}{c}\mathrm{x+y+z}\\ \mathrm{x+z}\\ \mathrm{y+z}\end{array}\right]$= $\left[\begin{array}{c}9\\ 5\\ 7\end{array}\right]$

Answer:

Solution:

(i) 4x + 3y = 5z

Substituting the given values, we get

4(4) + 3y = 5z

16 + 3y = 5z

3y = 5z - 16

y = (5z - 16)/3

(ii) x + y + 2 = 6

Substituting the value of y from step (i), we get

x + (5z - 16)/3 + 2 = 6

x + 5z - 16 + 6 = 18

x + 5z = 34

(iii) x + y + z = 9

Substituting the value of y from step (i), we get

x + (5z - 16)/3 + z = 9

x + 5z - 16 + 3z = 27

4z = 27 - x - 16

z = (27 - x - 16)/4

Substituting the value of z in equation (ii), we get

x + 5(27 - x - 16)/4 = 34

4x + 135 - 5x - 80 = 136

x = 25

Substituting the value of x in equation (i) and (iii), we get

y = (5(27) - 16)/3 = 33

z = (27 - 25 - 16)/4 = 3

Hence, x = 25, y = 33 and z = 3

Question:

If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elements?

Answer:

If a matrix has 24 elements, it can have an order of 2 x 12, 3 x 8, 4 x 6, 6 x 4, 8 x 3, or 12 x 2.

If a matrix has 13 elements, it can have an order of 1 x 13, 2 x 6.5, 3 x 4.5, 4 x 3.25, 5 x 2.6, 6 x 2.16, or 13 x 1.

Question:

Find the value of a,b,c and d from the equation: <math xmlns = “http://www.w3.org/1998/Math/MathML"[a-b2a+c2a-b3c+d ]=$\left[\begin{array}{cc}-1& 5\\ 0& 13\end{array}\right]$

Answer:

Step 1: Rewrite the equation as a system of linear equations.

a - b = -1 2a + c = 5 2a - b = 0 3c + d = 13

Step 2: Solve for a in the first and third equations by adding them together.

a - b + 2a - b = -1 + 0 3a = -1 a = -1/3

Step 3: Substitute the value of a in the second equation and solve for c.

2(-1/3) + c = 5 -2/3 + c = 5 c = 5 + 2/3

Step 4: Substitute the values of a and c in the fourth equation and solve for d.

3(5 + 2/3) + d = 13 15 + 2 + d = 13 d = 1

Question:

Construct a 3×4 matrix, whose elements are given by (i) $\begin{array}{}{a}_{\mathrm{ij}}\end{array}$=1​/2∣−3i+j∣ (ii) $\begin{array}{}{a}_{\mathrm{ij}}\end{array}$=2i−j

Answer:

Solution:

(i) A = [1/2|-3i+j|, 1/2|-3i+j|, 1/2|-3i+j|] [1/2|-3i+j|, 1/2|-3i+j|, 1/2|-3i+j|] [1/2|-3i+j|, 1/2|-3i+j|, 1/2|-3i+j|]

(ii) A = [2i-j, 2i-j, 2i-j] [2i-j, 2i-j, 2i-j] [2i-j, 2i-j, 2i-j]