Question:

If A and B are square matrices of the same order such that AB=BA, then prove by induction that $\begin{array}{}{\mathrm{AB}}^{\mathrm{n}}\end{array}$=$\begin{array}{}{B}^{\mathrm{n}}A\end{array}$. Further, prove that $\begin{array}{}{\mathrm{(AB)}}^{\mathrm{n}}\end{array}$=$\begin{array}{}{A}^{\mathrm{n}}{B}^{\mathrm{n}}\end{array}$ for all n∈N.

Answer:

1. Base case: Let n = 1. Then $\begin{array}{}{\mathrm{AB}}^1\end{array}$=AB and $\begin{array}{}{B}^{1}A\end{array}$=BA. Since AB = BA, the base case is true.

2. Assume that $\begin{array}{}{\mathrm{AB}}^{\mathrm{n}}\end{array}$=$\begin{array}{}{B}^{\mathrm{n}}A\end{array}$ for some n∈N.

3. Prove that $\begin{array}{}{\mathrm{AB}}^{\mathrm{n+1}}\end{array}$=$\begin{array}{}{B}^{\mathrm{n+1}}A\end{array}$.

By the induction hypothesis, $\begin{array}{}{\mathrm{AB}}^{\mathrm{n}}\end{array}$=$\begin{array}{}{B}^{\mathrm{n}}A\end{array}$.

Therefore, $\begin{array}{}{\mathrm{AB}}^{\mathrm{n+1}}\end{array}$=AB$\begin{array}{}{B}^{\mathrm{n}}A\end{array}$ =BA$\begin{array}{}{B}^{\mathrm{n}}A\end{array}$ =$\begin{array}{}{B}^{\mathrm{n+1}}A\end{array}$

4. Prove that $\begin{array}{}{\mathrm{(AB)}}^{\mathrm{n}}\end{array}$=<math xmlns = “http://www.w3.

Question:

If the matrix A is both symmetric and skew symmetric, then A A is a diagonal matrix B A is a zero matrix C A is a square matrix D None of these

Answer:

A. A A is a diagonal matrix.

B. B A is a zero matrix.

C. C A is a square matrix.

D. None of these.

Question:

Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

1. Recall that a matrix A is symmetric if A = A’ and a matrix A is skew symmetric if A = -A'.

2. Now, if A is symmetric, then B’AB = (B’A)B = (AB)‘B = (A’B)‘B = A’(BB’) = A’B = (A’B)’ = (B’A)’ = (B’AB)’.

3. If A is skew symmetric, then B’AB = (B’A)B = (AB)‘B = (-A’B)‘B = -(A’B)B = -A’(BB’) = -A’B = -(A’B)’ = -(B’A)’ = -(B’AB)’.

4. Therefore, B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Question:

If A=$\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$, then prove $\begin{array}{}{A}^{\mathrm{n}}\end{array}$=$\left[\begin{array}{cc}\mathrm{1+2n}& \mathrm{-4n}\\ \mathrm{n}& \mathrm{1-2n}\end{array}\right]$ where n is any positive integer

Answer:

Proof:

We will use mathematical induction to prove the statement.

Base case: Let n = 1.

$\begin{array}{}{A}^1\end{array}$ = $\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

= $\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

= $\left[\begin{array}{cc}\mathrm{1+2}& -4\\ 1& \mathrm{1-2}\end{array}\right]$

= $\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

Hence, the statement is true for n = 1.

Inductive step: Assume the statement is true for some positive integer n = k.

$\begin{array}{}{A}^{\mathrm{k}}\end{array}$ = $\left[\begin{array}{cc}\mathrm{1+2k}& \mathrm{-4k}\\ \mathrm{k}& \mathrm{1-2k}\end{array}\right]$

Now,

$\begin{array}{}{A}^{\mathrm{k+1}}\end{array}$ = ${\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]}^{}\mathrm{k+1}$

=

Question:

Find the matrix X so that X$\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]$=$\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$

Answer:

1. Begin by writing the equation as a system of equations:

x11 + 2x12 + 3x13 = -7 4x11 + 5x12 + 6x13 = -8 3x11 + 4x12 + 6x13 = -9

2. Use Gaussian Elimination to solve for the variables:

Step 1: Subtract 4 times the first equation from the second equation: -3x11 - 2x12 - 3x13 = -15

Step 2: Subtract 3 times the first equation from the third equation: -2x11 - x12 - 3x13 = -12

Step 3: Subtract twice the second equation from the third equation: -2x11 - x12 - 3x13 = -12 x12 = -2

Step 4: Substitute x12 = -2 into the second equation: -3x11 + 4 = -15 x11 = -5

Step 5: Substitute x11 = -5 and x12 = -2 into the first equation: 3x13 = -2 x13 = -2/3

3. The solution to the system of equations is x11 = -5, x12 = -2, and x13 = -2/3. Therefore, the matrix X is:

X = $\left[\begin{array}{ccc}-5& -2& \mathrm{-2/3}\\ 4& 5& 6\end{array}\right]$

Question:

If A and B are symmetric matrices, prove that AB−BA is a skew symmetric matrix.

Answer:

1. A skew-symmetric matrix is a matrix whose transpose is equal to its negative.

2. Since A and B are symmetric matrices, then their transpose is equal to themselves.

3. Therefore, AB−BA = (AB)T − (BA)T = BTAT − BAT = −ATBT + BAT = −(ATBT − BAT)

4. Since the transpose of a skew-symmetric matrix is equal to its negative, then AB−BA is a skew symmetric matrix.

Question:

Why do you think Prem wants to tell the story of the reptiles to the people of his village?

Answer:

1. Prem may want to tell the story of the reptiles to the people of his village because he believes it is an important part of his culture and heritage that should be shared and preserved.

2. He may also want to educate his fellow villagers about the reptiles, so that they can appreciate their beauty and importance to the environment.

3. Additionally, Prem may be hoping to use the story of the reptiles to inspire the people of his village to take action to protect and conserve the species.

Question:

For what values of x. $\left[\begin{array}{ccc}1& 2& 1\end{array}\right]$$\left[\begin{array}{ccc}1& 2& 0\\ 2& 0& 1\\ 1& 0& 2\end{array}\right]$​$\left[\begin{array}{c}0\\ 2\\ \mathrm{X}\end{array}\right]$=0?

Answer:

1. Write the augmented matrix by adding the columns of the two matrices: $\left[\begin{array}{cccc}1& 2& 1& 0\\ 2& 0& 1& 2\\ 1& 0& 2& \mathrm{X}\end{array}\right]$

2. Use row operations to reduce the matrix to reduced row-echelon form: $\left[\begin{array}{cccc}1& 2& 1& 0\\ 0& -4& -2& 2\\ 0& 0& 0& \mathrm{X+2}\end{array}\right]$

3. The value of x is x=-2.

Question:

If A is square matrix such that $\begin{array}{}{A}^2\end{array}$=A, then $\begin{array}{}{\mathrm{(I+A)}}^3\end{array}$−7A is equal to A A B I−A C I D 3A

Answer:

Step 1: $\begin{array}{}{\mathrm{(I+A)}}^3\end{array}$−7A

= $\begin{array}{}{\mathrm{(I+A)}}^2\end{array}$ + $\begin{array}{}A \cdot 2\end{array}$ − $\begin{array}{}A \cdot 7\end{array}$

Step 2: $\begin{array}{}{\mathrm{(I+A)}}^2\end{array}$ + $\begin{array}{}A \cdot 2\end{array}$ − $\begin{array}{}A \cdot 7\end{array}$

= $\begin{array}{}{\mathrm{(I+A)}}^2\end{array}$ + $\begin{array}{}A \cdot 2\end{array}$ − $\begin{array}{}A \cdot 5 \cdot A\end{array}$

= $\begin{array}{}{\mathrm{(I+A)}}^2\end{array}$ −

Question:

A manufacturer produces three products x,y,z which he sells in two markets. Annual sales are indicated below: Market Products I 10,000 2,000 18,000 II 6,000 20,000 8,000 (a) If unit sale prices of x,y and z are Rs.2.50, Rs.1.50 and Rs.1.00, respectively, find the total revenue in each market with the help of matrix algebra. (b) If the unit costs of the above three commodities are Rs.2.00,Rs.1.00 and 50 paise respectively. Find the gross profit.

Answer:

Answer: (a) Total revenue in each market can be found with the help of matrix algebra. Let A= [2.50 1.50 1.00]

Market I: A x [10,000 2,000 18,000]T = [50,000 30,000 18,000]T = Rs.98,000

Market II: A x [6,000 20,000 8,000]T = [30,000 30,000 8,000]T = Rs.68,000

(b) Gross Profit in each market can be found with the help of matrix algebra. Let B= [2.00 1.00 0.50]

Market I: B x [10,000 2,000 18,000]T = [20,000 2,000 9,000]T = Rs.31,000

Market II: B x [6,000 20,000 8,000]T = [12,000 20,000 4,000]T = Rs.36,000

Question:

If A=$\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$, prove that $\begin{array}{}{A}^{\mathrm{n}}\end{array}$=$\left[\begin{array}{}{3}^{\mathrm{n-1}}{3}^{\mathrm{n-1}}{3}^{\mathrm{n-1}}\\ 3^{\mathrm{n-1}}{3}^{\mathrm{n-1}}{3}^{\mathrm{n-1}}\\ 3^{\mathrm{n-1}}{3}^{\mathrm{n-1}}{3}^{\mathrm{n-1}}\end{array}\right]$,n∈N

Answer:

Proof:

We will prove this statement by induction on n.

Base Case:

For n = 1, A = $\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$

$\begin{array}{}{A}^1\end{array}$ = $\left[\begin{array}{}{3}^{\mathrm{1-1}}{3}^{\mathrm{1-1}}{3}^{\mathrm{1-1}}\\ 3^{\mathrm{1-1}}{3}^{\mathrm{1-1}}{3}^{\mathrm{1-1}}\\ 3^{\mathrm{1-1}}{3}^{\mathrm{1-1}}{3}^{\mathrm{1-1}}\end{array}\right]$ = $\left[\begin{array}{}111\\ 111\\ 111\end{array}\right]$

which is equal to A.

Inductive Step:

Assume that for some n = k, $\begin{array}{}{A}^{\mathrm{k}}\end{array}$ = $\left[\begin{array}{}{3}^{\mathrm{k-1}}{3}^{\mathrm{k-1}}{3}^{\mathrm{k-1}}\\ 3^{\mathrm{k-1}}{3}^{\mathrm{k-1}}{3}^{\mathrm{k-1}}\\ 3^{\mathrm{k-1}}{3}^{\mathrm{k-1}}{3}^{\mathrm{k-1}}\end{array}\right]$

Question:

Find x, if $\left[\begin{array}{ccc}\mathrm{X}& -5& -1\end{array}\right]$$\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]$$\left[\begin{array}{c}\mathrm{X}\\ 4\\ 1\end{array}\right]$=0

Answer:

1. Multiply the first matrix by the scalar X: $\left[\begin{array}{ccc}\mathrm{X}& \mathrm{-5X}& \mathrm{-X}\end{array}\right]$

2. Add the second matrix to the result of the first step: $\left[\begin{array}{ccc}\mathrm{X}& \mathrm{-5X}& \mathrm{-X}\\ 1& 2& 3\end{array}\right]$

3. Set the third matrix equal to the result of the second step: $\left[\begin{array}{ccc}\mathrm{X}& \mathrm{-5X}& \mathrm{-X}\\ 1& 2& 3\\ 4& 1& 0\end{array}\right]$

4. Solve for X: $\left[\begin{array}{ccc}\mathrm{X}& \mathrm{-5X}& \mathrm{-X}\\ 1& 2& 3\\ 4& 1& 0\end{array}\right]$

X = -1

Question:

Find the values of x,y,z if the matrix A=$\left[\begin{array}{ccc}1& \mathrm{2y}& \mathrm{z}\\ \mathrm{x}& \mathrm{y}& \mathrm{-z}\\ \mathrm{x}& \mathrm{-y}& \mathrm{z}\end{array}\right]$ satisfy the equation A′A=I.

Answer:

1. We need to find A’A. A’A = $\left[\begin{array}{ccc}1& \mathrm{x}& \mathrm{x}\\ \mathrm{2y}& \mathrm{y}& \mathrm{-y}\\ \mathrm{z}& \mathrm{-z}& \mathrm{z}\end{array}\right]$

2. We need to find the inverse of A’A A’A^-1 = $\left[\begin{array}{ccc}1& \frac{1}{\mathrm{2y}}& \frac{1}{\mathrm{z}}\\ \frac{1}{\mathrm{x}}& 1& \frac{1}{\mathrm{-z}}\\ \frac{1}{\mathrm{x}}& \frac{1}{\mathrm{-y}}& 1\end{array}\right]$

3. We can now solve for x,y,z: x = 1/x y = 1/y z = 1/z

Question:

Solve system of linear equations , using matrix method. 2x−y=−2 3x+4y=3

Answer:

Step 1: Rewrite the equations in matrix form.

[2 -1] [x] = [-2] [3 4] [y] [3]

Step 2: Multiply the inverse of the coefficient matrix by the constant matrix.

[2 -1] [x] [1 1/2] [x] [1/2] [3 4] [y] x [0 1/4] [y] = [3/4]

Step 3: Solve for x and y.

x = 1/2 y = 3/4

Question:

Let A=$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, show that $\begin{array}{}{\mathrm{(aI+bA)}}^{\mathrm{n}}\end{array}$=$\begin{array}{}{a}^{\mathrm{n}}I\end{array}$+$\begin{array}{}{\mathrm{na}}^{\mathrm{n-1}}\mathrm{bA}\end{array}$, where I is the identity matrix of order 2 and n∈N

Answer:

Answer:

1. A = $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

2. I = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

3. aI+bA = $\left[\begin{array}{cc}a+b0& b1\\ 0+b0& a0\end{array}\right]$

4. (aI+bA)^n = $\begin{array}{}{\mathrm{(aI+bA)}}^{\mathrm{n}}\end{array}$

5. (aI+bA)^n = $\begin{array}{}{a}^{\mathrm{n}}I+{\mathrm{na}}^{\mathrm{n-1}}\mathrm{bA}\end{array}$