Question:

Solve the equation for x,y,z and t if 2$\left[\begin{array}{cc}\mathrm{x}& \mathrm{z}\\ \mathrm{y}& \mathrm{t}\end{array}\right]$+3$\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]$=3$\left[\begin{array}{cc}3& 5\\ 4& 6\end{array}\right]$

Answer:

Step 1: Subtract 3[1 -1 0 2] from both sides of the equation.

2$\left[\begin{array}{cc}\mathrm{x}& \mathrm{z}\\ \mathrm{y}& \mathrm{t}\end{array}\right]$ = 0$\left[\begin{array}{cc}3& 5\\ 4& 6\end{array}\right]$

Step 2: Solve for x,y,z, and t by isolating each variable.

x = 3

y = 4

z = 5

t = 6

Question:

Find X and Y, if (i) X+Y=$\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]$ and X−Y=$\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$ (ii) 2X+3Y=$\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]$ and 3X+2Y=$\left[\begin{array}{cc}2& -2\\ -1& 5\end{array}\right]$

Answer:

Solution:

(i) To solve X+Y=$\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]$ and X−Y=$\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$

We can solve this equation by adding the equations:

2X = $\left[\begin{array}{cc}10& 0\\ 2& 8\end{array}\right]$

And subtracting the equations:

2Y = $\left[\begin{array}{cc}3& 0\\ -2& 2\end{array}\right]$

Therefore, X = $\left[\begin{array}{cc}5& 0\\ 1& 4\end{array}\right]$ and Y = $\left[\begin{array}{cc}2& 0\\ -1& 2\end{array}\right]$

(ii) To solve 2X+3Y = $\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]$ and 3X+2Y = $\left[\begin{array}{cc}2& -2\\ -1& 5\end{array}\right]$

We can solve this equation by subtracting the equations:

X =

Question:

If A=$\left[\begin{array}{c}102\\ 021\\ 203\end{array}\right]$, prove that $\begin{array}{}{A}^3\end{array}$−6$\begin{array}{}{A}^2\end{array}$+7A+2I=0

Answer:

1. Expand A3: $\begin{array}{}{A}^3=\begin{array}{c}102030201\\ 20301202\\ 30201203\end{array}\end{array}$

2. Expand A2: $\begin{array}{}{A}^2=\begin{array}{c}102021\\ 202102\\ 21302\end{array}\end{array}$

3. Expand A: $\begin{array}{}A=\begin{array}{c}102\\ 021\\ 203\end{array}\end{array}$

4. Substitute A3, A2, and A into the equation: $\begin{array}{}{A}^3-A^2+A+I=0\end{array}$

$\begin{array}{}102030201-102021+102+2=0\end{array}$

5. Simplify the equation: $\begin{array}{}30200+7A+2=0\end{array}$

Question:

Let A=$\left[\begin{array}{cc}2& 4\\ 3& 2\end{array}\right]$,B=$\left[\begin{array}{cc}1& 3\\ -2& 5\end{array}\right]$,C=$\left[\begin{array}{cc}-2& 5\\ 3& 4\end{array}\right]$ Find each of the following (i) A+B (ii) A−B (iii) 3A−C (iv) AB (v) BA

Answer:

(i) A+B = $\left[\begin{array}{cc}3& 7\\ 1& 7\end{array}\right]$

(ii) A−B = $\left[\begin{array}{cc}1& 1\\ 5& -3\end{array}\right]$

(iii) 3A−C = $\left[\begin{array}{cc}6& 17\\ 9& -2\end{array}\right]$

(iv) AB = $\left[\begin{array}{cc}-2& 14\\ -11& 17\end{array}\right]$

(v) BA = $\left[\begin{array}{cc}-5& -1\\ 8& 17\end{array}\right]$

Question:

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer:

Let A = \begin{bmatrix} 10 & 8 & 10 \ 80 & 60 & 40 \end{bmatrix}

A = \begin{bmatrix} 10\times80 & 8\times60 & 10\times40 \ 80 & 60 & 40 \end{bmatrix}

A = \begin{bmatrix} 800 & 480 & 400 \ 80 & 60 & 40 \end{bmatrix}

Total amount = 800 + 480 + 400 = 1680

Question:

Compute the indicated products (i) $\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{-b}& \mathrm{a}\end{array}\right]$$\left[\begin{array}{cc}\mathrm{a}& \mathrm{-b}\\ \mathrm{b}& \mathrm{a}\end{array}\right]$ (ii) $\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$$\left[\begin{array}{ccc}2& 3& 4\end{array}\right]$ (iii) $\left[\begin{array}{cc}1& -2\\ 2& 3\end{array}\right]$ $\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right]$(iv) $\left[\begin{array}{c}234\\ 345\\ 456\end{array}\right]$ $\left[\begin{array}{ccc}1& -3& 5\\ 0& 2& 4\\ 3& 0& 5\end{array}\right]$ (v) $\left[\begin{array}{cc}2& 1\\ 3& 2\\ -1& 1\end{array}\right]$$\left[\begin{array}{ccc}1& 0& 1\\ -1& 2& 1\end{array}\right]$ (vi) $\left[\begin{array}{ccc}3& -1& 3\\ -1& 0& 2\end{array}\right]$$\left[\begin{array}{cc}2& -3\\ 1& 0\\ 3& 1\end{array}\right]$

Answer:

(i) $\left[\begin{array}{cc}\mathrm{a}\cdot \mathrm{b}& \mathrm{-b}\cdot \mathrm{a}\end{array}\right]$ = $\left[\begin{array}{c}\mathrm{a}²-\mathrm{b}²\end{array}\right]$

(ii) $\left[\begin{array}{cc}1\cdot 2\cdot 3& 2\cdot 3\cdot 4\end{array}\right]$ = $\left[\begin{array}{c}2²3²4\end{array}\right]$

(iii) $\left[\begin{array}{cc}1\cdot -2& 2\cdot 3\end{array}\right]$$\left[\begin{array}{cc}1\cdot 2\cdot 3& 2\cdot 3\cdot 1\end{array}\right]$ = $\left[\begin{array}{c}-2²3²1\end{array}\right]$

(iv)

Question:

Compute the following: (i) $\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{-b}& \mathrm{a}\end{array}\right]$+$\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{b}& \mathrm{a}\end{array}\right]$ (ii) $\left[\begin{array}{cc}{a}^2{\mathrm{+b}}^2& b^2{\mathrm{+c}}^2\\ a^2{\mathrm{+c}}^2& a^2{\mathrm{+b}}^2\end{array}\right]$+$\left[\begin{array}{cc}\mathrm{2ab}& \mathrm{2bc}\\ \mathrm{-2ac}& \mathrm{-2ab}\end{array}\right]$ (iii) $\left[\begin{array}{ccc}-1& 4& -6\\ 8& 5& 16\\ 2& 8& 5\end{array}\right]$+$\left[\begin{array}{ccc}1& 2& 76\\ 8& 0& 5\\ 3& 2& 4\end{array}\right]$ (iv) $\left[\begin{array}{cc}{\cos }^{2}x& {\sin }^{2}x\\ {\sin }^{2}x& {\cos }^{2}x\end{array}\right]$+$\left[\begin{array}{cc}{\sin }^{2}x& {\cos }^{2}x\\ {\cos }^{2}x& {\sin }^{2}x\end{array}\right]$

Answer:

(i) $\left[\begin{array}{cc}\mathrm{2a}& 0\\ 0& \mathrm{2a}\end{array}\right]$

(ii) $\left[\begin{array}{cc}{a}^2{\mathrm{+b}}^2+c^2& b^2+c^2+a^2\\ a^2+c^2+b^2& a^2{\mathrm{+b}}^2+c^2\end{array}\right]$

(iii) $\left[\begin{array}{ccc}-7& 10& 70\\ 8& 5& 21\\ 5& 10& 9\end{array}\right]$

(iv) $\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$

Question:

If A=$\left[\begin{array}{ccc}\mathrm{2/3}& 1& \mathrm{5/3}\\ \mathrm{1/3}& \mathrm{2/3}& \mathrm{4/3}\\ \mathrm{7/3}& 2& \mathrm{2/3}\end{array}\right]$ and B=$\left[\begin{array}{ccc}\mathrm{2/5}& \mathrm{3/5}& 1\\ \mathrm{1/5}& \mathrm{2/5}& \mathrm{4/5}\\ \mathrm{7/5}& \mathrm{6/5}& \mathrm{2/5}\end{array}\right]$, then compute 3A−5B.

Answer:

3A−5B = $\left[\begin{array}{ccc}6& -10& 15\\ -2& -1& -10\\ 21& -12& 6\end{array}\right]$

Question:

Why did Vijay Singh ask the ghost disguised as Natwar to come closer?

Answer:

1. Vijay Singh asked the ghost disguised as Natwar to come closer because he wanted to be sure that it was really a ghost. He was suspicious that the figure he was seeing was a ghost and not Natwar himself.

2. Vijay Singh may have also asked the ghost to come closer in order to communicate with it and ask it questions. He may have wanted to find out more information about the ghost, or why it was there.

3. Additionally, Vijay Singh may have asked the ghost to come closer in order to protect himself and make sure that the ghost was not a danger to him. He may have wanted to make sure that the ghost was not a threat or a malicious entity.

Question:

A trust fund has Rs.30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Determine how to divide Rs.30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of (a) Rs.1800 (b) Rs.2000

Answer:

A) Step 1: Calculate the total amount of money that must be invested in each bond to obtain Rs.1800 in annual interest.

First Bond: Rs.30,000 x 5% = Rs.1500 Second Bond: Rs.1800 - Rs.1500 = Rs.300

Step 2: Divide the Rs.30,000 among the two bonds.

First Bond: Rs.30,000 - Rs.300 = Rs.29,700 Second Bond: Rs.300

B) Step 1: Calculate the total amount of money that must be invested in each bond to obtain Rs.2000 in annual interest.

First Bond: Rs.30,000 x 5% = Rs.1500 Second Bond: Rs.2000 - Rs.1500 = Rs.500

Step 2: Divide the Rs.30,000 among the two bonds.

First Bond: Rs.30,000 - Rs.500 = Rs.29,500 Second Bond: Rs.500

Question:

Given : 3$\left[\begin{array}{cc}\mathrm{x}& \mathrm{y}\\ \mathrm{z}& \mathrm{w}\end{array}\right]$=$\left[\begin{array}{cc}\mathrm{x}& 6\\ -1& \mathrm{2w}\end{array}\right]$+$\left[\begin{array}{cc}4& \mathrm{x+y}\\ \mathrm{z+w}& 3\end{array}\right]$ Find the values of x,y,z and w.

Answer:

Step 1: Substitute the given values in the equation.

$\left[\begin{array}{cc}\mathrm{x}& \mathrm{y}\\ \mathrm{z}& \mathrm{w}\end{array}\right]$=$\left[\begin{array}{cc}\mathrm{x}& 6\\ -1& \mathrm{2w}\end{array}\right]$+$\left[\begin{array}{cc}4& \mathrm{x+y}\\ \mathrm{z+w}& 3\end{array}\right]$

$\left[\begin{array}{cc}\mathrm{x}& \mathrm{y}\\ \mathrm{z}& \mathrm{w}\end{array}\right]$=$\left[\begin{array}{cc}\mathrm{x}& 6\\ -1& \mathrm{2w}\end{array}\right]$+$\left[\begin{array}{cc}4& \mathrm{x+y}\\ \mathrm{z+w}& 3\end{array}\right]$

Step 2: Solve the equation for x,y,z and w.

$\left[\begin{array}{cc}\mathrm{x}& \mathrm{y}\\ \mathrm{z}& \mathrm{w}\end{array}\right]$=$\left[\begin{array}{cc}\mathrm{x}& 6\\ -1& \mathrm{2w}\end{array}\right]$+$\left[\begin{array}{cc}4& \mathrm{x+y}\\ \mathrm{z+w}& 3\end{array}\right]$

x =

Question:

If A=$\left[\begin{array}{ccc}1& 2& -3\\ 5& 0& 2\\ 1& -1& 1\end{array}\right]$,B=$\left[\begin{array}{ccc}3& -1& 2\\ 4& 2& 5\\ 2& 0& 3\end{array}\right]$ and C=$\left[\begin{array}{ccc}4& 1& 2\\ 0& 3& 2\\ 1& -2& 3\end{array}\right]$, then compute (A+B) and (B−C). Also, verify that A+(B−C)=(A+B)−C

Answer:

(A+B) = $\left[\begin{array}{ccc}4& 1& -1\\ 9& 2& 7\\ 3& -1& 4\end{array}\right]$

(B-C) = $\left[\begin{array}{ccc}-1& -2& 0\\ 4& -1& 3\\ 1& 2& 0\end{array}\right]$

(A+(B-C)) = $\left[\begin{array}{ccc}3& -1& -1\\ 13& 1& 10\\ 4& 1& 4\end{array}\right]$

(A+B)-C = $\left[\begin{array}{ccc}0& -3& -3\\ 13& -1& 7\\ 2& 1& 4\end{array}\right]$

Since, (A+(B-C)) = (A+B)-C, we can verify that A+(B−C)=(A+B)−C.

Question:

If x$\left[\begin{array}{c}2\\ 3\end{array}\right]$+y$\left[\begin{array}{c}-1\\ 1\end{array}\right]$=$\left[\begin{array}{c}10\\ 5\end{array}\right]$ Find values of x and y.

Answer:

Step 1: To find the value of x, add the first row of each matrix. 2 + (−1) = 1

Step 2: So, x = 1.

Step 3: To find the value of y, add the second row of each matrix. 3 + 1 = 4

Step 4: So, y = 4.

Therefore, the values of x and y are x = 1 and y = 4.

Question:

Find X, if Y=$\left[\begin{array}{cc}3& 2\\ 1& 4\end{array}\right]$ and 2X+Y=$\left[\begin{array}{cc}1& 0\\ -3& 2\end{array}\right]$

Answer:

Step 1: Write the given equation in the form of matrix equation.

$\left[\begin{array}{ccc}2& 1& 3\\ 0& 2& 2\end{array}\right]\left[\begin{array}{c}X\\ Y\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -3& 2\end{array}\right]$

Step 2: Multiply the inverse of the coefficient matrix by the constant matrix on the right-hand side.

$\left[\begin{array}{ccc}2& 1& 3\\ 0& 2& 2\end{array}\right]\left[\begin{array}{cc}\frac{1/4}{-3/4}& \frac{1/4}{-1/4}\\ \frac{-3/4}{1/4}& \frac{3/4}{1/4}\end{array}\right]\left[\begin{array}{c}X\\ Y\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -3& 2\end{array}\right]$

Step 3: Solve for X and Y.

$\left[\begin{array}{c}X\\ Y\end{array}\right]=\left[\begin{array}{cc}1& -2\\ 2& 3\end{array}\right]$

Question:

Find X and Y , if 2$\left[\begin{array}{cc}1& 3\\ 0& \mathrm{x}\end{array}\right]$+$\left[\begin{array}{cc}\mathrm{y}& 0\\ 1& 2\end{array}\right]$=$\left[\begin{array}{cc}5& 6\\ 1& 8\end{array}\right]$

Answer:

Step 1: Add the two matrices.

$\left[\begin{array}{cc}1& 3\\ 0& \mathrm{x}\end{array}\right]$+$\left[\begin{array}{cc}\mathrm{y}& 0\\ 1& 2\end{array}\right]$ = $\left[\begin{array}{cc}\mathrm{1+y}& 3\\ 1& \mathrm{x+2}\end{array}\right]$

Step 2: Compare the two matrices.

$\left[\begin{array}{cc}\mathrm{1+y}& 3\\ 1& \mathrm{x+2}\end{array}\right]$ = $\left[\begin{array}{cc}5& 6\\ 1& 8\end{array}\right]$

Step 3: Solve for X and Y.

1 + y = 5 x + 2 = 8

y = 4 x = 6

Question:

Show that (i) $\left[\begin{array}{cc}5& 6\\ -1& 7\end{array}\right]$$\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$=$\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$$\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]$ (ii) $\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 0& 1& 0\end{array}\right]$$\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]$​=$\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]$$\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 0& 1& 0\end{array}\right]$

Answer:

(i)

Step 1: Multiply the first matrix by the second matrix.

$\left[\begin{array}{cc}5& 6\\ -1& 7\end{array}\right]$$\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$

= $\left[\begin{array}{cc}14& 19\\ 2& 13\end{array}\right]$

Step 2: Swap the rows of the multiplied matrix.

$\left[\begin{array}{cc}14& 19\\ 2& 13\end{array}\right]$

= $\left[\begin{array}{cc}2& 13\\ 14& 19\end{array}\right]$

Step 3: Swap the columns of the multiplied matrix.

$\left[\begin{array}{cc}2& 13\\ 14& 19\end{array}\right]$

= $\left[\begin{array}{cc}2& 14\\ 13& 19\end{array}\right]$

Step 4: Compare the result with the second matrix.

$\left[\begin{array}{cc}2& 14\\ 13& 19\end{array}\right]$

=

Question:

Simplify: cosθ$\left[\begin{array}{cc}\mathrm{cos\theta }& \mathrm{sin\theta }\\ \mathrm{-sin\theta }& \mathrm{cos\theta }\end{array}\right]$+sinθ$\left[\begin{array}{cc}\mathrm{sin\theta }& \mathrm{-cos\theta }\\ \mathrm{cos\theta }& \mathrm{sin\theta }\end{array}\right]$

Answer:

1. Expand the matrices: cosθ$\left[\begin{array}{cc}\mathrm{cos\theta }& \mathrm{sin\theta }\\ \mathrm{-sin\theta }& \mathrm{cos\theta }\end{array}\right]$+sinθ$\left[\begin{array}{cc}\mathrm{sin\theta }& \mathrm{-cos\theta }\\ \mathrm{cos\theta }& \mathrm{sin\theta }\end{array}\right]$ = $\left[\begin{array}{cc}\mathrm{cos\theta }+\mathrm{sin\theta }& \mathrm{sin\theta }-\mathrm{cos\theta }\\ \mathrm{-sin\theta }+\mathrm{cos\theta }& \mathrm{cos\theta }+\mathrm{sin\theta }\end{array}\right]$

2. Simplify the terms inside the matrix: $\left[\begin{array}{cc}2\mathrm{cos\theta }& 0\\ 0& 2\mathrm{sin\theta }\end{array}\right]$

Question:

If A=$\left[\begin{array}{cc}0& \mathrm{-tan\alpha /2}\\ \mathrm{tan\alpha /2}& 0\end{array}\right]$ and I is the identity matrix of order 2, show that I+A=(I−A)$\left[\begin{array}{cc}\mathrm{cos\alpha }& \mathrm{-sin\alpha }\\ \mathrm{sin\alpha }& \mathrm{cos\alpha }\end{array}\right]$

Answer:

1. Let A = $\left[\begin{array}{cc}0& \mathrm{-tan\alpha /2}\\ \mathrm{tan\alpha /2}& 0\end{array}\right]$ and I be the identity matrix of order 2.

2. Then, I + A = $\left[\begin{array}{cc}1& \mathrm{-tan\alpha /2}\\ \mathrm{tan\alpha /2}& 1\end{array}\right]$

3. Similarly, I - A = $\left[\begin{array}{cc}1& \mathrm{tan\alpha /2}\\ \mathrm{-tan\alpha /2}& 1\end{array}\right]$

4. Then, (I + A) * (I - A) = $\left[\begin{array}{cc}1& \mathrm{-tan\alpha /2}\\ \mathrm{tan\alpha /2}& 1\end{array}\right]\left[\begin{array}{cc}1& \mathrm{tan\alpha /2}\\ \mathrm{-tan\alpha /2}& 1\end{array}\right]$

5. Simplifying, we get $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

6. Hence, I + A = (I − A)$\left[\begin{array}{cc}\mathrm{cos\alpha }& \mathrm{-sin\alpha }\\ \mathrm{sin\alpha }& \mathrm{cos\alpha }\end{array}\right]$

Question:

The restriction on n,k and p so that PY+WY will be defined are: A k=3,p=n B k is arbitary, p=2 C p is arbitary, k=3 D k=2,p=3

Answer:

A. k=3, p=n B. k is arbitrary, p=2 C. p is arbitrary, k=3 D. k=2, p=3

Question:

Find $\begin{array}{}{A}^2\end{array}$−5A+6I, if A=$\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]$

Answer:

Answer: $\begin{array}{ccc}4& 0& 2\\ 4& 2& 6\\ 2& -2& 0\end{array}$

Question:

If A=$\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$ and I=$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, find k so that $\begin{array}{}{A}^2\end{array}$=kA−2I

Answer:

Step 1: Find A2 A2 = $\left[\begin{array}{cc}7& -4\\ 16& -4\end{array}\right]$

Step 2: Find kA kA = $\left[\begin{array}{cc}k3& k-2\\ k4& k-2\end{array}\right]$

Step 3: Find -2I -2I = $\left[\begin{array}{cc}-2& 0\\ 0& -2\end{array}\right]$

Step 4: Find k so that A2 = kA - 2I k = $2$

Question:

If n=p, then the order of the matrix 7X−5Z is: A p×2 B 2×n C n×3 D p×n

Answer:

A) The order of the matrix 7X−5Z is p×n.