Chapter 2 - Matrices EXERCISE 2.2 [PAGES 46 - 47]

Exercise 2.2 | Q 1.1 | Page 46

QUESTION

If \( A = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} \) and \( C = \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \), Show that \( A + B = B + A \)

SOLUTION

L.H.S: \( A + B = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} \)

\[ = \begin{bmatrix} 2+(-1) & -3+2 \\ 5+2 & -4+2 \\ -6+0 & 1+3 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} \quad ....(i) \]

R.H.S: \( B + A = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} \)

\[ = \begin{bmatrix} -1+2 & 2+(-3) \\ 2+5 & 2+(-4) \\ 0+(-6) & 3+1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} \quad ....(ii) \]

From (i) and (ii), we get
\( A + B = B + A \).

Exercise 2.2 | Q 1.2 | Page 46

QUESTION

SOLUTION

L.H.S: \( (A + B) + C \)

\[ = \left( \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} \right) + \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 2-1 & -3+2 \\ 5+2 & -4+2 \\ -6+0 & 1+3 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 1+4 & -1+3 \\ 7-1 & -2+4 \\ -6-2 & 4+1 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 6 & 2 \\ -8 & 5 \end{bmatrix} \quad ....(i) \]

R.H.S: \( A + (B + C) \)

\[ = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \left( \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \right) \] \[ = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} -1+4 & 2+3 \\ 2-1 & 2+4 \\ 0-2 & 3+1 \end{bmatrix} \] \[ = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 5 \\ 1 & 6 \\ -2 & 4 \end{bmatrix} \] \[ = \begin{bmatrix} 2+3 & -3+5 \\ 5+1 & -4+6 \\ -6-2 & 1+4 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 6 & 2 \\ -8 & 5 \end{bmatrix} \quad ....(ii) \]

From (i) and (ii), we get
\( (A + B) + C = A + (B + C) \).

Exercise 2.2 | Q 2 | Page 46

QUESTION

If \( A = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & -3 \\ 4 & -7 \end{bmatrix} \), then find the matrix \( A - 2B + 6I \), where I is the unit matrix of order 2.

SOLUTION

\( A - 2B + 6I \)

\[ = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix} - 2 \begin{bmatrix} 1 & -3 \\ 4 & -7 \end{bmatrix} + 6 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix} - \begin{bmatrix} 2 & -6 \\ 8 & -14 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} \] \[ = \begin{bmatrix} 1-2+6 & -2-(-6)+0 \\ 5-8+0 & 3-(-14)+6 \end{bmatrix} \] \[ = \begin{bmatrix} 1-2+6 & -2+6+0 \\ 5-8+0 & 3+14+6 \end{bmatrix} \] \[ = \begin{bmatrix} 5 & 4 \\ -3 & 23 \end{bmatrix} \]

Exercise 2.2 | Q 3 | Page 46

QUESTION

If \( A = \begin{bmatrix} 1 & 2 & -3 \\ -3 & 7 & -8 \\ 0 & -6 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 9 & -1 & 2 \\ -4 & 2 & 5 \\ 4 & 0 & -3 \end{bmatrix} \) then find the matrix C such that \( A + B + C \) is a zero matrix.

SOLUTION

\( A + B + C \) is a zero matrix.
∴ \( A + B + C = 0 \)
∴ \( C = -(A + B) \)

\[ = - \left( \begin{bmatrix} 1 & 2 & -3 \\ -3 & 7 & -8 \\ 0 & -6 & 1 \end{bmatrix} + \begin{bmatrix} 9 & -1 & 2 \\ -4 & 2 & 5 \\ 4 & 0 & -3 \end{bmatrix} \right) \] \[ = - \begin{bmatrix} 1+9 & 2-1 & -3+2 \\ -3-4 & 7+2 & -8+5 \\ 0+4 & -6+0 & 1-3 \end{bmatrix} \] \[ = - \begin{bmatrix} 10 & 1 & -1 \\ -7 & 9 & -3 \\ 4 & -6 & -2 \end{bmatrix} \] \[ = \begin{bmatrix} -10 & -1 & 1 \\ 7 & -9 & 3 \\ -4 & 6 & 2 \end{bmatrix} \]

Exercise 2.2 | Q 4 | Page 46

QUESTION

If \( A = \begin{bmatrix} 1 & -2 \\ 3 & -5 \\ -6 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} -1 & -2 \\ 4 & 2 \\ 1 & 5 \end{bmatrix} \) and \( C = \begin{bmatrix} 2 & 4 \\ -1 & -4 \\ -3 & 6 \end{bmatrix} \), find the matrix X such that \( 3A - 4B + 5X = C \).

SOLUTION

\( 3A - 4B + 5X = C \)
∴ \( 5X = C + 4B - 3A \)

\[ = \begin{bmatrix} 2 & 4 \\ -1 & -4 \\ -3 & 6 \end{bmatrix} + 4 \begin{bmatrix} -1 & -2 \\ 4 & 2 \\ 1 & 5 \end{bmatrix} - 3 \begin{bmatrix} 1 & -2 \\ 3 & -5 \\ -6 & 0 \end{bmatrix} \] \[ = \begin{bmatrix} 2 & 4 \\ -1 & -4 \\ -3 & 6 \end{bmatrix} + \begin{bmatrix} -4 & -8 \\ 16 & 8 \\ 4 & 20 \end{bmatrix} - \begin{bmatrix} 3 & -6 \\ 9 & -15 \\ -18 & 0 \end{bmatrix} \] \[ = \begin{bmatrix} 2-4-3 & 4-8-(-6) \\ -1+16-9 & -4+8-(-15) \\ -3+4-(-18) & 6+20-0 \end{bmatrix} \] \[ = \begin{bmatrix} 2-4-3 & 4-8+6 \\ -1+16-9 & -4+8+15 \\ -3+4+18 & 6+20-0 \end{bmatrix} \] \[ 5X = \begin{bmatrix} -5 & 2 \\ 6 & 19 \\ 19 & 26 \end{bmatrix} \] \[ X = \frac{1}{5} \begin{bmatrix} -5 & 2 \\ 6 & 19 \\ 19 & 26 \end{bmatrix} \] \[ X = \begin{bmatrix} -1 & \frac{2}{5} \\ \frac{6}{5} & \frac{19}{5} \\ \frac{19}{5} & \frac{26}{5} \end{bmatrix} \]

Exercise 2.2 | Q 5 | Page 46

QUESTION

If \( A = \begin{bmatrix} 5 & 1 & -4 \\ 3 & 2 & 0 \end{bmatrix} \), find \( (A^T)^T \).

SOLUTION

\( A = \begin{bmatrix} 5 & 1 & -4 \\ 3 & 2 & 0 \end{bmatrix} \)

\[ A^T = \begin{bmatrix} 5 & 3 \\ 1 & 2 \\ -4 & 0 \end{bmatrix} \] \[ (A^T)^T = \begin{bmatrix} 5 & 1 & -4 \\ 3 & 2 & 0 \end{bmatrix} = A \]

Exercise 2.2 | Q 6 | Page 46

QUESTION

If \( A = \begin{bmatrix} 7 & 3 & 1 \\ -2 & -4 & 1 \\ 5 & 9 & 1 \end{bmatrix} \), find \( (A^T)^T \).

SOLUTION

\( A = \begin{bmatrix} 7 & 3 & 1 \\ -2 & -4 & 1 \\ 5 & 9 & 1 \end{bmatrix} \)

\[ A^T = \begin{bmatrix} 7 & -2 & 5 \\ 3 & -4 & 9 \\ 1 & 1 & 1 \end{bmatrix} \] \[ (A^T)^T = \begin{bmatrix} 7 & 3 & 1 \\ -2 & -4 & 1 \\ 5 & 9 & 1 \end{bmatrix} = A \]

Exercise 2.2 | Q 7 | Page 47

QUESTION

Find a, b, c, if \( \begin{bmatrix} 1 & \frac{3}{5} & a \\ b & -5 & -7 \\ -4 & c & 0 \end{bmatrix} \) is a symmetric matrix.

SOLUTION

Let \( A = \begin{bmatrix} 1 & \frac{3}{5} & a \\ b & -5 & -7 \\ -4 & c & 0 \end{bmatrix} \)

Then \( A^T = \begin{bmatrix} 1 & b & -4 \\ \frac{3}{5} & -5 & c \\ a & -7 & 0 \end{bmatrix} \)

Since A is a symmetric matrix, \( A = A^T \).

\[ \begin{bmatrix} 1 & \frac{3}{5} & a \\ b & -5 & -7 \\ -4 & c & 0 \end{bmatrix} = \begin{bmatrix} 1 & b & -4 \\ \frac{3}{5} & -5 & c \\ a & -7 & 0 \end{bmatrix} \]

By equality of matrices, we get:

\( a = -4 \)
\( b = \frac{3}{5} \)
\( c = -7 \)

Exercise 2.2 | Q 8 | Page 47

QUESTION

Find x, y, z if \( \begin{bmatrix} 0 & -5i & x \\ y & 0 & z \\ \frac{3}{2} & -\sqrt{2} & 0 \end{bmatrix} \) is a skew symmetric matrix.

SOLUTION

Let \( A = \begin{bmatrix} 0 & -5i & x \\ y & 0 & z \\ \frac{3}{2} & -\sqrt{2} & 0 \end{bmatrix} \)

Then \( A^T = \begin{bmatrix} 0 & y & \frac{3}{2} \\ -5i & 0 & -\sqrt{2} \\ x & z & 0 \end{bmatrix} \)

Since A is a skew-symmetric matrix, \( A^T = -A \).
So, \( A = -A^T \).

\[ \begin{bmatrix} 0 & -5i & x \\ y & 0 & z \\ \frac{3}{2} & -\sqrt{2} & 0 \end{bmatrix} = - \begin{bmatrix} 0 & y & \frac{3}{2} \\ -5i & 0 & -\sqrt{2} \\ x & z & 0 \end{bmatrix} \] \[ \begin{bmatrix} 0 & -5i & x \\ y & 0 & z \\ \frac{3}{2} & -\sqrt{2} & 0 \end{bmatrix} = \begin{bmatrix} 0 & -y & -\frac{3}{2} \\ 5i & 0 & \sqrt{2} \\ -x & -z & 0 \end{bmatrix} \]

By equality of matrices, we get:

\( -5i = -y \Rightarrow y = 5i \)
\( x = -\frac{3}{2} \)
\( z = \sqrt{2} \)

Thus, \( x = -\frac{3}{2} \), \( y = 5i \), \( z = \sqrt{2} \).

Exercise 2.2 | Q 9.1 | Page 47

QUESTION

For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither.

\[ \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \]

SOLUTION

Let \( A = \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \)

\[ A^T = \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \]

Since \( A^T = A \), the matrix A is symmetric.

Exercise 2.2 | Q 9.2 | Page 47

QUESTION

For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither.

\[ \begin{bmatrix} 2 & 5 & 1 \\ -5 & 4 & 6 \\ -1 & -6 & 3 \end{bmatrix} \]

SOLUTION

Let \( A = \begin{bmatrix} 2 & 5 & 1 \\ -5 & 4 & 6 \\ -1 & -6 & 3 \end{bmatrix} \)

\[ A^T = \begin{bmatrix} 2 & -5 & -1 \\ 5 & 4 & -6 \\ 1 & 6 & 3 \end{bmatrix} \]

Comparing A and A^T:
\( A \neq A^T \)

Now consider -A^T:

\[ -A^T = \begin{bmatrix} -2 & 5 & 1 \\ -5 & -4 & 6 \\ -1 & -6 & -3 \end{bmatrix} \]

Comparing A and -A^T:
\( A \neq -A^T \)

Since \( A \neq A^T \) and \( A \neq -A^T \), the matrix A is neither symmetric nor skew-symmetric.

Exercise 2.2 | Q 9.3 | Page 47

QUESTION

For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither.

\[ \begin{bmatrix} 0 & 1+2i & i-2 \\ -1-2i & 0 & -7 \\ 2-i & 7 & 0 \end{bmatrix} \]

SOLUTION

Let \( A = \begin{bmatrix} 0 & 1+2i & i-2 \\ -1-2i & 0 & -7 \\ 2-i & 7 & 0 \end{bmatrix} \)

\[ A^T = \begin{bmatrix} 0 & -1-2i & 2-i \\ 1+2i & 0 & 7 \\ i-2 & -7 & 0 \end{bmatrix} \]

Now consider -A:

\[ -A = - \begin{bmatrix} 0 & 1+2i & i-2 \\ -1-2i & 0 & -7 \\ 2-i & 7 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -(1+2i) & -(i-2) \\ -(-1-2i) & 0 & -(-7) \\ -(2-i) & -7 & 0 \end{bmatrix} \] \[ -A = \begin{bmatrix} 0 & -1-2i & 2-i \\ 1+2i & 0 & 7 \\ -2+i & -7 & 0 \end{bmatrix} \]

We see that \( A^T = -A \).
(Note: The element `i-2` in A^T matches `-(-i+2) = 2-i` if the comparison is made with -A. Let's re-check: \(A^T_{13} = 2-i\). \((-A)_{13} = -(i-2) = 2-i\). Correct. \(A^T_{31} = i-2\). \((-A)_{31} = -(2-i) = i-2\). Correct.)

Since \( A^T = -A \), the matrix A is skew-symmetric.

Exercise 2.2 | Q 10 | Page 47

QUESTION

Construct the matrix \( A = [a_{ij}]_{3 \times 3} \) where \( a_{ij} = i - j \). State whether A is symmetric or skew-symmetric.

SOLUTION

\( A = [a_{ij}]_{3 \times 3} \)

\[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \]

Given, \( a_{ij} = i - j \)
∴ \( a_{11} = 1 - 1 = 0, a_{12} = 1 - 2 = -1, a_{13} = 1 - 3 = -2 \)
\( a_{21} = 2 - 1 = 1, a_{22} = 2 - 2 = 0, a_{23} = 2 - 3 = -1 \)
\( a_{31} = 3 - 1 = 2, a_{32} = 3 - 2 = 1, a_{33} = 3 - 3 = 0 \)

\[ A = \begin{bmatrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} \] \[ A^T = \begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{bmatrix} \] \[ -A = - \begin{bmatrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{bmatrix} \]

Since \( A^T = -A \), the matrix A is skew-symmetric.

Exercise 2.2 | Q 11 | Page 47

QUESTION

Solve the following equations for X and Y, if \( 3X - Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \( X - 3Y = \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \).

SOLUTION

Given equations are:
\( 3X - Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \quad ....(i) \)
\( X - 3Y = \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \quad ....(ii) \)

Multiply equation (i) by 3:
\( 9X - 3Y = 3 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \quad ....(iii) \)

Subtract equation (ii) from equation (iii):
\( (9X - 3Y) - (X - 3Y) = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \)
\( 8X = \begin{bmatrix} 3-0 & -3-(-1) \\ -3-0 & 3-(-1) \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -3 & 4 \end{bmatrix} \)

\[ X = \frac{1}{8} \begin{bmatrix} 3 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{8} & \frac{-2}{8} \\ \frac{-3}{8} & \frac{4}{8} \end{bmatrix} = \begin{bmatrix} \frac{3}{8} & -\frac{1}{4} \\ -\frac{3}{8} & \frac{1}{2} \end{bmatrix} \]

Multiply equation (ii) by 3:
\( 3X - 9Y = 3 \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & -3 \\ 0 & -3 \end{bmatrix} \quad ....(iv) \)

Subtract equation (iv) from equation (i):
\( (3X - Y) - (3X - 9Y) = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -3 \\ 0 & -3 \end{bmatrix} \)
\( 8Y = \begin{bmatrix} 1-0 & -1-(-3) \\ -1-0 & 1-(-3) \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} \)

\[ Y = \frac{1}{8} \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} \frac{1}{8} & \frac{2}{8} \\ \frac{-1}{8} & \frac{4}{8} \end{bmatrix} = \begin{bmatrix} \frac{1}{8} & \frac{1}{4} \\ -\frac{1}{8} & \frac{1}{2} \end{bmatrix} \]

Exercise 2.2 | Q 12 | Page 47

QUESTION

Find matrices A and B, if \( 2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \) and \( A - 2B = \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \).

SOLUTION

Given equations are:
\( 2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \quad ....(i) \)
\( A - 2B = \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \quad ....(ii) \)

Multiply equation (ii) by 2:
\( 2A - 4B = 2 \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} = \begin{bmatrix} 6 & 4 & 16 \\ -4 & 2 & -14 \end{bmatrix} \quad ....(iii) \)

Subtract equation (iii) from equation (i):
\( (2A - B) - (2A - 4B) = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} - \begin{bmatrix} 6 & 4 & 16 \\ -4 & 2 & -14 \end{bmatrix} \)
\( 3B = \begin{bmatrix} 6-6 & -6-4 & 0-16 \\ -4-(-4) & 2-2 & 1-(-14) \end{bmatrix} = \begin{bmatrix} 0 & -10 & -16 \\ 0 & 0 & 15 \end{bmatrix} \)

\[ B = \frac{1}{3} \begin{bmatrix} 0 & -10 & -16 \\ 0 & 0 & 15 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{10}{3} & -\frac{16}{3} \\ 0 & 0 & 5 \end{bmatrix} \]

Multiply equation (i) by 2:
\( 4A - 2B = 2 \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 12 & -12 & 0 \\ -8 & 4 & 2 \end{bmatrix} \quad ....(iv) \)

Subtract equation (ii) from equation (iv):
\( (4A - 2B) - (A - 2B) = \begin{bmatrix} 12 & -12 & 0 \\ -8 & 4 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \)
\( 3A = \begin{bmatrix} 12-3 & -12-2 & 0-8 \\ -8-(-2) & 4-1 & 2-(-7) \end{bmatrix} = \begin{bmatrix} 9 & -14 & -8 \\ -6 & 3 & 9 \end{bmatrix} \)

\[ A = \frac{1}{3} \begin{bmatrix} 9 & -14 & -8 \\ -6 & 3 & 9 \end{bmatrix} = \begin{bmatrix} 3 & -\frac{14}{3} & -\frac{8}{3} \\ -2 & 1 & 3 \end{bmatrix} \]

Exercise 2.2 | Q 13 | Page 47

QUESTION

Find x and y, if \( \begin{bmatrix} 2x+y & -1 & 1 \\ 3 & 4y & 4 \end{bmatrix} + \begin{bmatrix} -1 & 6 & 4 \\ 3 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 5 \\ 6 & 18 & 7 \end{bmatrix} \).

SOLUTION

\[ \begin{bmatrix} 2x+y & -1 & 1 \\ 3 & 4y & 4 \end{bmatrix} + \begin{bmatrix} -1 & 6 & 4 \\ 3 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 5 \\ 6 & 18 & 7 \end{bmatrix} \] \[ \begin{bmatrix} (2x+y)+(-1) & -1+6 & 1+4 \\ 3+3 & 4y+0 & 4+3 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 5 \\ 6 & 18 & 7 \end{bmatrix} \] \[ \begin{bmatrix} 2x+y-1 & 5 & 5 \\ 6 & 4y & 7 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 5 \\ 6 & 18 & 7 \end{bmatrix} \]

By equality of matrices, we get:
\( 2x + y - 1 = 3 \Rightarrow 2x + y = 4 \quad ....(i) \)
\( 4y = 18 \Rightarrow y = \frac{18}{4} = \frac{9}{2} \)

Substitute \( y = \frac{9}{2} \) into equation (i):
\( 2x + \frac{9}{2} = 4 \)
\( 2x = 4 - \frac{9}{2} = \frac{8-9}{2} = -\frac{1}{2} \)
\( x = -\frac{1}{4} \)

∴ \( x = -\frac{1}{4} \) and \( y = \frac{9}{2} \).

Exercise 2.2 | Q 14 | Page 47

QUESTION

If \( \begin{bmatrix} 2a+b & 3a-b \\ c+2d & 2c-d \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \), find a, b, c and d.

SOLUTION

\[ \begin{bmatrix} 2a+b & 3a-b \\ c+2d & 2c-d \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \]

By equality of matrices, we get:
\( 2a + b = 2 \quad ....(i) \)
\( 3a - b = 3 \quad ....(ii) \)
\( c + 2d = 4 \quad ....(iii) \)
\( 2c - d = -1 \quad ....(iv) \)

Adding (i) and (ii):
\( (2a+b) + (3a-b) = 2+3 \)
\( 5a = 5 \Rightarrow a = 1 \)

Substitute \( a=1 \) into (i):
\( 2(1) + b = 2 \Rightarrow 2 + b = 2 \Rightarrow b = 0 \)

Multiply (iv) by 2:
\( 4c - 2d = -2 \quad ....(v) \)

Add (iii) and (v):
\( (c+2d) + (4c-2d) = 4 + (-2) \)
\( 5c = 2 \Rightarrow c = \frac{2}{5} \)

Substitute \( c=\frac{2}{5} \) into (iii):
\( \frac{2}{5} + 2d = 4 \)
\( 2d = 4 - \frac{2}{5} = \frac{20-2}{5} = \frac{18}{5} \)
\( d = \frac{18}{10} = \frac{9}{5} \)

∴ \( a=1, b=0, c=\frac{2}{5}, d=\frac{9}{5} \).

Exercise 2.2 | Q 15.1 | Page 47

QUESTION

There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects - Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B.
July sales (in Rupees): Physics Chemistry Mathematics
\( A = \begin{bmatrix} 5600 & 6750 & 8500 \\ 6650 & 7055 & 8905 \end{bmatrix} \begin{matrix} \text{Suresh} \\ \text{Ganesh} \end{matrix} \)
August Sales (in Rupees):
\( B = \begin{bmatrix} 6650 & 7055 & 8905 \\ 7000 & 7500 & 10200 \end{bmatrix} \begin{matrix} \text{Suresh} \\ \text{Ganesh} \end{matrix} \)
Find the increase in sales in Rupees from July to August 2017.

SOLUTION

Increase in sales = August Sales (B) - July Sales (A)

\[ B - A = \begin{bmatrix} 6650 & 7055 & 8905 \\ 7000 & 7500 & 10200 \end{bmatrix} - \begin{bmatrix} 5600 & 6750 & 8500 \\ 6650 & 7055 & 8905 \end{bmatrix} \] \[ = \begin{bmatrix} 6650-5600 & 7055-6750 & 8905-8500 \\ 7000-6650 & 7500-7055 & 10200-8905 \end{bmatrix} \] \[ = \begin{bmatrix} 1050 & 305 & 405 \\ 350 & 445 & 1295 \end{bmatrix} \begin{matrix} \text{Suresh} \\ \text{Ganesh} \end{matrix} \]

Increase in sales from July to August 2017:
For Suresh:
Physics: ₹1050
Chemistry: ₹305
Mathematics: ₹405
For Ganesh:
Physics: ₹350
Chemistry: ₹445
Mathematics: ₹1295

Exercise 2.2 | Q 15.2 | Page 47

QUESTION

There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects - Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B.
July sales (in Rupees): Physics Chemistry Mathematics
\( A = \begin{bmatrix} 5600 & 6750 & 8500 \\ 6650 & 7055 & 8905 \end{bmatrix} \begin{matrix} \text{Suresh} \\ \text{Ganesh} \end{matrix} \)
August Sales (in Rupees):
\( B = \begin{bmatrix} 6650 & 7055 & 8905 \\ 7000 & 7500 & 10200 \end{bmatrix} \begin{matrix} \text{Suresh} \\ \text{Ganesh} \end{matrix} \)
If both book shops get 10% profit in the month of August 2017, find the profit for each book seller in each subject in that month.

SOLUTION

Profit is 10% of August sales (Matrix B).
Profit Matrix = \( 0.10 \times B \)

\[ = 0.10 \times \begin{bmatrix} 6650 & 7055 & 8905 \\ 7000 & 7500 & 10200 \end{bmatrix} \] \[ = \begin{bmatrix} 0.10 \times 6650 & 0.10 \times 7055 & 0.10 \times 8905 \\ 0.10 \times 7000 & 0.10 \times 7500 & 0.10 \times 10200 \end{bmatrix} \] \[ = \begin{bmatrix} 665 & 705.50 & 890.50 \\ 700 & 750 & 1020 \end{bmatrix} \begin{matrix} \text{Suresh} \\ \text{Ganesh} \end{matrix} \]

Profit for each bookseller in August 2017:
For Suresh:
Physics: ₹665
Chemistry: ₹705.50
Mathematics: ₹890.50
For Ganesh:
Physics: ₹700
Chemistry: ₹750
Mathematics: ₹1020

HSC Mathematics Full Solution

Concept: Algebra of Matrices

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board

Chapter 2 Matrices

Exercise 2.2 [PAGES 46 - 47]