Maharashtra Board Class 12 Maths Part 1 Chapter 2 Matrices 2.2 Solutions

Question 1. 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
(i) \( A + B = B + A \)
(ii) \( (A + B) + C = A + (B + C) \)
Answer:
(i) To show that \( A + B = B + A \):
Let's find the Left Hand Side (LHS):
\( A + B = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} \)
\( \implies A + B = \begin{bmatrix} 2 + (-1) & -3 + 2 \\ 5 + 2 & -4 + 2 \\ -6 + 0 & 1 + 3 \end{bmatrix} \)
\( \implies A + B = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} \) ——— (1)

Now, let's find the Right Hand Side (RHS):
\( B + A = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} \)
\( \implies B + A = \begin{bmatrix} -1 + 2 & 2 + (-3) \\ 2 + 5 & 2 + (-4) \\ 0 + (-6) & 3 + 1 \end{bmatrix} \)
\( \implies B + A = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} \) ——— (2)
From (1) and (2), we get:
\( A + B = B + A \)
Hence, the commutative property of matrix addition is verified.

(ii) To show that \( (A + B) + C = A + (B + C) \):
Let's find the Left Hand Side (LHS):
\( (A + B) + C = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \)
\( \implies (A + B) + C = \begin{bmatrix} 1 + 4 & -1 + 3 \\ 7 + (-1) & -2 + 4 \\ -6 + (-2) & 4 + 1 \end{bmatrix} \)
\( \implies (A + B) + C = \begin{bmatrix} 5 & 2 \\ 6 & 2 \\ -8 & 5 \end{bmatrix} \) ——— (3)

Now, let's find the Right Hand Side (RHS):
First, calculate \( B + C \):
\( B + C = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \)
\( \implies B + C = \begin{bmatrix} -1 + 4 & 2 + 3 \\ 2 + (-1) & 2 + 4 \\ 0 + (-2) & 3 + 1 \end{bmatrix} \)
\( \implies B + C = \begin{bmatrix} 3 & 5 \\ 1 & 6 \\ -2 & 4 \end{bmatrix} \)
Now, calculate \( A + (B + C) \):
\( A + (B + C) = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 5 \\ 1 & 6 \\ -2 & 4 \end{bmatrix} \)
\( \implies A + (B + C) = \begin{bmatrix} 2 + 3 & -3 + 5 \\ 5 + 1 & -4 + 6 \\ -6 + (-2) & 1 + 4 \end{bmatrix} \)
\( \implies A + (B + C) = \begin{bmatrix} 5 & 2 \\ 6 & 2 \\ -8 & 5 \end{bmatrix} \) ——— (4)
From (3) and (4), we get:
\( (A + B) + C = A + (B + C) \)
Hence, the associative property of matrix addition is verified.
In simple words: Adding matrices is just like adding regular numbers. It does not matter which order you add them in, or how you group them together; you will always get the exact same final answer.

🎯 Exam Tip: Show every step of the addition clearly, especially when dealing with negative numbers, to avoid simple calculation errors that can cost you marks.

Question 1. If \( A = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} \), and \( C = \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \), verify that:
(i) \( A + B = B + A \)
(ii) \( (A + B) + C = A + (B + C) \)
Answer:
(i) To verify \( A + B = B + A \):
LHS \( = A + B \)
\( = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 2-1 & -3+2 \\ 5+2 & -4+2 \\ -6+0 & 1+3 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} \) --- (1)

RHS \( = B + A \)
\( = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} -1+2 & 2-3 \\ 2+5 & 2-4 \\ 0-6 & 3+1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} \) --- (2)
From (1) and (2), LHS = RHS.
Therefore, \( A + B = B + A \). This shows that matrix addition is commutative.

(ii) To verify \( (A + B) + C = A + (B + C) \):
LHS \( = (A + B) + C \)
From part (i), we have \( A + B = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} \).
\( \therefore (A + B) + C = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1+4 & -1+3 \\ 7-1 & -2+4 \\ -6-2 & 4+1 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & 2 \\ 6 & 2 \\ -8 & 5 \end{bmatrix} \) --- (1)

RHS \( = A + (B + C) \)
First, let us find \( B + C \):
\( B + C = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} -1+4 & 2+3 \\ 2-1 & 2+4 \\ 0-2 & 3+1 \end{bmatrix} \)
\( = \begin{bmatrix} 3 & 5 \\ 1 & 6 \\ -2 & 4 \end{bmatrix} \)
Now, \( A + (B + C) = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 5 \\ 1 & 6 \\ -2 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 2+3 & -3+5 \\ 5+1 & -4+6 \\ -6-2 & 1+4 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & 2 \\ 6 & 2 \\ -8 & 5 \end{bmatrix} \) --- (2)
From (1) and (2), LHS = RHS.
Therefore, \( (A + B) + C = A + (B + C) \). This confirms the associative property of matrix addition.
In simple words: Adding matrices is just like adding regular numbers. It does not matter which order you add them in, or how you group them together; you will always get the exact same final answer.

🎯 Exam Tip: When adding matrices, always add the corresponding elements in the exact same positions. Clearly label your equations as (1) and (2) to show the examiner that both sides of the proof are equal.

Question 2. 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.
Answer: Given matrices are:
\( A = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -3 \\ 4 & -7 \end{bmatrix} \)
Since \( I \) is the unit matrix of order 2, we have:
\( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Now, we calculate \( A - 2B + 6I \):
\( A - 2B + 6I = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix} - 2\begin{bmatrix} 1 & -3 \\ 4 & -7 \end{bmatrix} + 6\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix} - \begin{bmatrix} 2 & -6 \\ 8 & -14 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} \)
\( = \begin{bmatrix} 1 - 2 + 6 & -2 - (-6) + 0 \\ 5 - 8 + 0 & 3 - (-14) + 6 \end{bmatrix} \)
\( = \begin{bmatrix} 1 - 2 + 6 & -2 + 6 + 0 \\ 5 - 8 + 0 & 3 + 14 + 6 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & 4 \\ -3 & 23 \end{bmatrix} \)
This resulting matrix is obtained by performing corresponding element-wise operations across all three matrices.
In simple words: To solve this, we multiply every number in matrix B by 2 and every number in the identity matrix I by 6. Then, we simply add and subtract the matching numbers in the same positions to get our final matrix.

🎯 Exam Tip: Be very careful with negative signs when subtracting matrices, especially when subtracting a negative number like \( -2 - (-6) \) which simplifies to \( -2 + 6 = 4 \).

Question 3. 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.
Answer: Given that \( A + B + C = 0 \), where \( 0 \) represents the zero matrix of order \( 3 \times 3 \).
Rearranging the equation to solve for \( C \):
\( \implies C = -A - B \)
\( = -\begin{bmatrix} 1 & 2 & -3 \\ -3 & 7 & -8 \\ 0 & -6 & 1 \end{bmatrix} - \begin{bmatrix} 9 & -1 & 2 \\ -4 & 2 & 5 \\ 4 & 0 & -3 \end{bmatrix} \)
\( = \begin{bmatrix} -1 & -2 & 3 \\ 3 & -7 & 8 \\ 0 & 6 & -1 \end{bmatrix} - \begin{bmatrix} 9 & -1 & 2 \\ -4 & 2 & 5 \\ 4 & 0 & -3 \end{bmatrix} \)
\( = \begin{bmatrix} -1 - 9 & -2 - (-1) & 3 - 2 \\ 3 - (-4) & -7 - 2 & 8 - 5 \\ 0 - 4 & 6 - 0 & -1 - (-3) \end{bmatrix} \)
\( = \begin{bmatrix} -10 & -2 + 1 & 1 \\ 3 + 4 & -9 & 3 \\ -4 & 6 & -1 + 3 \end{bmatrix} \)
\( \implies C = \begin{bmatrix} -10 & -1 & 1 \\ 7 & -9 & 3 \\ -4 & 6 & 2 \end{bmatrix} \)
This calculation ensures that the sum of all three matrices results in a matrix containing only zero elements.
In simple words: We want to find a matrix C that cancels out matrices A and B so their sum is zero. We do this by changing the signs of all numbers in A and then subtracting B from it.

🎯 Exam Tip: Always double-check your final matrix by adding \( A + B + C \) together; if every element in the sum is 0, your answer is 100% correct.

Question 4. 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 \( 3A - 4B + 5X = C \).
Answer: Given the matrix equation:
\( 3A - 4B + 5X = C \)
\( \implies 5X = C - 3A + 4B \)
\( \implies 5X = \begin{bmatrix} 2 & 4 \\ -1 & -4 \\ -3 & 6 \end{bmatrix} - 3\begin{bmatrix} 1 & -2 \\ 3 & -5 \\ -6 & 0 \end{bmatrix} + 4\begin{bmatrix} -1 & -2 \\ 4 & 2 \\ 1 & 5 \end{bmatrix} \)
\( \implies 5X = \begin{bmatrix} 2 & 4 \\ -1 & -4 \\ -3 & 6 \end{bmatrix} - \begin{bmatrix} 3 & -6 \\ 9 & -15 \\ -18 & 0 \end{bmatrix} + \begin{bmatrix} -4 & -8 \\ 16 & 8 \\ 4 & 20 \end{bmatrix} \)
\( \implies 5X = \begin{bmatrix} 2 - 3 + (-4) & 4 - (-6) - 8 \\ -1 - 9 + 16 & -4 - (-15) + 8 \\ -3 - (-18) + 4 & 6 - 0 + 20 \end{bmatrix} \)
\( \implies 5X = \begin{bmatrix} -5 & 2 \\ 6 & 19 \\ 19 & 26 \end{bmatrix} \)
\( \implies X = \frac{1}{5} \begin{bmatrix} -5 & 2 \\ 6 & 19 \\ 19 & 26 \end{bmatrix} \)
\( \implies X = \begin{bmatrix} -1 & \frac{2}{5} \\ \frac{6}{5} & \frac{19}{5} \\ \frac{19}{5} & \frac{26}{5} \end{bmatrix} \)
This final matrix represents the unique solution that satisfies the given matrix equation.
In simple words: To find the unknown matrix \( X \), we rearrange the equation just like normal algebra to get \( 5X \) on one side. Then, we perform the matrix addition and subtraction, and finally divide every number in the resulting matrix by 5.

🎯 Exam Tip: Be extremely careful with signs when subtracting negative numbers, such as \( 4 - (-6) = 10 \). Double-check each element calculation individually to avoid simple arithmetic errors.

Question 5. If \( A = \begin{bmatrix} 5 & 1 & -4 \\ 3 & 2 & 0 \end{bmatrix} \), find \( (A^T)^T \).
Answer: Given matrix:
\( A = \begin{bmatrix} 5 & 1 & -4 \\ 3 & 2 & 0 \end{bmatrix} \)
The transpose of matrix \( A \), denoted as \( A^T \), is obtained by swapping its rows and columns.
\( \implies A^T = \begin{bmatrix} 5 & 3 \\ 1 & 2 \\ -4 & 0 \end{bmatrix} \)
Taking the transpose of \( A^T \) again:
\( \implies (A^T)^T = \begin{bmatrix} 5 & 1 & -4 \\ 3 & 2 & 0 \end{bmatrix} = A \)
This demonstrates the fundamental property that transposing a matrix twice returns it to its original form.
In simple words: Transposing a matrix means turning its rows into columns. If you do this twice, you swap the rows and columns back to their original positions, ending up with the exact same matrix you started with.

🎯 Exam Tip: Remember the property \( (A^T)^T = A \). In exams, you can write down this property directly to verify your step-by-step transpose calculations.

Question 6. If \( A = \begin{bmatrix} 7 & 3 & 1 \\ -2 & -4 & 1 \\ 5 & 9 & 1 \end{bmatrix} \), find \( (A^T)^T \).
Answer: Given matrix is \( A = \begin{bmatrix} 7 & 3 & 1 \\ -2 & -4 & 1 \\ 5 & 9 & 1 \end{bmatrix} \). Taking the transpose of matrix \( A \) by swapping its rows and columns, we obtain the first transpose matrix.
\( \implies A^T = \begin{bmatrix} 7 & -2 & 5 \\ 3 & -4 & 9 \\ 1 & 1 & 1 \end{bmatrix} \)
Taking the transpose of \( A^T \) again, we swap its rows and columns back to their original positions.
\( \implies (A^T)^T = \begin{bmatrix} 7 & 3 & 1 \\ -2 & -4 & 1 \\ 5 & 9 & 1 \end{bmatrix} = A \)
In simple words: Taking the transpose of a matrix means swapping its rows and columns. If you do this twice, you end up right back where you started with the original matrix.

🎯 Exam Tip: Remember the standard property \( (A^T)^T = A \). You can use this property to quickly verify your final answer without re-calculating.

Question 7. Find a, b, c if \( \begin{bmatrix} 1 & \frac{3}{5} & a \\ b & -5 & -7 \\ -4 & c & 0 \end{bmatrix} \) is a symmetric matrix.
Answer: Let \( A = \begin{bmatrix} 1 & \frac{3}{5} & a \\ b & -5 & -7 \\ -4 & c & 0 \end{bmatrix} \). Since \( A \) is a symmetric matrix, we know that \( A = A^T \), which means the matrix is equal to its own transpose.
\( \implies \begin{bmatrix} 1 & \frac{3}{5} & a \\ b & -5 & -7 \\ -4 & c & 0 \end{bmatrix} = \begin{bmatrix} 1 & b & -4 \\ \frac{3}{5} & -5 & c \\ a & -7 & 0 \end{bmatrix} \)
By comparing the corresponding elements of these two equal matrices, we get:
\( \implies b = \frac{3}{5} \)
\( \implies a = -4 \)
\( \implies c = -7 \)
Therefore, the required values are \( a = -4 \), \( b = \frac{3}{5} \), and \( c = -7 \).
In simple words: A symmetric matrix is like a mirror across its main diagonal. This means the numbers opposite each other across the diagonal must be exactly equal, allowing us to easily find the missing values.

🎯 Exam Tip: For a symmetric matrix, always use the relation \( a_{ij} = a_{ji} \) to directly equate corresponding elements across the main diagonal and save time.

Since, A is a symmetric matrix, \( a_{ij} = a_{ji} \) for all i and j
\( \therefore a_{13} = a_{31}, a_{12} = a_{21} \) and \( a_{23} = a_{32} \)
\( \therefore a = -4, \frac{3}{5} = b \) and \( -7 = c \)
\( \therefore a = -4, b = \frac{3}{5} \) and \( c = -7 \).

Alternative Method:
Let \( A = \begin{bmatrix} 1 & \frac{3}{5} & a \\ b & -5 & -7 \\ -4 & c & 0 \end{bmatrix} \)
Then \( A^T = \begin{bmatrix} 1 & b & -4 \\ \frac{3}{5} & -5 & -7 \\ a & -7 & 0 \end{bmatrix} \)
Since, A is symmetric matrix, \( A = A^T \)

\( \implies \begin{bmatrix} 1 & \frac{3}{5} & a \\ b & -5 & -7 \\ -4 & c & 0 \end{bmatrix} = \begin{bmatrix} 1 & b & -4 \\ \frac{3}{5} & -5 & -7 \\ a & -7 & 0 \end{bmatrix} \)
By equality of matrices,
\( a = -4, b = \frac{3}{5} \) and \( c = -7 \).

Question 8. 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.
Answer:
Let \( A = \begin{bmatrix} 0 & -5i & x \\ y & 0 & z \\ \frac{3}{2} & -\sqrt{2} & 0 \end{bmatrix} \)
Since, A is a skew-symmetric matrix, \( A^T = -A \).
Therefore, \( a_{ji} = -a_{ij} \) for all \( i \) and \( j \).
Comparing the elements, we get:
\( y = -(-5i) \)

\( \implies y = 5i \)
\( \frac{3}{2} = -x \)

\( \implies x = -\frac{3}{2} \)
\( -\sqrt{2} = -z \)

\( \implies z = \sqrt{2} \)
Thus, the values are \( x = -\frac{3}{2} \), \( y = 5i \), and \( z = \sqrt{2} \). We can verify this by checking that the diagonal elements are indeed all zero.
In simple words: In a skew-symmetric matrix, the elements across the main diagonal are negatives of each other. By matching these opposite positions, we can easily find the missing values of x, y, and z.

🎯 Exam Tip: Remember that for any skew-symmetric matrix, the diagonal elements are always zero, and \( a_{ji} = -a_{ij} \). Double-check your signs when equating these elements to avoid simple calculation errors.

Question 9. For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither:
(i) \( \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 0 & 1+2i & i-2 \\ -1-2i & 0 & -7 \\ 2-i & 7 & 0 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 1 & 2 & -1 \\ 3 & -2 & 5 \end{bmatrix} \)
Answer:
(i) Let \( A = \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \)
The transpose of matrix \( A \), denoted by \( A^T \), is obtained by interchanging its rows and columns to create a new reflected grid:
\( A^T = \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \)
Since \( A^T = A \), the given matrix is a symmetric matrix.

(ii) Let \( B = \begin{bmatrix} 0 & 1+2i & i-2 \\ -1-2i & 0 & -7 \\ 2-i & 7 & 0 \end{bmatrix} \)
The transpose of matrix \( B \), denoted by \( B^T \), is:
\( B^T = \begin{bmatrix} 0 & -1-2i & 2-i \\ 1+2i & 0 & 7 \\ i-2 & -7 & 0 \end{bmatrix} \)
Now, let's find \( -B \) by changing the sign of every element in the matrix:
\( -B = \begin{bmatrix} 0 & -(1+2i) & -(i-2) \\ -(-1-2i) & 0 & -(-7) \\ -(2-i) & -7 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1-2i & 2-i \\ 1+2i & 0 & 7 \\ i-2 & -7 & 0 \end{bmatrix} \)
Since \( B^T = -B \), the given matrix is a skew-symmetric matrix.

(iii) Let \( C = \begin{bmatrix} 1 & 2 & -1 \\ 3 & -2 & 5 \end{bmatrix} \)
The transpose of matrix \( C \), denoted by \( C^T \), is:
\( C^T = \begin{bmatrix} 1 & 3 \\ 2 & -2 \\ -1 & 5 \end{bmatrix} \)
Since \( C \) is a non-square matrix of order \( 2 \times 3 \), its transpose \( C^T \) has order \( 3 \times 2 \). Thus, \( C^T \neq C \) and \( C^T \neq -C \). Therefore, the given matrix is neither symmetric nor skew-symmetric.
In simple words: To find the transpose, we swap the rows and columns of a matrix. If the transpose is exactly the same as the original matrix, it is symmetric; if it is the negative of the original, it is skew-symmetric; otherwise, it is neither.

🎯 Exam Tip: Remember that a skew-symmetric matrix must always be a square matrix with all its diagonal elements equal to zero. This is a quick way to identify or verify your answer!

Question 1. Classify the following matrices as symmetric, skew-symmetric, or neither:
(i) \( \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 2 & 5 & 1 \\ -5 & 4 & 6 \\ -1 & -6 & 3 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 0 & 1 + 2i & i - 2 \\ -1 - 2i & 0 & -7 \\ 2 - i & 7 & 0 \end{bmatrix} \)
Answer:
(i) Let \( A = \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \)
Then \( A^T = \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix} \)
Since \( A = A^T \), \( A \) is a symmetric matrix. This classification helps us understand the underlying symmetry properties of square matrices.

(ii) Let \( B = \begin{bmatrix} 2 & 5 & 1 \\ -5 & 4 & 6 \\ -1 & -6 & 3 \end{bmatrix} \)
Then \( B^T = \begin{pmatrix} 2 & -5 & -1 \\ 5 & 4 & -6 \\ 1 & 6 & 3 \end{pmatrix} \)
\( \implies B \neq B^T \)
Also, \( -B^T = -\begin{pmatrix} 2 & -5 & -1 \\ 5 & 4 & -6 \\ 1 & 6 & 3 \end{pmatrix} = \begin{pmatrix} -2 & 5 & 1 \\ -5 & -4 & 6 \\ -1 & -6 & -3 \end{pmatrix} \)
\( \implies B \neq -B^T \)
Hence, \( B \) is neither a symmetric nor a skew-symmetric matrix.

(iii) Let \( C = \begin{bmatrix} 0 & 1 + 2i & i - 2 \\ -1 - 2i & 0 & -7 \\ 2 - i & 7 & 0 \end{bmatrix} \)
Then \( C^T = \begin{bmatrix} 0 & -1 - 2i & 2 - i \\ 1 + 2i & 0 & 7 \\ i - 2 & -7 & 0 \end{bmatrix} \)
Also, \( -C^T = -\begin{bmatrix} 0 & -1 - 2i & 2 - i \\ 1 + 2i & 0 & 7 \\ i - 2 & -7 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 + 2i & i - 2 \\ -1 - 2i & 0 & -7 \\ 2 - i & 7 & 0 \end{bmatrix} \)
Since \( C = -C^T \), \( C \) is a skew-symmetric matrix.
In simple words: A matrix is symmetric if it is equal to its transpose, meaning the rows and columns are identical. It is skew-symmetric if its transpose is equal to its negative, and if it satisfies neither condition, it is neither.

🎯 Exam Tip: To quickly identify a skew-symmetric matrix, check if all diagonal elements are zero and the non-diagonal elements across the diagonal are negatives of each other.

Question 10. Construct the matrix \( A = [a_{ij}]_{3 \times 3} \), where \( a_{ij} = i - j \). State whether A is symmetric or skew-symmetric.
Answer:
The general \( 3 \times 3 \) matrix is represented as:
\[ A = [a_{ij}]_{3 \times 3} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \]
Given that \( a_{ij} = i - j \) for all \( i \) and \( j \).
Let us calculate each element of the matrix:
\( a_{11} = 1 - 1 = 0 \)
\( a_{12} = 1 - 2 = -1 \)
\( a_{13} = 1 - 3 = -2 \)
\( a_{21} = 2 - 1 = 1 \)
\( a_{22} = 2 - 2 = 0 \)
\( a_{23} = 2 - 3 = -1 \)
\( a_{31} = 3 - 1 = 2 \)
\( a_{32} = 3 - 2 = 1 \)
\( a_{33} = 3 - 3 = 0 \)

Substituting these values into the matrix, we get:
\[ A = \begin{bmatrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} \]
To determine if \( A \) is symmetric or skew-symmetric, we find its transpose \( A^T \):
\[ A^T = \begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{bmatrix} \]
Factoring out a negative sign from \( A^T \):
\[ -A^T = \begin{bmatrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} = A \]

\( \implies A^T = -A \)
Since \( a_{ij} = -a_{ji} \) for all \( i \) and \( j \), the matrix \( A \) is a skew-symmetric matrix.
In simple words: We construct the matrix by subtracting the column index from the row index for each position. Since the resulting matrix is equal to the negative of its transpose, it is classified as a skew-symmetric matrix.

🎯 Exam Tip: Remember that the diagonal elements of any skew-symmetric matrix must always be zero. Checking this first is a quick way to verify your answer.

Question 11. Solve the following equations for X and Y, if \( 3X - Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \( X - 3Y = \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \).
Answer:
Let \( A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \).
The given system of equations is:
\( 3X - Y = A \) --- (i)
\( X - 3Y = B \) --- (ii)
Multiplying equation (i) by 3, we get:
\( 9X - 3Y = 3A \) --- (iii)
Subtracting equation (ii) from equation (iii):
\( (9X - 3Y) - (X - 3Y) = 3A - B \)

\( \implies 8X = 3 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \)

\( \implies 8X = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \)

\( \implies 8X = \begin{bmatrix} 3 - 0 & -3 - (-1) \\ -3 - 0 & 3 - (-1) \end{bmatrix} \)

\( \implies 8X = \begin{bmatrix} 3 & -2 \\ -3 & 4 \end{bmatrix} \)

\( \implies X = \frac{1}{8} \begin{bmatrix} 3 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{8} & -\frac{1}{4} \\ -\frac{3}{8} & \frac{1}{2} \end{bmatrix} \)

Now, multiplying equation (ii) by 3, we get:
\( 3X - 9Y = 3B \) --- (iv)
Subtracting equation (iv) from equation (i):
\( (3X - Y) - (3X - 9Y) = A - 3B \)

\( \implies 8Y = A - 3B \)

\( \implies 8Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} - 3 \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \)

\( \implies 8Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -3 \\ 0 & -3 \end{bmatrix} \)

\( \implies 8Y = \begin{bmatrix} 1 - 0 & -1 - (-3) \\ -1 - 0 & 1 - (-3) \end{bmatrix} \)

\( \implies 8Y = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} \)

\( \implies Y = \frac{1}{8} \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} \frac{1}{8} & \frac{1}{4} \\ -\frac{1}{8} & \frac{1}{2} \end{bmatrix} \)

Thus, the solutions are \( X = \begin{bmatrix} \frac{3}{8} & -\frac{1}{4} \\ -\frac{3}{8} & \frac{1}{2} \end{bmatrix} \) and \( Y = \begin{bmatrix} \frac{1}{8} & \frac{1}{4} \\ -\frac{1}{8} & \frac{1}{2} \end{bmatrix} \).
In simple words: We solve these matrix equations just like regular simultaneous algebraic equations by multiplying to match coefficients and then subtracting to isolate and find the unknown matrices X and Y.

🎯 Exam Tip: Be extremely careful with signs when subtracting negative numbers inside matrices, as a single arithmetic slip will throw off the entire final matrix.

Question 1. Find the matrices \( X \) and \( Y \) if \( 3X - Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \( X - 3Y = \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \).
Answer: Let the given equations be:
\( 3X - Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \quad \text{--- (1)} \)
\( X - 3Y = \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \quad \text{--- (2)} \)
Multiplying equation (1) by 3, we get:
\( 9X - 3Y = 3 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \quad \text{--- (3)} \)
Subtracting equation (2) from (3), we get:
\( 8X = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} \)
\( 8X = \begin{bmatrix} 3 - 0 & -3 - (-1) \\ -3 - 0 & 3 - (-1) \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -3 & 4 \end{bmatrix} \)
\( \implies X = \frac{1}{8} \begin{bmatrix} 3 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{8} & -\frac{1}{4} \\ -\frac{3}{8} & \frac{1}{2} \end{bmatrix} \)
Now, substitute the value of \( X \) in equation (1):
\( 3 \begin{bmatrix} \frac{3}{8} & -\frac{1}{4} \\ -\frac{3}{8} & \frac{1}{2} \end{bmatrix} - Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\( \implies Y = \begin{bmatrix} \frac{9}{8} & -\frac{3}{4} \\ -\frac{9}{8} & \frac{3}{2} \end{bmatrix} - \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\( \implies Y = \begin{bmatrix} \frac{9}{8} - 1 & -\frac{3}{4} - (-1) \\ -\frac{9}{8} - (-1) & \frac{3}{2} - 1 \end{bmatrix} \)
\( \implies Y = \begin{bmatrix} \frac{1}{8} & \frac{1}{4} \\ -\frac{1}{8} & \frac{1}{2} \end{bmatrix} \)
Hence, \( X = \begin{bmatrix} \frac{3}{8} & -\frac{1}{4} \\ -\frac{3}{8} & \frac{1}{2} \end{bmatrix} \) and \( Y = \begin{bmatrix} \frac{1}{8} & \frac{1}{4} \\ -\frac{1}{8} & \frac{1}{2} \end{bmatrix} \). This systematic elimination method allows us to solve simultaneous matrix equations just like regular algebraic equations.
In simple words: To find the unknown matrices, we multiply the first equation by 3 so that the Y terms match, subtract the equations to solve for X, and then plug X back into the first equation to find Y.

🎯 Exam Tip: When multiplying a matrix by a scalar, remember to multiply every single element inside the matrix. Double-check your signs when subtracting negative fractions to avoid simple calculation errors.

Question 12. Find matrices A and B, if \( 2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \) and \( A - 2B = \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \)
Answer:
Given equations are:
\( 2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \) ---(i)
\( A - 2B = \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \) ---(ii)

To eliminate \( A \), multiply equation (ii) by 2:
\( 2(A - 2B) = 2 \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \)

\( \implies 2A - 4B = \begin{bmatrix} 6 & 4 & 16 \\ -4 & 2 & -14 \end{bmatrix} \) ---(iii)

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

\( \implies 3B = \begin{bmatrix} 6 - 6 & -6 - 4 & 0 - 16 \\ -4 + 4 & 2 - 2 & 1 + 14 \end{bmatrix} \)

\( \implies 3B = \begin{bmatrix} 0 & -10 & -16 \\ 0 & 0 & 15 \end{bmatrix} \)

\( \implies B = \frac{1}{3} \begin{bmatrix} 0 & -10 & -16 \\ 0 & 0 & 15 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{10}{3} & -\frac{16}{3} \\ 0 & 0 & 5 \end{bmatrix} \)

Now, substitute the value of matrix \( B \) in equation (ii) to find \( A \):
\( A = 2B + \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \)

\( \implies A = 2 \begin{bmatrix} 0 & -\frac{10}{3} & -\frac{16}{3} \\ 0 & 0 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \)

\( \implies A = \begin{bmatrix} 0 & -\frac{20}{3} & -\frac{32}{3} \\ 0 & 0 & 10 \end{bmatrix} + \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix} \)

\( \implies A = \begin{bmatrix} 0 + 3 & -\frac{20}{3} + 2 & -\frac{32}{3} + 8 \\ 0 - 2 & 0 + 1 & 10 - 7 \end{bmatrix} \)

\( \implies A = \begin{bmatrix} 3 & -\frac{14}{3} & -\frac{8}{3} \\ -2 & 1 & 3 \end{bmatrix} \)

Thus, the required matrices are \( A = \begin{bmatrix} 3 & -\frac{14}{3} & -\frac{8}{3} \\ -2 & 1 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & -\frac{10}{3} & -\frac{16}{3} \\ 0 & 0 & 5 \end{bmatrix} \).
In simple words: To find the unknown matrices, we treat the equations just like regular simultaneous equations. We multiply one equation to match the coefficients, subtract them to find matrix B, and then substitute B back to solve for matrix A.

🎯 Exam Tip: When multiplying a matrix by a scalar (like multiplying by 2), make sure to multiply every single element inside the matrix to avoid simple calculation errors.

Question 13. Find x and y, if \[ \begin{bmatrix} 2x + y & -1 & 1 \\ 3 & 4y & 4 \end{bmatrix} + \begin{bmatrix} -1 & 6 & 4 \\ 3 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 5 \\ 6 & 18 & 7 \end{bmatrix} \]
Answer: Given matrix equation: \[ \begin{bmatrix} 2x + y & -1 & 1 \\ 3 & 4y & 4 \end{bmatrix} + \begin{bmatrix} -1 & 6 & 4 \\ 3 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 5 \\ 6 & 18 & 7 \end{bmatrix} \] By adding the corresponding elements of the two matrices on the left-hand side:
\( \implies \) \[ \begin{bmatrix} 2x + y - 1 & -1 + 6 & 1 + 4 \\ 3 + 3 & 4y + 0 & 4 + 3 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 5 \\ 6 & 18 & 7 \end{bmatrix} \]
\( \implies \) \[ \begin{bmatrix} 2x + y - 1 & 5 & 5 \\ 6 & 4y & 7 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 5 \\ 6 & 18 & 7 \end{bmatrix} \] Comparing corresponding elements of equal matrices is a fundamental step in solving matrix equations. By equality of matrices, we get:
\( 2x + y - 1 = 3 \) --- (1)
and \( 4y = 18 \) --- (2) From equation (2):
\( y = \frac{18}{4} \)
\( \implies y = \frac{9}{2} \) Substituting \( y = \frac{9}{2} \) in equation (1):
\( 2x + \frac{9}{2} - 1 = 3 \)
\( \implies 2x + \frac{7}{2} = 3 \)
\( \implies 2x = 3 - \frac{7}{2} \)
\( \implies 2x = -\frac{1}{2} \)
\( \implies x = -\frac{1}{4} \) Therefore, the values are \( x = -\frac{1}{4} \) and \( y = \frac{9}{2} \).
In simple words: To find the unknown values, we first add the matching positions of the two left matrices together. Then, we set these simplified parts equal to the numbers in the same positions on the right side to solve for x and y.

🎯 Exam Tip: Always write down the step showing the addition of individual elements before simplifying, as examiners look for this to award step-by-step marks.

Question 14. 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.
Answer:
Given,
\( \begin{bmatrix} 2a + b & 3a - b \\ c + 2d & 2c - d \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \)
By equality of matrices, we can equate the corresponding elements of the two matrices to form a system of linear equations.
\( 2a + b = 2 \) ..... (1)
\( 3a - b = 3 \) ...... (2)
\( c + 2d = 4 \) ...... (3)
\( 2c - d = -1 \) ...... (4)
Adding (1) and (2), we get
\( 5a = 5 \)
\( \therefore a = 1 \)
Substituting \( a = 1 \) in (1), we get
\( 2(1) + b = 2 \)
\( \dots b = 0 \)
Multiplying equation (4) by 2, we get
\( 4c - 2d = -2 \) ...... (5)
Adding (3) and (5), we get
\( 5c = 2 \)
\( \therefore c = \frac{2}{5} \)
Substituting \( c = \frac{2}{5} \) in (4), we get
\( 2\left(\frac{2}{5}\right) - d = -1 \)
\( \therefore d = \frac{4}{5} + 1 = \frac{9}{5} \)
Hence, \( a = 1 \), \( b = 0 \), \( c = \frac{2}{5} \) and \( d = \frac{9}{5} \).
In simple words: When two matrices are equal, their corresponding elements in the same positions must be equal. We write down equations for each position, solve them step-by-step, and find the values of the unknown variables.

🎯 Exam Tip: Always state the reason "By equality of matrices" before equating the elements, and write down the equation numbers clearly to make your steps easy for the examiner to follow.

Question 15. There are two book shops owned by Suresh and Ganesh. Their sales (in Rs.) 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 Rs.), Physics, Chemistry, Mathematics
\( A = \begin{bmatrix} 5600 & 6750 & 8500 \\ 6650 & 7055 & 8905 \end{bmatrix} \) First Row Suresh / Second Row Ganesh
August Sales (in Rs.), Physics, Chemistry, Mathematics
\( B = \begin{bmatrix} 6650 & 7055 & 8905 \\ 7000 & 7500 & 10200 \end{bmatrix} \) First Row Suresh / Second Row Ganesh
(i) Find the increase in sales in Rs. from July to August 2017.
(ii) If both book shops get 10% profit in the month of August 2017, find the profit for each bookseller in each subject in that month.
Answer:
(i) The increase in sales (in Rs.) from July to August 2017 is obtained by subtracting the matrix A from B.
\( B - A = \begin{bmatrix} 6650 & 7055 & 8905 \\ 7000 & 7500 & 10200 \end{bmatrix} - \begin{bmatrix} 5600 & 6750 & 8500 \\ 6650 & 7055 & 8905 \end{bmatrix} \)
\( = \begin{bmatrix} 6650 - 5600 & 7055 - 6750 & 8905 - 8500 \\ 7000 - 6650 & 7500 - 7055 & 10200 - 8905 \end{bmatrix} \)
\( = \begin{bmatrix} 1050 & 305 & 405 \\ 350 & 445 & 1295 \end{bmatrix} \)
Hence, the increase in sales (in Rs.) from July to August 2017 for:
Suresh book shop: Rs. 1050 in Physics, Rs. 305 in Chemistry, and Rs. 405 in Mathematics.
Ganesh book shop: Rs. 350 in Physics, Rs. 445 in Chemistry, and Rs. 1295 in Mathematics.

(ii) The profit in the month of August 2017 is 10% of the August sales (Matrix B). This calculation helps the owners evaluate their monthly financial performance.
Profit Matrix \( = 10\% \text{ of } B = 0.1 \times B \)
\( = 0.1 \times \begin{bmatrix} 6650 & 7055 & 8905 \\ 7000 & 7500 & 10200 \end{bmatrix} \)
\( = \begin{bmatrix} 0.1 \times 6650 & 0.1 \times 7055 & 0.1 \times 8905 \\ 0.1 \times 7000 & 0.1 \times 7500 & 0.1 \times 10200 \end{bmatrix} \)
\( = \begin{bmatrix} 665 & 705.5 & 890.5 \\ 700 & 750 & 1020 \end{bmatrix} \)
Hence, the profit (in Rs.) for each bookseller in each subject in August 2017 is:
Suresh book shop: Rs. 665 in Physics, Rs. 705.50 in Chemistry, and Rs. 890.50 in Mathematics.
Ganesh book shop: Rs. 700 in Physics, Rs. 750 in Chemistry, and Rs. 1020 in Mathematics.
In simple words: To find the increase in sales, we subtract the July sales matrix from the August sales matrix element by element. To find the 10% profit for August, we multiply every value in the August sales matrix by 0.1.

🎯 Exam Tip: When performing matrix subtraction or scalar multiplication, make sure to perform calculations on corresponding elements carefully to avoid simple arithmetic mistakes.

Question 1. (ii) Both the book shops get 10% profit in August 2017. Find the profit for each bookseller in each subject.
Answer: The profit for each bookseller in each subject in August 2017 is obtained by the scalar multiplication of matrix B by 10%,
i.e. \( \frac{10}{100} = \frac{1}{10} \)
Now, \( \frac{1}{10} B = \frac{1}{10} \begin{pmatrix} 6650 & 7055 & 8905 \\ 7000 & 7500 & 10200 \end{pmatrix} \)
\( = \begin{pmatrix} 665 & 705.5 & 890.5 \\ 700 & 750 & 1020 \end{pmatrix} \)
Hence, the profit for Suresh book shop is Rs. 665 in Physics, Rs. 705.50 in Chemistry, and Rs. 890.50 in Mathematics, and for Ganesh book shop is Rs. 700 in Physics, Rs. 750 in Chemistry, and Rs. 1020 in Mathematics.
In simple words: To find a 10% profit, we multiply every number in our original matrix by \( \frac{1}{10} \). This scales down each value to show the exact profit earned for each subject.

🎯 Exam Tip: When performing scalar multiplication on a matrix, make sure to multiply every single element inside the matrix by the scalar value and write down the final units clearly.