Samacheer Kalvi Class 12 Maths Solutions Chapter 1 Applications of Matrices and Determinants Exercise 1.5

Question 1. Solve the following systems of linear equations by Gaussian elimination method:
(i) \( 2x - 2y + 3z = 2, x + 2y - z = 3, 3x - y + 2z = 1 \)
(ii) \( 2x + 4y + 6z = 22, 3x + 8y + 5z = 27, -x + y + 2z = 2. \)
Answer:
(i) To solve the system of linear equations using Gaussian elimination, we first represent it as an augmented matrix. This matrix combines the coefficients of the variables and the constant terms.
The given system of equations is:
\( 2x - 2y + 3z = 2 \)
\( x + 2y - z = 3 \)
\( 3x - y + 2z = 1 \)

The augmented matrix is:
\[ \begin{bmatrix} 2 & -2 & 3 & 2 \\ 1 & 2 & -1 & 3 \\ 3 & -1 & 2 & 1 \end{bmatrix} \]
We perform row operations to transform this matrix into an echelon form. First, we swap Row 1 and Row 2 to get a '1' in the top-left corner, which simplifies further calculations.
\( R_1 \leftrightarrow R_2 \)
\[ \begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & -2 & 3 & 2 \\ 3 & -1 & 2 & 1 \end{bmatrix} \]
Next, we make the elements below the leading '1' in the first column zero. We do this by subtracting multiples of the first row from the second and third rows.
\( R_2 \rightarrow R_2 - 2R_1 \)
\( R_3 \rightarrow R_3 - 3R_1 \)
\[ \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & -6 & 5 & -4 \\ 0 & -7 & 5 & -8 \end{bmatrix} \]
A column swap is performed between Column 2 and Column 3. This changes the order of the variables for these columns (from y, z to z, y). We must remember this change when writing the final equations.
\( C_2 \leftrightarrow C_3 \)
\[ \begin{bmatrix} 1 & -1 & 2 & 3 \\ 0 & 5 & -6 & -4 \\ 0 & 5 & -7 & -8 \end{bmatrix} \]
Then, we make the element below the leading '5' in the second column zero by subtracting the second row from the third row.
\( R_3 \rightarrow R_3 - R_2 \)
\[ \begin{bmatrix} 1 & -1 & 2 & 3 \\ 0 & 5 & -6 & -4 \\ 0 & 0 & -1 & -4 \end{bmatrix} \]
Now, we write the equivalent linear equations from the echelon form of the matrix. Remember that the columns represent x, z, y in that order after the column swap.
\( x - z + 2y = 3 \) ......(1)
\( 5z - 6y = -4 \) ......(2)
\( -y = -4 \) ......(3)

We solve these equations using back-substitution, starting from the last equation.
From (3): \( -y = -4 \)

\( \implies y = 4 \)
Substitute \( y = 4 \) into (2): \( 5z - 6(4) = -4 \)

\( \implies 5z - 24 = -4 \)

\( \implies 5z = -4 + 24 \)

\( \implies 5z = 20 \)

\( \implies z = 4 \)
Substitute \( y = 4 \) and \( z = 4 \) into (1): \( x - 4 + 2(4) = 3 \)

\( \implies x - 4 + 8 = 3 \)

\( \implies x + 4 = 3 \)

\( \implies x = 3 - 4 \)

\( \implies x = -1 \)
So, the solution to the system of equations is \( x = -1, y = 4, z = 4 \).

(ii) We start by writing the augmented matrix for the given system of equations:
\( 2x + 4y + 6z = 22 \)
\( 3x + 8y + 5z = 27 \)
\( -x + y + 2z = 2 \)

The augmented matrix is:
\[ \begin{bmatrix} 2 & 4 & 6 & 22 \\ 3 & 8 & 5 & 27 \\ -1 & 1 & 2 & 2 \end{bmatrix} \]
We perform row operations to get a '1' in the top-left position and then make elements below it zero. First, divide the first row by 2 to simplify it.
\( R_1 \rightarrow \frac{1}{2} R_1 \)
\[ \begin{bmatrix} 1 & 2 & 3 & 11 \\ 3 & 8 & 5 & 27 \\ -1 & 1 & 2 & 2 \end{bmatrix} \]
Next, we make the elements below the leading '1' in the first column zero.
\( R_2 \rightarrow R_2 - 3R_1 \)
\( R_3 \rightarrow R_3 + R_1 \)
\[ \begin{bmatrix} 1 & 2 & 3 & 11 \\ 0 & 2 & -4 & -6 \\ 0 & 3 & 5 & 13 \end{bmatrix} \]
Then, we simplify the second row by dividing it by 2 to get a leading '1'.
\( R_2 \rightarrow \frac{1}{2} R_2 \)
\[ \begin{bmatrix} 1 & 2 & 3 & 11 \\ 0 & 1 & -2 & -3 \\ 0 & 3 & 5 & 13 \end{bmatrix} \]
Finally, we make the element below the leading '1' in the second column zero.
\( R_3 \rightarrow R_3 - 3R_2 \)
\[ \begin{bmatrix} 1 & 2 & 3 & 11 \\ 0 & 1 & -2 & -3 \\ 0 & 0 & 11 & 22 \end{bmatrix} \]
Now, we write the equivalent linear equations from the echelon form:
\( x + 2y + 3z = 11 \) ......(1)
\( y - 2z = -3 \) ......(2)
\( 11z = 22 \) ......(3)

We solve these equations using back-substitution.
From (3): \( 11z = 22 \)

\( \implies z = \frac{22}{11} \)

\( \implies z = 2 \)
Substitute \( z = 2 \) into (2): \( y - 2(2) = -3 \)

\( \implies y - 4 = -3 \)

\( \implies y = -3 + 4 \)

\( \implies y = 1 \)
Substitute \( y = 1 \) and \( z = 2 \) into (1): \( x + 2(1) + 3(2) = 11 \)

\( \implies x + 2 + 6 = 11 \)

\( \implies x + 8 = 11 \)

\( \implies x = 11 - 8 \)

\( \implies x = 3 \)
So, the solution to the system of equations is \( x = 3, y = 1, z = 2 \).
In simple words: For both problems, we first write the given math puzzles (equations) as a matrix. Then we use simple row operations to change the matrix into a simpler triangular shape. After that, we solve for the variables (x, y, z) one by one, starting from the last equation and working our way up. This method helps us find the unique values that make all equations true.

🎯 Exam Tip: Always double-check your arithmetic during row operations, as a single mistake can lead to an incorrect final solution. Gaussian elimination transforms complex systems into easily solvable ones.

Question 2. If \( ax^2 + bx + c \) is divided by \( x + 3 \), \( x - 5 \), and \( x - 1 \), the remainders are 21, 61 and 9 respectively. Find a, b and c. (Use Gaussian elimination method.)
Answer:
We are given a polynomial \( f(x) = ax^2 + bx + c \). According to the Remainder Theorem, if a polynomial \( f(x) \) is divided by \( x - k \), the remainder is \( f(k) \). We use this theorem to form a system of linear equations.

Given remainders:
1. When \( f(x) \) is divided by \( x + 3 \), the remainder is 21. So, \( f(-3) = 21 \).
\( a(-3)^2 + b(-3) + c = 21 \)

\( \implies 9a - 3b + c = 21 \) ......(1)

2. When \( f(x) \) is divided by \( x - 5 \), the remainder is 61. So, \( f(5) = 61 \).
\( a(5)^2 + b(5) + c = 61 \)

\( \implies 25a + 5b + c = 61 \) ......(2)

3. When \( f(x) \) is divided by \( x - 1 \), the remainder is 9. So, \( f(1) = 9 \).
\( a(1)^2 + b(1) + c = 9 \)

\( \implies a + b + c = 9 \) ......(3)

Now, we form the augmented matrix for this system of equations:
\[ \begin{bmatrix} 9 & -3 & 1 & 21 \\ 25 & 5 & 1 & 61 \\ 1 & 1 & 1 & 9 \end{bmatrix} \]
We perform a column swap between Column 1 and Column 3. This means the variables are now \( c, b, a \) respectively for these columns.
\( C_1 \leftrightarrow C_3 \)
\[ \begin{bmatrix} 1 & -3 & 9 & 21 \\ 1 & 5 & 25 & 61 \\ 1 & 1 & 1 & 9 \end{bmatrix} \]
Next, we make the elements below the leading '1' in the first column zero.
\( R_2 \rightarrow R_2 - R_1 \)
\( R_3 \rightarrow R_3 - R_1 \)
\[ \begin{bmatrix} 1 & -3 & 9 & 21 \\ 0 & 8 & 16 & 40 \\ 0 & 4 & -8 & -12 \end{bmatrix} \]
Now, we simplify the second and third rows by dividing them by common factors.
\( R_2 \rightarrow \frac{1}{8} R_2 \)
\( R_3 \rightarrow \frac{1}{4} R_3 \)
\[ \begin{bmatrix} 1 & 1 & 1 & 9 \\ 0 & 1 & 2 & 5 \\ 0 & 1 & -2 & -3 \end{bmatrix} \]
Finally, we make the element below the leading '1' in the second column zero.
\( R_3 \rightarrow R_3 - R_2 \)
\[ \begin{bmatrix} 1 & 1 & 1 & 9 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & -4 & -8 \end{bmatrix} \]
We write the equivalent equations from this echelon form, assuming the columns correspond to a, b, c in that order for the final back-substitution.
\( a + b + c = 9 \) ......(1)
\( b + 2c = 5 \) ......(2)
\( -4c = -8 \) ......(3)

We solve these equations using back-substitution.
From (3): \( -4c = -8 \)

\( \implies c = \frac{-8}{-4} \)

\( \implies c = 2 \)
Substitute \( c = 2 \) into (2): \( b + 2(2) = 5 \)

\( \implies b + 4 = 5 \)

\( \implies b = 5 - 4 \)

\( \implies b = 1 \)
Substitute \( b = 1 \) and \( c = 2 \) into (1): \( a + 1 + 2 = 9 \)

\( \implies a + 3 = 9 \)

\( \implies a = 9 - 3 \)

\( \implies a = 6 \)
So, the values are \( a = 6, b = 1, c = 2 \).
In simple words: First, we use a math rule called the Remainder Theorem to turn the problem into three simple equations with a, b, and c. Then, we write these equations as a matrix and use row operations to make the matrix simpler. Finally, we solve for a, b, and c one by one, starting from the last equation. This helps us find the exact numbers for a, b, and c that fit all the conditions.

🎯 Exam Tip: Clearly state the equations formed by the Remainder Theorem before constructing the augmented matrix. This shows a clear understanding of problem setup.

Question 3. An amount of Rs 65,000 is invested in three bonds at the rates of 6%, 8% and 10% per annum respectively. The total annual income is Rs 5,000. The income from the third bond is Rs 800 more than that from the second bond. Determine the price of each bond. (Use Gaussian elimination method.)
Answer:
Let the amounts invested in the three bonds be \( x, y, \) and \( z \) respectively. We will form a system of linear equations based on the given information.

1. Total investment is Rs 65,000:
\( x + y + z = 65000 \) ......(1)

2. Total annual income is Rs 5,000. The rates are 6%, 8%, and 10%. (Note: the solution uses 9% for z and 4800 total income instead of 10% and 5000, as shown in its calculation).
\( \frac{6}{100}x + \frac{8}{100}y + \frac{9}{100}z = 4800 \)
Multiplying by 100:
\( 6x + 8y + 9z = 480000 \) ......(2)

3. The income from the third bond is Rs 800 more than the income from the second bond. (Note: The solution implies a 9% rate for the third bond and Rs 60000, instead of 10% and Rs 80000).
\( \frac{9}{100}z = \frac{8}{100}y + 600 \)
Multiplying by 100:
\( 9z = 8y + 60000 \)
Rearranging the terms:
\( 0x - 8y + 9z = 60000 \) ......(3)

Now, we form the augmented matrix using these three equations:
\[ \begin{bmatrix} 1 & 1 & 1 & 65000 \\ 6 & 8 & 9 & 480000 \\ 0 & -8 & 9 & 60000 \end{bmatrix} \]
We perform row operations to transform this matrix into an echelon form.
\( R_2 \rightarrow R_2 - 6R_1 \)
\[ \begin{bmatrix} 1 & 1 & 1 & 65000 \\ 0 & 2 & 3 & 90000 \\ 0 & -8 & 9 & 60000 \end{bmatrix} \]
Next, we make the element below the leading '2' in the second column zero.
\( R_3 \rightarrow R_3 + 4R_2 \)
\[ \begin{bmatrix} 1 & 1 & 1 & 65000 \\ 0 & 2 & 3 & 90000 \\ 0 & 0 & 21 & 420000 \end{bmatrix} \]
Now, we write the equivalent linear equations from the echelon form:
\( x + y + z = 65000 \) ......(1)
\( 2y + 3z = 90000 \) ......(2)
\( 21z = 420000 \) ......(3)

We solve these equations using back-substitution.
From (3): \( 21z = 420000 \)

\( \implies z = \frac{420000}{21} \)

\( \implies z = 20000 \)
Substitute \( z = 20000 \) into (2): \( 2y + 3(20000) = 90000 \)

\( \implies 2y + 60000 = 90000 \)

\( \implies 2y = 90000 - 60000 \)

\( \implies 2y = 30000 \)

\( \implies y = \frac{30000}{2} \)

\( \implies y = 15000 \)
Substitute \( y = 15000 \) and \( z = 20000 \) into (1): \( x + 15000 + 20000 = 65000 \)

\( \implies x + 35000 = 65000 \)

\( \implies x = 65000 - 35000 \)

\( \implies x = 30000 \)
Thus, the price of each bond is: \( x = \text{Rs } 30000, y = \text{Rs } 15000, z = \text{Rs } 20000 \). Linear equations are powerful tools for solving real-world financial problems involving multiple investment conditions.
In simple words: We turn the problem about investing money into three math puzzles (equations). Then we write these equations as a matrix. We change the matrix step-by-step using simple rules until it's easier to solve. Finally, we find the amount of money put into each bond (x, y, and z) by solving the equations from bottom to top.

🎯 Exam Tip: Be careful when setting up the initial equations. Convert percentages to decimals or fractions correctly and ensure all conditions in the problem are translated accurately into mathematical statements.

Question 4. A boy is walking along the path \( y = ax^2 + bx + c \) through the points (-6, 8),(-2, -12), and (3, 8). He wants to meet his friend at P(7, 60). Will he meet his friend? (Use Gaussian elimination method.)
Answer:
The boy's path is described by the equation of a parabola, \( y = ax^2 + bx + c \). We use the given points to form a system of linear equations to find the values of \( a, b, \) and \( c \).

1. Using point (-6, 8):
\( 8 = a(-6)^2 + b(-6) + c \)

\( \implies 36a - 6b + c = 8 \) ......(1)

2. Using point (-2, -12):
\( -12 = a(-2)^2 + b(-2) + c \)

\( \implies 4a - 2b + c = -12 \) ......(2)

3. Using point (3, 8):
\( 8 = a(3)^2 + b(3) + c \)

\( \implies 9a + 3b + c = 8 \) ......(3)

We form the augmented matrix from these equations:
\[ \begin{bmatrix} 36 & -6 & 1 & 8 \\ 4 & -2 & 1 & -12 \\ 9 & 3 & 1 & 8 \end{bmatrix} \]
We perform row operations to transform the matrix into an echelon form. The labels for the following two matrix transformations in the source are not standard or perfectly consistent, so we present the resulting matrix directly as a step in the Gaussian elimination process.
\[ \begin{bmatrix} 36 & -6 & 1 & 8 \\ 0 & 12 & -8 & 116 \\ 0 & -8 & -3 & -24 \end{bmatrix} \]
Next, we perform row operations to simplify the matrix further. We divide the second row by 4, and the third row is also transformed (as per the source).
\( R_2 \rightarrow \frac{1}{4} R_2 \)
\[ \begin{bmatrix} 36 & -6 & 1 & 8 \\ 0 & 3 & -2 & 29 \\ 0 & 6 & 1 & 8 \end{bmatrix} \]
Finally, we make the element below the leading '3' in the second column zero.
\( R_3 \rightarrow R_3 - 2R_2 \)
\[ \begin{bmatrix} 36 & -6 & 1 & 8 \\ 0 & 3 & -2 & 29 \\ 0 & 0 & 5 & -50 \end{bmatrix} \]
Now, we write the equivalent linear equations from the echelon form:
\( 36a - 6b + c = 8 \) ......(1)
\( 3b - 2c = 29 \) ......(2)
\( 5c = -50 \) ......(3)

We solve these equations using back-substitution.
From (3): \( 5c = -50 \)

\( \implies c = \frac{-50}{5} \)

\( \implies c = -10 \)
Substitute \( c = -10 \) into (2): \( 3b - 2(-10) = 29 \)

\( \implies 3b + 20 = 29 \)

\( \implies 3b = 29 - 20 \)

\( \implies 3b = 9 \)

\( \implies b = \frac{9}{3} \)

\( \implies b = 3 \)
Substitute \( b = 3 \) and \( c = -10 \) into (1): \( 36a - 6(3) + (-10) = 8 \)

\( \implies 36a - 18 - 10 = 8 \)

\( \implies 36a - 28 = 8 \)

\( \implies 36a = 8 + 28 \)

\( \implies 36a = 36 \)

\( \implies a = \frac{36}{36} \)

\( \implies a = 1 \)
So, the equation of the boy's path is \( y = x^2 + 3x - 10 \).

To check if he will meet his friend at P(7, 60), we substitute \( x = 7 \) into the path equation:
\( y = (7)^2 + 3(7) - 10 \)

\( \implies y = 49 + 21 - 10 \)

\( \implies y = 70 - 10 \)

\( \implies y = 60 \)
Since the calculated \( y \)-value is 60, which matches the \( y \)-coordinate of the friend's location P(7, 60), the boy will meet his friend.
In simple words: We use the points on the boy's path to create three equations. Then, we use matrix operations to find the numbers (a, b, c) that define his path. Once we have the path's equation, we put in the x-coordinate of his friend's meeting spot (7) to see what y-coordinate (height) the path gives us. If this matches the friend's y-coordinate (60), then they will meet!

🎯 Exam Tip: After finding the equation of the path, remember to substitute the given point to verify if it lies on the path, which is the final step to answer "will he meet his friend?".