Samacheer Kalvi Class 12 Maths Solutions Chapter 1 Applications of Matrices and Determinants Exercise 1.2

Question 1. Find the rank of the following matrices by minor method:
(i) \( \begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix} \)
(ii) \( \begin{bmatrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 1 & -2 & -1 & 0 \\ 3 & -6 & -3 & 1 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{bmatrix} \)
(v) \( \begin{bmatrix} 0 & 1 & 2 & 1 \\ 0 & 2 & 4 & 3 \\ 8 & 1 & 0 & 2 \end{bmatrix} \)
Answer:
(i) The given matrix A is a 2x2 matrix, so its rank can be at most 2. To find the rank, we first check the determinant of the 2x2 matrix.
\[ |A| = \begin{vmatrix} 2 & -4 \\ -1 & 2 \end{vmatrix} = (2 \times 2) - (-4 \times -1) = 4 - 4 = 0 \]
Since the determinant of the 2x2 matrix is 0, the rank cannot be 2. Next, we check if there is any 1x1 minor (a single element) that is not zero. For example, the element '2' in the top-left corner is not zero. This means at least one 1x1 minor is non-zero, so the rank of the matrix is 1. The rank indicates the maximum number of linearly independent row or column vectors.
(ii) The given matrix A is a 3x2 matrix, meaning it has 3 rows and 2 columns. The maximum possible rank for this matrix is the smaller of the number of rows and columns, which is 2. To check if the rank is 2, we look for at least one 2x2 minor (a sub-matrix of size 2x2) whose determinant is not zero.
Let's choose the minor from the first two rows:
\[ \begin{vmatrix} -1 & 3 \\ 4 & -7 \end{vmatrix} = (-1 \times -7) - (3 \times 4) = 7 - 12 = -5 \]
Since the determinant of this minor is -5, which is not equal to 0, we can confirm that the rank of the matrix is 2. The rank of a matrix helps understand the fundamental structure of linear equations related to it.
(iii) The given matrix A has 2 rows and 4 columns, so its order is 2x4. The maximum possible rank for this matrix is the minimum of its dimensions, which is 2. To check if the rank is 2, we need to find at least one 2x2 sub-matrix (minor) whose determinant is not zero.
Let's try a few 2x2 minors:
Minor 1 (from columns 1 and 2):
\[ \begin{vmatrix} 1 & -2 \\ 3 & -6 \end{vmatrix} = (1 \times -6) - (-2 \times 3) = -6 - (-6) = -6 + 6 = 0 \]
Minor 2 (from columns 2 and 3):
\[ \begin{vmatrix} -2 & -1 \\ -6 & -3 \end{vmatrix} = (-2 \times -3) - (-1 \times -6) = 6 - 6 = 0 \]
Minor 3 (from columns 3 and 4):
\[ \begin{vmatrix} -1 & 0 \\ -3 & 1 \end{vmatrix} = (-1 \times 1) - (0 \times -3) = -1 - 0 = -1 \]
Since the determinant of the third minor is -1, which is not zero, the rank of the matrix is 2. The rank helps us understand how many unique information-carrying rows or columns a matrix has.
(iv) The given matrix A is a 3x3 matrix. This means its rank can be at most 3. To find the rank, we first calculate the determinant of the entire 3x3 matrix (which is its highest order minor).
\[ \begin{vmatrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{vmatrix} \]
Let's expand the determinant using the first row:
\( = 1 \times ((4 \times -1) - (-6 \times 1)) - (-2) \times ((2 \times -1) - (-6 \times 5)) + 3 \times ((2 \times 1) - (4 \times 5)) \)
\( = 1(-4 + 6) + 2(-2 + 30) + 3(2 - 20) \)
\( = 1(2) + 2(28) + 3(-18) \)
\( = 2 + 56 - 54 \)
\( = 4 \)
Since the determinant of the 3x3 matrix is 4, which is not zero, the rank of the matrix is 3. A non-zero determinant for the highest order minor directly gives the rank.
(v) The given matrix A is a 3x4 matrix. Its rank can be at most 3 (the smaller of 3 rows and 4 columns). To find the rank, we need to look for a 3x3 minor (a 3x3 sub-matrix) whose determinant is not zero.
Let's consider the minor formed by the first three columns:
\[ \begin{vmatrix} 0 & 1 & 2 \\ 0 & 2 & 4 \\ 8 & 1 & 0 \end{vmatrix} \]
Expand along the first column:
\( = 0 \times (minor) - 0 \times (minor) + 8 \times \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} \)
\( = 8 \times ((1 \times 4) - (2 \times 2)) \)
\( = 8 \times (4 - 4) = 8 \times 0 = 0 \)
Since this minor's determinant is 0, we must look for another 3x3 minor. Let's try the minor formed by columns 2, 3, and 4 (and all three rows):
\[ \begin{vmatrix} 1 & 2 & 1 \\ 2 & 4 & 3 \\ 1 & 0 & 2 \end{vmatrix} \]
Expand along the first row:
\( = 1 \times ((4 \times 2) - (3 \times 0)) - 2 \times ((2 \times 2) - (3 \times 1)) + 1 \times ((2 \times 0) - (4 \times 1)) \)
\( = 1(8 - 0) - 2(4 - 3) + 1(0 - 4) \)
\( = 1(8) - 2(1) + 1(-4) \)
\( = 8 - 2 - 4 \)
\( = 2 \)
Since the determinant of this 3x3 minor is 2, which is not zero, the rank of the matrix is 3. Finding any non-zero minor of the highest possible order is key to determining rank.
In simple words: The rank of a matrix tells you how many rows or columns are truly unique and not just combinations of others. You find it by checking determinants of square sub-matrices. The largest size of a sub-matrix that has a non-zero determinant is the rank.

🎯 Exam Tip: When using the minor method, always start by checking the determinant of the largest possible square sub-matrix. If it's non-zero, that's your rank; otherwise, look for non-zero determinants in smaller sub-matrices.

Question 2. Find the rank of the following matrices by row reduction method:
(i) \( \begin{bmatrix} 1 & 1 & 1 & 3 \\ 2 & -1 & 3 & 4 \\ 5 & -1 & 7 & 11 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \\ 1 & -2 & 3 \\ 1 & -1 & 1 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 3 & -8 & 5 & 2 \\ 2 & -5 & 1 & 4 \\ -1 & 2 & 3 & -2 \end{bmatrix} \)
Answer:
(i) To find the rank of matrix A using the row reduction method, we transform it into row echelon form. The rank will then be the number of non-zero rows.
First, the given matrix is:
\[ A = \begin{bmatrix} 1 & 1 & 1 & 3 \\ 2 & -1 & 3 & 4 \\ 5 & -1 & 7 & 11 \end{bmatrix} \]
We perform row operations to make the elements below the leading 1 in the first column zero:
\( R_2 \rightarrow R_2 - 2R_1 \)
\( R_3 \rightarrow R_3 - 5R_1 \)
This gives us:
\[ \begin{bmatrix} 1 & 1 & 1 & 3 \\ 2 - 2(1) & -1 - 2(1) & 3 - 2(1) & 4 - 2(3) \\ 5 - 5(1) & -1 - 5(1) & 7 - 5(1) & 11 - 5(3) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & -3 & 1 & -2 \\ 0 & -6 & 2 & -4 \end{bmatrix} \]
Next, we make the element below the leading -3 in the second column zero:
\( R_3 \rightarrow R_3 - 2R_2 \)
This gives us:
\[ \begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & -3 & 1 & -2 \\ 0 - 2(0) & -6 - 2(-3) & 2 - 2(1) & -4 - 2(-2) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & -3 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]
This matrix is now in row echelon form. We can see there are two rows that are not entirely zero. Thus, the rank of matrix A is 2. Row echelon form simplifies matrices, making calculations like rank easier.
(ii) To find the rank of matrix A using row reduction, we convert it to row echelon form. The number of non-zero rows in this form gives the rank.
The initial matrix is:
\[ A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \\ 1 & -2 & 3 \\ 1 & -1 & 1 \end{bmatrix} \]
First, we make elements below the leading 1 in the first column zero:
\( R_2 \rightarrow R_2 - 3R_1 \)
\( R_3 \rightarrow R_3 - R_1 \)
\( R_4 \rightarrow R_4 - R_1 \)
\[ \begin{bmatrix} 1 & 2 & -1 \\ 0 & -7 & 5 \\ 0 & -4 & 4 \\ 0 & -3 & 2 \end{bmatrix} \]
To simplify the numbers, we can multiply the new rows \( R_2, R_3, R_4 \) by -1 if desired, as shown in the source:
\( R_2 \rightarrow (-1)R_2 \)
\( R_3 \rightarrow (-1)R_3 \)
\( R_4 \rightarrow (-1)R_4 \)
\[ \begin{bmatrix} 1 & 2 & -1 \\ 0 & 7 & -5 \\ 0 & 4 & -4 \\ 0 & 3 & -2 \end{bmatrix} \]
Next, we aim to make elements below the leading 7 in the second column zero. It's often easier to work with common multiples:
\( R_3 \rightarrow 7R_3 - 4R_2 \)
\( R_4 \rightarrow 7R_4 - 3R_2 \)
\[ \begin{bmatrix} 1 & 2 & -1 \\ 0 & 7 & -5 \\ 0 & 0 & -8 \\ 0 & 0 & 1 \end{bmatrix} \]
To get a leading 1 in the third row, divide \( R_3 \) by -8:
\( R_3 \rightarrow \frac{R_3}{-8} \)
\[ \begin{bmatrix} 1 & 2 & -1 \\ 0 & 7 & -5 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \]
Finally, make the element below the leading 1 in the third column zero:
\( R_4 \rightarrow R_4 - R_3 \)
\[ \begin{bmatrix} 1 & 2 & -1 \\ 0 & 7 & -5 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \]
This matrix is now in row echelon form. There are three non-zero rows. Therefore, the rank of matrix A is 3. This method is effective for large matrices.
(iii) To find the rank of matrix A using row reduction, we convert it to row echelon form. The rank is determined by the number of non-zero rows.
The initial matrix is:
\[ A = \begin{bmatrix} 3 & -8 & 5 & 2 \\ 2 & -5 & 1 & 4 \\ -1 & 2 & 3 & -2 \end{bmatrix} \]
It's often helpful to start with a leading 1. We can swap \( R_1 \) and \( R_3 \):
\( R_1 \leftrightarrow R_3 \)
\[ \begin{bmatrix} -1 & 2 & 3 & -2 \\ 2 & -5 & 1 & 4 \\ 3 & -8 & 5 & 2 \end{bmatrix} \]
Then multiply the new \( R_1 \) by -1 to get a positive leading 1:
\( R_1 \rightarrow -R_1 \)
\[ \begin{bmatrix} 1 & -2 & -3 & 2 \\ 2 & -5 & 1 & 4 \\ 3 & -8 & 5 & 2 \end{bmatrix} \]
Next, make the elements below the leading 1 in the first column zero:
\( R_2 \rightarrow R_2 - 2R_1 \)
\( R_3 \rightarrow R_3 - 3R_1 \)
\[ \begin{bmatrix} 1 & -2 & -3 & 2 \\ 0 & -1 & 7 & 0 \\ 0 & -2 & 14 & -4 \end{bmatrix} \]
Now, make the element below the leading -1 in the second column zero:
\( R_3 \rightarrow R_3 - 2R_2 \)
\[ \begin{bmatrix} 1 & -2 & -3 & 2 \\ 0 & -1 & 7 & 0 \\ 0 & 0 & 0 & -4 \end{bmatrix} \]
This matrix is in row echelon form. We can see that there are three rows that are not entirely zero. Therefore, the rank of matrix A is 3. Row operations are powerful tools for matrix analysis.
In simple words: Row reduction changes a matrix step-by-step into a simpler form called row echelon form. In this form, the number of rows that are not all zeros tells you the matrix's rank. This method helps simplify large matrices.

🎯 Exam Tip: Remember that swapping rows, multiplying a row by a non-zero scalar, and adding a multiple of one row to another are the only allowed elementary row operations. Use them carefully to get to row echelon form.

Question 3. Find the inverse of each of the following by Gauss-Jordan method:
(i) \( \begin{bmatrix} 2 & -1 \\ 5 & -2 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{bmatrix} \)
Answer:
(i) To find the inverse of matrix A using the Gauss-Jordan method, we form an augmented matrix \( [A | I_2] \) and perform row operations until the left side becomes the identity matrix \( I_2 \). The right side will then be \( A^{-1} \).
The given matrix is:
\[ A = \begin{bmatrix} 2 & -1 \\ 5 & -2 \end{bmatrix} \]
Form the augmented matrix:
\[ [A | I_2] = \begin{bmatrix} 2 & -1 & | & 1 & 0 \\ 5 & -2 & | & 0 & 1 \end{bmatrix} \]
First, we want a leading 1 in the first row, first column.
\( R_1 \rightarrow \frac{1}{2}R_1 \)
\[ \begin{bmatrix} 1 & -\frac{1}{2} & | & \frac{1}{2} & 0 \\ 5 & -2 & | & 0 & 1 \end{bmatrix} \]
Next, make the element below the leading 1 zero.
\( R_2 \rightarrow R_2 - 5R_1 \)
\[ \begin{bmatrix} 1 & -\frac{1}{2} & | & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & | & -\frac{5}{2} & 1 \end{bmatrix} \]
Now, create a leading 1 in the second row, second column.
\( R_2 \rightarrow 2R_2 \)
\[ \begin{bmatrix} 1 & -\frac{1}{2} & | & \frac{1}{2} & 0 \\ 0 & 1 & | & -5 & 2 \end{bmatrix} \]
Finally, make the element above the leading 1 in the second column zero.
\( R_1 \rightarrow R_1 + \frac{1}{2}R_2 \)
\[ \begin{bmatrix} 1 & 0 & | & -2 & 1 \\ 0 & 1 & | & -5 & 2 \end{bmatrix} \]
The left side is now the identity matrix. So, the inverse of A is the matrix on the right side:
\[ A^{-1} = \begin{bmatrix} -2 & 1 \\ -5 & 2 \end{bmatrix} \]
(ii) To find the inverse of matrix A using the Gauss-Jordan method, we set up an augmented matrix \( [A | I_3] \) and use row operations to transform A into the identity matrix \( I_3 \). The right side will then become \( A^{-1} \).
The given matrix is:
\[ A = \begin{bmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{bmatrix} \]
Form the augmented matrix:
\[ [A | I_3] = \begin{bmatrix} 1 & -1 & 0 & | & 1 & 0 & 0 \\ 1 & 0 & -1 & | & 0 & 1 & 0 \\ 6 & -2 & -3 & | & 0 & 0 & 1 \end{bmatrix} \]
First, we make elements below the leading 1 in the first column zero:
\( R_2 \rightarrow R_2 - R_1 \)
\( R_3 \rightarrow R_3 - 6R_1 \)
\[ \begin{bmatrix} 1 & -1 & 0 & | & 1 & 0 & 0 \\ 0 & 1 & -1 & | & -1 & 1 & 0 \\ 0 & 4 & -3 & | & -6 & 0 & 1 \end{bmatrix} \]
Next, make the element below the leading 1 in the second column zero:
\( R_3 \rightarrow R_3 - 4R_2 \)
\[ \begin{bmatrix} 1 & -1 & 0 & | & 1 & 0 & 0 \\ 0 & 1 & -1 & | & -1 & 1 & 0 \\ 0 & 0 & 1 & | & -2 & -4 & 1 \end{bmatrix} \]
Now, make elements above the leading 1 in the third column zero:
\( R_2 \rightarrow R_2 + R_3 \)
\[ \begin{bmatrix} 1 & -1 & 0 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & -3 & -3 & 1 \\ 0 & 0 & 1 & | & -2 & -4 & 1 \end{bmatrix} \]
Finally, make the element above the leading 1 in the second column zero:
\( R_1 \rightarrow R_1 + R_2 \)
\[ \begin{bmatrix} 1 & 0 & 0 & | & -2 & -3 & 1 \\ 0 & 1 & 0 & | & -3 & -3 & 1 \\ 0 & 0 & 1 & | & -2 & -4 & 1 \end{bmatrix} \]
The left side is now the identity matrix. Thus, the inverse of A is:
\[ A^{-1} = \begin{bmatrix} -2 & -3 & 1 \\ -3 & -3 & 1 \\ -2 & -4 & 1 \end{bmatrix} \]
(iii) To find the inverse of matrix A using the Gauss-Jordan method, we create an augmented matrix \( [A | I_3] \) and apply row operations to transform the left part into the identity matrix \( I_3 \). The right part will then be the inverse \( A^{-1} \).
The given matrix is:
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{bmatrix} \]
Form the augmented matrix:
\[ [A | I_3] = \begin{bmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 2 & 5 & 3 & | & 0 & 1 & 0 \\ 1 & 0 & 8 & | & 0 & 0 & 1 \end{bmatrix} \]
First, make elements below the leading 1 in the first column zero:
\( R_2 \rightarrow R_2 - 2R_1 \)
\( R_3 \rightarrow R_3 - R_1 \)
\[ \begin{bmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & -3 & | & -2 & 1 & 0 \\ 0 & -2 & 5 & | & -1 & 0 & 1 \end{bmatrix} \]
Next, make the element below the leading 1 in the second column zero:
\( R_3 \rightarrow R_3 + 2R_2 \)
\[ \begin{bmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & -3 & | & -2 & 1 & 0 \\ 0 & 0 & -1 & | & -5 & 2 & 1 \end{bmatrix} \]
Now, create a leading 1 in the third row, third column.
\( R_3 \rightarrow -R_3 \)
\[ \begin{bmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & -3 & | & -2 & 1 & 0 \\ 0 & 0 & 1 & | & 5 & -2 & -1 \end{bmatrix} \]
Next, make elements above the leading 1 in the third column zero:
\( R_2 \rightarrow R_2 + 3R_3 \)
\( R_1 \rightarrow R_1 - 3R_3 \)
\[ \begin{bmatrix} 1 & 2 & 0 & | & -14 & 6 & 3 \\ 0 & 1 & 0 & | & 13 & -5 & -3 \\ 0 & 0 & 1 & | & 5 & -2 & -1 \end{bmatrix} \]
Finally, make the element above the leading 1 in the second column zero:
\( R_1 \rightarrow R_1 - 2R_2 \)
\[ \begin{bmatrix} 1 & 0 & 0 & | & -40 & 16 & 9 \\ 0 & 1 & 0 & | & 13 & -5 & -3 \\ 0 & 0 & 1 & | & 5 & -2 & -1 \end{bmatrix} \]
The left side is now the identity matrix. So, the inverse of A is:
\[ A^{-1} = \begin{bmatrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{bmatrix} \]
In simple words: The Gauss-Jordan method helps find the inverse of a matrix. You start by writing the matrix next to an identity matrix. Then, you use row operations to turn the original matrix into an identity matrix. Whatever appears on the right side is the inverse. This method essentially "un-does" the original matrix.

🎯 Exam Tip: Practice the elementary row operations (swapping rows, multiplying a row by a constant, adding one row to another) very carefully to avoid calculation errors, especially with signs and fractions. Accuracy in each step is crucial for the Gauss-Jordan method.