Samacheer Kalvi Class 11 Business Maths Solutions Chapter 1 Matrices and Determinants Exercise 1.5

Choose the Correct Answer.

Question 1. The value of \( \left|\begin{array}{III} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{array}\right|=0 \) is
(a) 0, -1
(b) 0, 1
(c) -1, 1
(d) -1, -1
Answer: (b) 0, 1
\( 0 - 1[x^2 - x] + 0 = 0 \)
\( \implies x^2 - x = 0 \)
\( \implies x(x - 1) = 0 \)
\( \implies x = 0 \text{ (or) } x = 1 \) These are the two possible values for x.
In simple words: To find the value of x, you need to solve the determinant equation. When you calculate the determinant, you get a simple equation for x, which gives two possible answers.

🎯 Exam Tip: Remember to calculate the determinant carefully, especially with variables, and then solve the resulting equation for all possible values.

Question 2. The value of \( \left|\begin{array}{III} 2 x+y & x & y \\ 2 y+z & y & z \\ 2 z+x & z & x \end{array}\right| \) is
(a) xyz
(b) x + y + z
(c) (empty option)
(d) 0
Answer: (d) 0
First, perform the column operation \( C_1 \rightarrow C_1 - C_3 \).
The determinant becomes: \( \left|\begin{array}{III} 2 x & x & y \\ 2 y & y & z \\ 2 z & z & x \end{array}\right| \)
Now, take 2 common from the first column \( C_1 \).
This results in \( 2 \left|\begin{array}{III} x & x & y \\ y & y & z \\ z & z & x \end{array}\right| \)
Since the first column \( C_1 \) and the second column \( C_2 \) are identical, the value of the determinant is 0. If any two columns (or rows) are the same, the determinant is zero.
In simple words: When two columns in a determinant are exactly the same, the total value of that determinant is always zero. This is a quick rule to solve certain determinant problems.

🎯 Exam Tip: Always look for column or row operations that can make two rows or columns identical or proportional to simplify determinants, as this often leads to a zero value quickly.

Question 3. The cofactor of -7 in the determinant \( \left|\begin{array}{rrr} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \) is
(a) -18
(b) 18
(c) -7
(d) 7
Answer: (b) 18
The element -7 is in the 3rd row and 3rd column (a₃₃).
To find its cofactor, we use the formula \( C_{ij} = (-1)^{i+j} M_{ij} \), where \( M_{ij} \) is the minor.
For -7, \( i=3, j=3 \), so \( (-1)^{3+3} = (-1)^6 = 1 \).
The minor \( M_{33} \) is the determinant obtained by removing the 3rd row and 3rd column:
\( M_{33} = \left|\begin{array}{rr} 2 & -3 \\ 6 & 0 \end{array}\right| \)
Calculate this minor: \( (2 \times 0) - (-3 \times 6) = 0 - (-18) = 18 \).
Therefore, the cofactor of -7 is \( 1 \times 18 = 18 \). The sign for cofactor \(C_{33}\) is positive because \(3+3=6\) is an even number.
In simple words: To find the cofactor of an element, first cover its row and column. Then, find the determinant of the smaller matrix that's left. Finally, multiply this answer by +1 or -1 based on its position in the original matrix (even-numbered position sum gets +, odd gets -).

🎯 Exam Tip: Always remember the sign convention \( (-1)^{i+j} \) when calculating cofactors. A common mistake is to forget this sign, leading to an incorrect result.

Question 4. If \( \Delta = \left|\begin{array}{III} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{array}\right| \), then \( \left|\begin{array}{III} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right| \) is
(a) \( \Delta \)
(b) \( -\Delta \)
(c) \( 3\Delta \)
(d) \( -3\Delta \)
Answer: (b) \( -\Delta \)
Let the given second determinant be \( D = \left|\begin{array}{III} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right| \).
We can get \( D \) from \( \Delta \) by swapping the first row \( R_1 \) with the second row \( R_2 \).
\( \Delta = \left|\begin{array}{III} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{array}\right| \)
If we swap \( R_1 \leftrightarrow R_2 \) in \( \Delta \), the determinant changes its sign.
So, \( \left|\begin{array}{III} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right| = - \left|\begin{array}{III} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{array}\right| \)
Therefore, \( D = -\Delta \). Swapping two rows always flips the sign of the determinant.
In simple words: When you swap any two rows or any two columns in a determinant, the value of the determinant changes its sign (from positive to negative, or negative to positive).

🎯 Exam Tip: Remember that swapping any two rows or columns in a determinant multiplies its value by -1. This property is crucial for quickly evaluating determinants after row/column interchanges.

Question 5. The value of the determinant \( \left|\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right|^{2} \) is
(a) abc
(b) 0
(c) a²b²c²
(d) -abc
Answer: (c) a²b²c²
The given matrix is a diagonal matrix. The determinant of a diagonal matrix (or a triangular matrix) is simply the product of its diagonal elements.
So, \( \left|\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right| = a \times b \times c = abc \).
We need to find the square of this determinant.
\( (abc)^2 = a^2b^2c^2 \). This property helps simplify calculations significantly.
In simple words: For a special type of matrix where numbers are only on the main slanted line (diagonal) and zeros everywhere else, its determinant is just those diagonal numbers multiplied together. Then, we square that result.

🎯 Exam Tip: For diagonal or triangular matrices, the determinant is the product of the diagonal elements. Use this shortcut to save time in calculations.

Question 6. If A is square matrix of order 3 then \( |kA| \) is:
(a) k|A|
(b) -k|A|
(c) k³|A|
(d) -k³|A|
Answer: (c) k³|A|
There is a property of determinants that states if A is a square matrix of order \( n \), then \( |kA| = k^n|A| \).
In this question, the order of matrix A is given as 3.
So, \( n=3 \).
Therefore, \( |kA| = k^3|A| \). This rule applies universally for any scalar k and square matrix A.
In simple words: When you multiply a whole matrix by a number 'k' and then find its determinant, it's the same as taking the original determinant and multiplying it by 'k' raised to the power of the matrix's size.

🎯 Exam Tip: Remember the formula \( |kA| = k^n|A| \) for an \( n \times n \) matrix A. This is a fundamental property for scalar multiplication of determinants.

Question 7. adj (AB) is equal to:
(a) adj A adj B
(c) adj B adj A
Answer: (c) adj B adj A
The property of the adjoint of a product of matrices states that for any two invertible square matrices A and B of the same order:
\( \text{adj (AB)} = \text{adj B} \text{ adj A} \).
This is similar to the property of inverse of a product: \( (AB)^{-1} = B^{-1} A^{-1} \). This order reversal is important in matrix algebra.
In simple words: When you find the adjoint of two multiplied matrices, you switch their order and then find the adjoint of each separately, then multiply them.

🎯 Exam Tip: Always remember that the adjoint of a product reverses the order of the matrices, similar to the inverse of a product: \( \text{adj(AB)} = \text{adj(B)adj(A)} \).

Question 8. The inverse matrix of \( \left(\begin{array}{cc} \frac{4}{5} & \frac{5}{12} \\ \frac{2}{5} & \frac{1}{2} \end{array}\right) \) is
(a) \( \frac{7}{30}\left(\begin{array}{cc} \frac{1}{2} & \frac{5}{12} \\ \frac{2}{5} & \frac{4}{5} \end{array}\right) \)
(b) \( \frac{7}{30}\left(\begin{array}{cc} \frac{1}{2} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{1}{5} \end{array}\right) \)
(c) \( \frac{30}{7}\left(\begin{array}{rr} \frac{1}{2} & \frac{5}{12} \\ \frac{2}{5} & \frac{4}{5} \end{array}\right) \)
(d) \( \frac{30}{7}\left(\begin{array}{rr} \frac{1}{2} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{4}{5} \end{array}\right) \)
Answer: (d) \( \frac{30}{7}\left(\begin{array}{rr} \frac{1}{2} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{4}{5} \end{array}\right) \)
Let \( A = \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) = \left(\begin{array}{cc} \frac{4}{5} & \frac{5}{12} \\ \frac{2}{5} & \frac{1}{2} \end{array}\right) \).
The determinant of A is \( |A| = ad - bc = \left(\frac{4}{5}\right)\left(\frac{1}{2}\right) - \left(\frac{5}{12}\right)\left(\frac{2}{5}\right) \)
\( = \frac{4}{10} - \frac{10}{60} = \frac{2}{5} - \frac{1}{6} \)
\( = \frac{12 - 5}{30} = \frac{7}{30} \).
The inverse of a 2x2 matrix is \( A^{-1} = \frac{1}{|A|} \text{adj}(A) \).
Where \( \text{adj}(A) = \left(\begin{array}{rr} d & -b \\ -c & a \end{array}\right) = \left(\begin{array}{rr} \frac{1}{2} & -\frac{5}{12} \\ -\frac{2}{5} & \frac{4}{5} \end{array}\right) \).
So, \( A^{-1} = \frac{1}{\frac{7}{30}} \left(\begin{array}{rr} \frac{1}{2} & -\frac{5}{12} \\ -\frac{2}{5} & \frac{4}{5} \end{array}\right) = \frac{30}{7} \left(\begin{array}{rr} \frac{1}{2} & -\frac{5}{12} \\ -\frac{2}{5} & \frac{4}{5} \end{array}\right) \). The calculation of the determinant and adjoint is crucial here.
In simple words: To find the inverse of a 2x2 matrix, you first calculate its determinant. Then, you swap the main diagonal elements, change the signs of the off-diagonal elements, and finally divide this new matrix by the determinant.

🎯 Exam Tip: Carefully calculate the determinant and then the adjoint matrix. A common error is mixing up the signs of the off-diagonal elements or failing to swap the main diagonal ones.

Question 9. If \( A = \left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \) such that \( ad – bc \neq 0 \) then \( A^{-1} \) is:
(a) \( \frac{1}{a d-b c}\left[\begin{array}{cc} d & b \\ -c & a \end{array}\right] \)
(b) \( \frac{1}{a d-b c}\left[\begin{array}{ll} d & b \\ c & a \end{array}\right] \)
(c) \( \frac{1}{a d-b c}\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \)
(d) \( \frac{1}{a d-b c}\left[\begin{array}{ll} d & -b \\ c & a \end{array}\right] \)
Answer: (c) \( \frac{1}{a d-b c}\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \)
For a 2x2 matrix \( A = \left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \):
First, find its determinant: \( |A| = ad - bc \). The condition \( ad - bc \neq 0 \) ensures that the inverse exists.
Next, find the adjoint of A: \( \text{adj}(A) = \left[\begin{array}{rr} d & -b \\ -c & a \end{array}\right] \). This involves swapping the main diagonal elements and changing the signs of the off-diagonal elements.
Finally, the inverse is \( A^{-1} = \frac{1}{|A|} \text{adj}(A) \).
So, \( A^{-1} = \frac{1}{ad - bc} \left[\begin{array}{rr} d & -b \\ -c & a \end{array}\right] \). This formula is a standard result in matrix algebra.
In simple words: The inverse of a 2x2 matrix is found by swapping the top-left and bottom-right numbers, changing the signs of the other two numbers, and then dividing the whole new matrix by the value of its determinant.

🎯 Exam Tip: Memorize the formula for the inverse of a 2x2 matrix as it's frequently tested. Pay close attention to the sign changes for the off-diagonal elements.

Question 10. The number of Hawkins-Simon conditions for the viability of input-output analysis is:
(a) 1
(b) 3
(c) 4
(d) 2
Answer: (d) 2
The Hawkins-Simon conditions are important criteria used in input-output analysis to ensure that the economic system is viable (i.e., it can produce a positive output).
There are two main conditions:
1. All the main diagonal elements of the (I - A) matrix must be positive.
2. The determinant of the (I - A) matrix must be positive.
These conditions ensure that production is possible and self-sustaining within the economy.
In simple words: For an economic model to work properly, there are two specific rules, called Hawkins-Simon conditions, that must be met. These rules make sure that the economy can actually produce things.

🎯 Exam Tip: Remember that there are exactly two Hawkins-Simon conditions for the viability of an input-output model, both involving the (I-A) matrix.

Question 11. The inventor of input-output analysis is:
(a) Sir Francis Galton
(b) Fisher
(c) Prof. Wassily W. Leontief
(d) Arthur Cayley
Answer: (c) Prof. Wassily W. Leontief
Input-output analysis is a technique used in economics to study the interdependence between different sectors of an economy.
Professor Wassily W. Leontief (1906-1999) was a Russian-American economist who developed this analytical framework.
He was awarded the Nobel Memorial Prize in Economic Sciences in 1973 for his development of the input-output method and for its application to important economic problems. His work provided a way to understand how industries interact.
In simple words: Input-output analysis, which helps us understand how different parts of an economy rely on each other, was created by an economist named Professor Wassily W. Leontief.

🎯 Exam Tip: Associate Wassily W. Leontief with input-output analysis; he is the key figure in its development and won a Nobel Prize for it.

Question 12. Which of the following matrix has no inverse?
(a) \( \left(\begin{array}{rr} -1 & 1 \\ 1 & -4 \end{array}\right) \)
(b) \( \left(\begin{array}{rr} 2 & -1 \\ -4 & 2 \end{array}\right) \)
(c) \( \left(\begin{array}{cc} \cos a & \sin a \\ -\sin a & \cos a \end{array}\right) \)
(d) \( \left(\begin{array}{rr} \sin a & \sin a \\ -\cos a & \cos a \end{array}\right) \)
Answer: (b) \( \left(\begin{array}{rr} 2 & -1 \\ -4 & 2 \end{array}\right) \)
A square matrix has no inverse if and only if its determinant is zero.
Let's calculate the determinant for each option:
(a) \( \left|\begin{array}{rr} -1 & 1 \\ 1 & -4 \end{array}\right| = (-1)(-4) - (1)(1) = 4 - 1 = 3 \neq 0 \). Inverse exists.
(b) \( \left|\begin{array}{rr} 2 & -1 \\ -4 & 2 \end{array}\right| = (2)(2) - (-1)(-4) = 4 - 4 = 0 \). Inverse does not exist.
(c) \( \left|\begin{array}{cc} \cos a & \sin a \\ -\sin a & \cos a \end{array}\right| = \cos^2 a - (-\sin^2 a) = \cos^2 a + \sin^2 a = 1 \neq 0 \). Inverse exists.
(d) \( \left|\begin{array}{rr} \sin a & \sin a \\ -\cos a & \cos a \end{array}\right| = (\sin a)(\cos a) - (\sin a)(-\cos a) = \sin a \cos a + \sin a \cos a = 2 \sin a \cos a = \sin (2a) \). This can be zero only for specific values of 'a', not always, so an inverse generally exists. The determinant for option (b) is exactly zero, meaning it never has an inverse.
In simple words: A matrix cannot be "un-done" or inverted if its special number called the determinant is zero. We check each option to see which one has a determinant of zero.

🎯 Exam Tip: The key to determining if a matrix has an inverse is to calculate its determinant. If the determinant is zero, the inverse does not exist.

Question 13. Inverse of \( \left(\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right) \) is:
(a) \( \left(\begin{array}{rr} 2 & -1 \\ -5 & 3 \end{array}\right) \)
(b) \( \left(\begin{array}{rr} -2 & 5 \\ 1 & -3 \end{array}\right) \)
(c) \( \left(\begin{array}{rr} 3 & -1 \\ -5 & -3 \end{array}\right) \)
(d) \( \left(\begin{array}{rr} -3 & 5 \\ 1 & -2 \end{array}\right) \)
Answer: (a) \( \left(\begin{array}{rr} 2 & -1 \\ -5 & 3 \end{array}\right) \)
Let \( A = \left(\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right) \).
First, calculate the determinant of A: \( |A| = (3 \times 2) - (1 \times 5) = 6 - 5 = 1 \).
Next, find the adjoint of A: \( \text{adj}(A) = \left(\begin{array}{rr} 2 & -1 \\ -5 & 3 \end{array}\right) \). (Swap main diagonal, change signs of off-diagonal).
The inverse \( A^{-1} = \frac{1}{|A|} \text{adj}(A) \).
Since \( |A| = 1 \), \( A^{-1} = \frac{1}{1} \left(\begin{array}{rr} 2 & -1 \\ -5 & 3 \end{array}\right) = \left(\begin{array}{rr} 2 & -1 \\ -5 & 3 \end{array}\right) \). A determinant of 1 makes the inverse calculation very straightforward.
In simple words: To get the inverse of this 2x2 matrix, you swap the numbers on the main slanted line and change the signs of the other two numbers. Since the determinant is 1, you don't need to divide by anything.

🎯 Exam Tip: When the determinant is 1, the inverse matrix is simply the adjoint matrix. This can save calculation time if you notice the determinant early.

Question 14. If \( A = \left(\begin{array}{rr} -1 & 2 \\ 1 & -4 \end{array}\right) \) then \( A (\text{adj} A) \) is:
(a) \( \left(\begin{array}{ll} -4 & -2 \\ -1 & -1 \end{array}\right) \)
(b) \( \left(\begin{array}{rr} 4 & -2 \\ -1 & 1 \end{array}\right) \)
(c) \( \left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right) \)
(d) \( \left(\begin{array}{ll} 0 & 2 \\ 2 & 0 \end{array}\right) \)
Answer: (c) \( \left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right) \)
A fundamental property of matrices states that for any square matrix A:
\( A (\text{adj} A) = (\text{adj} A) A = |A| I \), where \( I \) is the identity matrix.
First, calculate the determinant of A:
\( |A| = \left|\begin{array}{rr} -1 & 2 \\ 1 & -4 \end{array}\right| = (-1)(-4) - (2)(1) = 4 - 2 = 2 \).
The identity matrix of order 2 is \( I = \left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \).
So, \( A (\text{adj} A) = |A| I = 2 \left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) = \left(\begin{array}{ll} 2 \times 1 & 2 \times 0 \\ 2 \times 0 & 2 \times 1 \end{array}\right) = \left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right) \). This property simplifies matrix multiplication significantly.
In simple words: When you multiply a matrix by its adjoint, the result is always a special matrix. This special matrix is made by taking the matrix's determinant and multiplying it by an identity matrix (which has ones on the diagonal and zeros elsewhere).

🎯 Exam Tip: Remember the identity \( A(\text{adj} A) = |A|I \). This property helps quickly find the product of a matrix and its adjoint without performing full matrix multiplication.

Question 15. If A and B non-singular matrix then, which of the following is incorrect?
(a) \( A^2 = I \) implies \( A^{-1} = A \)
(b) \( I^{-1} = I \)
(c) If AX = B then X = B⁻¹A
(d) If A is square matrix of order 3 then \( |\text{adj} A| = |A|^2 \)
Answer: (c) If AX = B then X = B⁻¹A
Let's check each option:
(a) If \( A^2 = I \), then multiplying by \( A^{-1} \) on both sides gives \( A^{-1}A^2 = A^{-1}I \implies A = A^{-1} \). This statement is Correct.
(b) The inverse of an identity matrix is itself, so \( I^{-1} = I \). This statement is Correct.
(c) If \( AX = B \), to solve for X, we multiply by \( A^{-1} \) on the left side of both equations: \( A^{-1}(AX) = A^{-1}B \implies (A^{-1}A)X = A^{-1}B \implies IX = A^{-1}B \implies X = A^{-1}B \). Therefore, \( X = B^{-1}A \) is Incorrect.
(d) For a square matrix A of order \( n \), the property is \( |\text{adj} A| = |A|^{n-1} \). If the order is 3, then \( n=3 \). So, \( |\text{adj} A| = |A|^{3-1} = |A|^2 \). This statement is Correct. The formula for adjoint determinant is a key identity.
Thus, the incorrect statement is (c).
In simple words: We are checking rules for matrices. The rule for solving a matrix equation like AX = B is to multiply by the inverse of A on the left side, so X should be A-inverse times B, not B-inverse times A.

🎯 Exam Tip: Pay close attention to the order of multiplication when dealing with matrix inverses, especially when solving equations like AX=B or XA=B. Matrix multiplication is not commutative.

Question 16. The value of \( \left|\begin{array}{rrr} 5 & 5 & 5 \\ 4 x & 4 y & 4 z \\ -3 x & -3 y & -3 z \end{array}\right| \) is:
(a) 5
(b) 4
(c) 0
(d) -3
Answer: (c) 0
Let the given determinant be D.
\( D = \left|\begin{array}{rrr} 5 & 5 & 5 \\ 4 x & 4 y & 4 z \\ -3 x & -3 y & -3 z \end{array}\right| \)
We can factor out common terms from rows:
Factor 5 from \( R_1 \): \( 5 \left|\begin{array}{rrr} 1 & 1 & 1 \\ 4 x & 4 y & 4 z \\ -3 x & -3 y & -3 z \end{array}\right| \)
Factor 4 from \( R_2 \): \( 5 \times 4 \left|\begin{array}{rrr} 1 & 1 & 1 \\ x & y & z \\ -3 x & -3 y & -3 z \end{array}\right| \)
Factor -3 from \( R_3 \): \( 5 \times 4 \times (-3) \left|\begin{array}{rrr} 1 & 1 & 1 \\ x & y & z \\ x & y & z \end{array}\right| \)
So, \( D = -60 \left|\begin{array}{rrr} 1 & 1 & 1 \\ x & y & z \\ x & y & z \end{array}\right| \).
Since \( R_2 \) and \( R_3 \) are identical, the value of the determinant is 0. If two rows are identical, the determinant is zero. Thus, the entire expression becomes \( -60 \times 0 = 0 \).
In simple words: If you can make two rows or two columns in a determinant exactly the same by taking out common numbers, then the determinant's value will always be zero.

🎯 Exam Tip: Always check for common factors in rows or columns and for identical or proportional rows/columns. These properties quickly lead to the determinant being zero, saving complex calculations.

Question 17. If A is an invertible matrix of order 2 then det \( (A^{-1}) \) be equal
(a) det (A)
(b) \( \frac{1}{{\text{det}}(A)} \)
(c) 1
(d) 0
Answer: (b) \( \frac{1}{{\text{det}}(A)} \)
For any invertible square matrix A, a fundamental property relates its determinant to the determinant of its inverse.
We know that \( AA^{-1} = I \), where \( I \) is the identity matrix.
Taking the determinant of both sides: \( |AA^{-1}| = |I| \).
Using the property \( |AB| = |A||B| \), we get \( |A||A^{-1}| = |I| \).
Since the determinant of an identity matrix is always 1, we have \( |A||A^{-1}| = 1 \).
Dividing by \( |A| \) (which is non-zero because A is invertible), we get \( |A^{-1}| = \frac{1}{|A|} \).
Thus, \( \text{det}(A^{-1}) = \frac{1}{\text{det}(A)} \). This relationship is consistent regardless of the matrix order.
In simple words: The determinant of an inverse matrix is simply 1 divided by the determinant of the original matrix. They are reciprocals of each other.

🎯 Exam Tip: Remember the property \( |\text{det}(A^{-1})| = \frac{1}{|\text{det}(A)|} \). This is a direct consequence of \( AA^{-1} = I \) and \( |AB| = |A||B| \), and is widely applicable.

Question 18. If A is 3 \( \times \) 3 matrix and \( |A| = 4 \) then \( |A^{-1}| \) is equal to:
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{16} \)
(d) 4
Answer: (a) \( \frac{1}{4} \)
Using the property learned in the previous question, for any invertible matrix A,
\( |A^{-1}| = \frac{1}{|A|} \).
Given that \( |A| = 4 \).
Substitute this value into the formula:
\( |A^{-1}| = \frac{1}{4} \). The order of the matrix (3x3) does not affect this particular relationship between the determinant of a matrix and its inverse.
In simple words: Since we know the determinant of the original matrix is 4, the determinant of its inverse is simply one divided by 4.

🎯 Exam Tip: Apply the rule \( |A^{-1}| = 1/|A| \) directly. The order of the matrix is extra information here and does not change this specific calculation.

Question 19. If A is a square matrix of order 3 and \( |A| = 3 \) then \( |\text{adj} A| \) is equal to:
(a) 81
(b) 27
(c) 3
(d) 9
Answer: (d) 9
For a square matrix A of order \( n \), the determinant of its adjoint is given by the formula:
\( |\text{adj} A| = |A|^{n-1} \).
In this question, the order of the matrix A is given as 3, so \( n=3 \).
The determinant of A is given as \( |A| = 3 \).
Substitute these values into the formula:
\( |\text{adj} A| = |3|^{3-1} = 3^2 = 9 \). This property is very useful for finding the adjoint determinant without first calculating the adjoint matrix.
In simple words: To find the determinant of the adjoint matrix, you take the determinant of the original matrix and raise it to a power. That power is one less than the size of the matrix.

🎯 Exam Tip: Remember the formula \( |\text{adj} A| = |A|^{n-1} \) for an \( n \times n \) matrix. It's a standard result for adjoints and determinants.

Question 20. The value of \( \left|\begin{array}{ccc} x & x^{2}-y z & 1 \\ y & y^{2}-z x & 1 \\ z & z^{2}-x y & 1 \end{array}\right| \) is:
(a) 1
(b) 0
(c) -1
(d) -xyz
Answer: (b) 0
Let the determinant be D.
\( D = \left|\begin{array}{ccc} x & x^{2}-y z & 1 \\ y & y^{2}-z x & 1 \\ z & z^{2}-x y & 1 \end{array}\right| \)
We can split the second column into two parts using a determinant property:
\( D = \left|\begin{array}{ccc} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right| - \left|\begin{array}{ccc} x & y z & 1 \\ y & z x & 1 \\ z & x y & 1 \end{array}\right| \)
Now, for the second determinant, multiply \( R_1 \) by \( x \), \( R_2 \) by \( y \), and \( R_3 \) by \( z \). To keep the determinant value unchanged, divide by \( xyz \).
\( D = \left|\begin{array}{ccc} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right| - \frac{1}{x y z} \left|\begin{array}{ccc} x^2 & x y z & x \\ y^2 & y z x & y \\ z^2 & z x y & z \end{array}\right| \)

\( \implies D = \left|\begin{array}{ccc} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right| - \frac{x y z}{x y z} \left|\begin{array}{ccc} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right| \)
The second determinant, after factoring out \( xyz \) from the second column and then rearranging columns, becomes identical to the first determinant.
\( \implies D = \left|\begin{array}{ccc} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right| - \left|\begin{array}{ccc} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array}\right| \)
\( \implies D = 0 \). The determinant properties allow for this kind of simplification.
In simple words: By splitting the determinant and using special rules to change its columns, we can see that the determinant subtracts itself, leading to an answer of zero.

🎯 Exam Tip: When elements in a column (or row) are sums or differences, try splitting the determinant into two simpler determinants. Also, remember that if two columns (or rows) become identical, the determinant is zero.

Question 21. If \( A = \left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right], \) then \( |2A| \) is equal to:
(a) 4 cos 2\( \theta \)
(b) 4
(c) 2
(d) 1
Answer: (b) 4
First, find the determinant of matrix A:
\( |A| = \left|\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right| = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) \)
\( = \cos^2 \theta + \sin^2 \theta \).
We know the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \).
So, \( |A| = 1 \).
Next, use the property \( |kA| = k^n|A| \), where n is the order of the matrix.
Here, the order of matrix A is 2 (a 2x2 matrix), so \( n=2 \).
We need to find \( |2A| \).
\( |2A| = 2^2 |A| = 4 |A| \).
Substitute the value of \( |A| \):
\( |2A| = 4 \times 1 = 4 \). This shows how scalar multiplication affects the determinant.
In simple words: First, find the determinant of the given matrix, which turns out to be 1. Then, use the rule that multiplying a matrix by a number 'k' changes its determinant by 'k' raised to the power of the matrix's size.

🎯 Exam Tip: Recognize trigonometric identities like \( \cos^2 \theta + \sin^2 \theta = 1 \) to simplify determinants. Also, remember the scalar multiplication rule \( |kA| = k^n|A| \).

Question 22. If \( \Delta = \left|\begin{array}{III} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| \) and Aij is cofactor of aij, then value of \( \Delta \) is given by:
(a) \( a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} \)
(b) \( a_{11}A_{11} + a_{12}A_{21} + a_{13}A_{31} \)
(c) \( a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13} \)
(d) \( a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31} \)
Answer: (d) \( a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31} \)
The value of a determinant can be found by expanding along any row or any column.
If expanding along a row, the formula is \( \Delta = a_{i1}A_{i1} + a_{i2}A_{i2} + a_{i3}A_{i3} \).
If expanding along a column, the formula is \( \Delta = a_{1j}A_{1j} + a_{2j}A_{2j} + a_{3j}A_{3j} \).
Let's examine the options:
(a) This is an expansion using elements of \( R_1 \) with cofactors of \( R_3 \). This would result in 0, not \( \Delta \).
(b) This mixes elements and cofactors from different positions in a way that is not a valid expansion for \( \Delta \).
(c) This mixes elements of \( R_2 \) with cofactors of \( R_1 \). This would also result in 0, not \( \Delta \).
(d) This is an expansion along the first column, \( C_1 \), using its elements \( (a_{11}, a_{21}, a_{31}) \) and their corresponding cofactors \( (A_{11}, A_{21}, A_{31}) \). This is a correct way to compute the determinant. The definition of a determinant is precisely this.
In simple words: To find the value of a determinant, you can pick any column (or row). Then, for each number in that column, multiply it by its matching "cofactor" and add all these products together.

🎯 Exam Tip: Understand the expansion of a determinant: it is the sum of the products of the elements of any row/column with their corresponding cofactors. Using elements from one row/column with cofactors from another (different) row/column will always result in a zero sum.

Question 23. If \( \left|\begin{array}{ll} x & 2 \\ 8 & 5 \end{array}\right|=0 \) then the value of x is:
(a) \( \frac{-5}{6} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{-16}{5} \)
(d) \( \frac{16}{5} \)
Answer: (d) \( \frac{16}{5} \)
Given the determinant equation: \( \left|\begin{array}{ll} x & 2 \\ 8 & 5 \end{array}\right|=0 \).
To calculate the determinant of a 2x2 matrix \( \left|\begin{array}{ll} a & b \\ c & d \end{array}\right| \), we use the formula \( ad - bc \).
Applying this to the given equation:
\( (x \times 5) - (2 \times 8) = 0 \)
\( 5x - 16 = 0 \).
Now, solve for x:
Add 16 to both sides: \( 5x = 16 \).
Divide by 5: \( x = \frac{16}{5} \). Solving this linear equation gives the unique value for x.
In simple words: To find 'x', multiply the numbers on the main diagonal, then subtract the product of the other two numbers. Since the determinant is zero, set the result to zero and solve for 'x'.

🎯 Exam Tip: Always set up the determinant equation correctly as \( ad - bc = 0 \). Be careful with signs when performing the multiplication and subtraction steps.

Question 24. If \( \left|\begin{array}{ll} 4 & 3 \\ 3 & 1 \end{array}\right| = -5 \) then the value of \( \left|\begin{array}{rr} 20 & 15 \\ 15 & 5 \end{array}\right| \) is:
(a) -5
(b) -125
(c) -25
(d) 0
Answer: (b) -125
Let the given determinant be \( D = \left|\begin{array}{rr} 20 & 15 \\ 15 & 5 \end{array}\right| \).
We can factor out common terms from the rows or columns.
Factor 5 from the first row \( R_1 \): \( \left|\begin{array}{rr} 20 & 15 \\ 15 & 5 \end{array}\right| = 5 \left|\begin{array}{rr} 4 & 3 \\ 15 & 5 \end{array}\right| \).
Now, factor 5 from the second row \( R_2 \): \( 5 \times 5 \left|\begin{array}{rr} 4 & 3 \\ 3 & 1 \end{array}\right| \).
So, \( D = 25 \left|\begin{array}{rr} 4 & 3 \\ 3 & 1 \end{array}\right| \). This factorization is a key determinant property.
We are given that \( \left|\begin{array}{ll} 4 & 3 \\ 3 & 1 \end{array}\right| = -5 \).
Substitute this value: \( D = 25 \times (-5) \).
\( D = -125 \). The ability to factor constants from rows/columns simplifies calculations.
In simple words: We can take out common numbers from each row of the second determinant. After taking out the common numbers, the remaining small determinant is the same as the first one given. Then, we just multiply the numbers together.

🎯 Exam Tip: Look for common factors within rows or columns that can be factored out. This often simplifies complex determinants into a known or easily calculable form.

Question 25. If any three rows or columns of a determinant are identical then the value of the determinant is:
(a) 0
(b) 2
(c) 1
(d) 3
Answer: (a) 0
A fundamental property of determinants states that if any two rows or any two columns of a determinant are identical (or proportional), then the value of the determinant is zero.
If three rows or columns are identical, it automatically means that at least two rows or columns are identical.
For example, if \( R_1, R_2, R_3 \) are identical, then \( R_1 = R_2 \) and \( R_2 = R_3 \). Since \( R_1 = R_2 \), the determinant will be 0.
This property is a powerful shortcut for evaluating determinants. The presence of identical rows or columns implies linear dependence between them.
In simple words: When a determinant has two or more rows (or columns) that are exactly the same, its total value is always zero. If three are the same, it means two are the same, so the rule still applies.

🎯 Exam Tip: Remember the core property: a determinant is zero if any two rows or columns are identical or linearly dependent. This is a very common scenario in determinant problems.

Choose the Correct Answer.

Question 1. The value of \( \begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0 \) is
(a) \( 0, -1 \)
(b) \( 0, 1 \)
(c) \( -1, 1 \)
(d) \( -1, -1 \)
Answer: (b) \( 0, 1 \)
To find the value, we calculate the determinant of the matrix. We expand it along the first row. The minor for the element '1' in the first row is found by removing its row and column, giving \( \begin{vmatrix} x & x \\ 1 & x \end{vmatrix} = x^2 - x \). So, the determinant calculation is \( 0 - 1(x^2 - x) + 0 = 0 \). This simplifies to \( -x^2 + x = 0 \), or \( x(x - 1) = 0 \). Solving this equation, we find that \( x \) can be 0 or 1. Determinants are often used to solve systems of linear equations and check for matrix invertibility.
In simple words: To find \( x \), we calculate the determinant. The calculation leads to the equation \( x^2 - x = 0 \). Solving this means \( x \) can be 0 or 1.

🎯 Exam Tip: When solving for \( x \) in a determinant equation, always remember to factorize the resulting polynomial to find all possible values of \( x \).

Question 2. The value of \( \begin{vmatrix} 2x+y & x & y \\ 2y+z & y & z \\ 2z+x & z & x \end{vmatrix} \) is
(a) \( xyz \)
(b) \( x + y + z \)
(c) \( 2x+2y+2z \)
(d) \( 0 \)
Answer: (d) \( 0 \)
To find the value of the determinant, we apply a column operation. We perform \( C_1 \rightarrow C_1 - C_3 \), meaning we subtract the elements of column 3 from the corresponding elements of column 1. \[ \begin{vmatrix} 2x+y-y & x & y \\ 2y+z-z & y & z \\ 2z+x-x & z & x \end{vmatrix} = \begin{vmatrix} 2x & x & y \\ 2y & y & z \\ 2z & z & x \end{vmatrix} \] Now, observe the new matrix. The first column is \( (2x, 2y, 2z) \) and the second column is \( (x, y, z) \). Clearly, the first column is exactly twice the second column (i.e., \( C_1 = 2C_2 \)). A property of determinants states that if any two columns (or rows) of a determinant are proportional, the value of the determinant is 0. This property greatly simplifies determinant calculations.
In simple words: We subtract column 3 from column 1. After this step, the first column becomes double the second column. If two columns in a determinant are related this way, the determinant's value is always zero.

🎯 Exam Tip: Look for opportunities to use determinant properties, such as row/column operations or proportionality, as they often simplify complex calculations to zero.

Question 3. The cofactor of -7 in the determinant \( \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \) is
(a) \( -18 \)
(b) \( 18 \)
(c) \( -7 \)
(d) \( 7 \)
Answer: (b) \( 18 \)
The element -7 is in the 3rd row and 3rd column, so its position is \( a_{33} \). The cofactor \( C_{ij} \) is calculated as \( (-1)^{i+j} \times M_{ij} \), where \( M_{ij} \) is the minor of the element. For \( a_{33} \), we have \( i=3 \) and \( j=3 \), so \( (-1)^{3+3} = (-1)^6 = 1 \). The minor \( M_{33} \) is the determinant of the submatrix obtained by removing the 3rd row and 3rd column: \[ M_{33} = \begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} \] Now, we calculate this minor: \( (2 \times 0) - (-3 \times 6) = 0 - (-18) = 18 \). Therefore, the cofactor of -7 is \( 1 \times 18 = 18 \). Understanding the sign rule is essential for finding cofactors correctly.
In simple words: To find the cofactor of -7, we find its position (row 3, column 3). We then cover that row and column and find the determinant of the smaller matrix left. The sign for this position is positive. The determinant of the smaller matrix is \( (2 \times 0) - (-3 \times 6) = 18 \).

🎯 Exam Tip: Remember the formula \( (-1)^{i+j}M_{ij} \) for cofactors, and be careful with the signs, especially for elements in odd \( i+j \) sum positions.

Question 4. If \( \Delta = \begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} \) then \( \begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix} \) is
(a) \( \Delta \)
(b) \( -\Delta \)
(c) \( 3\Delta \)
(d) \( -3\Delta \)
Answer: (b) \( -\Delta \)
We are given the determinant \( \Delta = \begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} \). We need to find the value of \( \begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix} \). If we compare the two determinants, we can see that the first two rows of \( \Delta \) have been interchanged to form the new determinant. Specifically, the first row of the new determinant `(3, 1, 2)` is the second row of \( \Delta \), and the second row of the new determinant `(1, 2, 3)` is the first row of \( \Delta \). The third row remains unchanged. A property of determinants states that if any two rows (or columns) of a determinant are interchanged, the sign of the determinant changes. Therefore, the value of the new determinant is \( -\Delta \). This property is fundamental for manipulating determinants.
In simple words: The new determinant is created by swapping the first two rows of the original determinant \( \Delta \). When you swap two rows in a determinant, its sign changes. So, the new determinant's value is the negative of \( \Delta \).

🎯 Exam Tip: Remember that swapping two rows or two columns in a determinant changes its sign. This is a crucial property for manipulating and evaluating determinants.

Question 5. The value of the determinant \( \left( \begin{vmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{vmatrix} \right)^2 \) is
(a) \( abc \)
(b) \( 0 \)
(c) \( a^2b^2c^2 \)
(d) \( -abc \)
Answer: (c) \( a^2b^2c^2 \)
First, we need to find the value of the determinant \( \begin{vmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{vmatrix} \). This is a diagonal matrix, which is a special type of matrix where all elements outside the main diagonal are zero. The determinant of a diagonal matrix is simply the product of its diagonal elements. So, \( \begin{vmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{vmatrix} = a \times b \times c = abc \). The question then asks for the square of this value. Therefore, \( (abc)^2 = a^2b^2c^2 \). This property simplifies calculations for special matrix forms.
In simple words: We first find the determinant of the matrix. Since numbers are only on the main diagonal, its determinant is just \( a \times b \times c \). Then, we square this result, which gives us \( a^2b^2c^2 \).

🎯 Exam Tip: For diagonal or triangular matrices, the determinant is simply the product of the elements on the main diagonal. Remember to apply the exponent after finding the determinant.

Question 6. If A is square matrix of order 3 then \( |kA| \) is:
(a) \( k|A| \)
(b) \( -k|A| \)
(c) \( k^3|A| \)
(d) \( -k^3|A| \)
Answer: (c) \( k^3|A| \)
This question uses a fundamental property of determinants concerning scalar multiplication. For any square matrix A of order \( n \), and any scalar \( k \), the determinant of the matrix \( kA \) is given by the formula \( |kA| = k^n|A| \). In this specific problem, the matrix A is of order 3, which means \( n = 3 \). Substituting \( n=3 \) into the formula, we get \( |kA| = k^3|A| \). This property is very important in matrix algebra for understanding how scaling a matrix affects its determinant.
In simple words: When you multiply a matrix by a number \( k \) and then find its determinant, the result is the original determinant multiplied by \( k \) raised to the power of the matrix's size. Since the matrix is of order 3, the answer is \( k^3 \) times the determinant of A.

🎯 Exam Tip: A common mistake is to write \( k|A| \) for \( |kA| \). Always remember the power \( n \) when scaling a determinant: \( |kA| = k^n|A| \), where \( n \) is the order of the matrix.

Question 7. \( \text{adj} (AB) \) is equal to:
(a) \( \text{adj} A \text{adj} B \)
(c) \( \text{adj} B \text{adj} A \)
(d) \( \text{adj} B^T \text{adj} A^T \)
Answer: (c) \( \text{adj} B \text{adj} A \)
This question asks for the adjoint of a product of two matrices, \( AB \). A key property in matrix algebra states that the adjoint of a product of two invertible matrices \( A \) and \( B \) is given by the product of their individual adjoints in reverse order. The formula is \( \text{adj}(AB) = (\text{adj} B) (\text{adj} A) \). This property is useful in various matrix calculations and highlights that the order of multiplication for adjoints is reversed, similar to how the inverse of a product is handled \( (AB)^{-1} = B^{-1}A^{-1} \).
In simple words: When you find the adjoint of two matrices multiplied together, you multiply their individual adjoints. But you must do it in the reverse order of the original multiplication, so it's \( \text{adj} B \) first, then \( \text{adj} A \).

🎯 Exam Tip: Remember the 'reversal law' for adjoints and inverses: \( \text{adj}(AB) = \text{adj} B \text{adj} A \) and \( (AB)^{-1} = B^{-1}A^{-1} \). This is a common point of error.

Question 8. The inverse matrix of \( \begin{pmatrix} \frac{4}{5} & \frac{5}{12} \\ \frac{2}{5} & \frac{1}{2} \end{pmatrix} \) is
(a) \( \frac{7}{30}\begin{pmatrix} \frac{1}{2} & \frac{5}{12} \\ \frac{2}{5} & \frac{4}{5} \end{pmatrix} \)
(b) \( \frac{7}{30}\begin{pmatrix} \frac{1}{2} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{1}{5} \end{pmatrix} \)
(c) \( \frac{30}{7}\begin{pmatrix} \frac{1}{2} & \frac{5}{12} \\ \frac{2}{5} & \frac{4}{5} \end{pmatrix} \)
(d) \( \frac{30}{7}\begin{pmatrix} \frac{1}{2} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{4}{5} \end{pmatrix} \)
Answer: (d) \( \frac{30}{7}\begin{pmatrix} \frac{1}{2} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{4}{5} \end{pmatrix} \)
Let the given matrix be \( A = \begin{pmatrix} \frac{4}{5} & \frac{5}{12} \\ \frac{2}{5} & \frac{1}{2} \end{pmatrix} \). To find the inverse of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), we use the formula \( A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \). First, calculate the determinant \( |A| = ad - bc \):
\( |A| = \left(\frac{4}{5}\right)\left(\frac{1}{2}\right) - \left(\frac{5}{12}\right)\left(\frac{2}{5}\right) \)
\( \implies |A| = \frac{4}{10} - \frac{10}{60} = \frac{2}{5} - \frac{1}{6} \)
\( \implies |A| = \frac{12 - 5}{30} = \frac{7}{30} \)
Next, find the adjoint matrix \( \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \):
\( \text{adj}(A) = \begin{pmatrix} \frac{1}{2} & -\frac{5}{12} \\ -\frac{2}{5} & \frac{4}{5} \end{pmatrix} \)
Finally, calculate the inverse:
\( A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{\frac{7}{30}} \begin{pmatrix} \frac{1}{2} & -\frac{5}{12} \\ -\frac{2}{5} & \frac{4}{5} \end{pmatrix} \)
\( \implies A^{-1} = \frac{30}{7} \begin{pmatrix} \frac{1}{2} & -\frac{5}{12} \\ -\frac{2}{5} & \frac{4}{5} \end{pmatrix} \)
This result matches option (d). Calculating the inverse is a systematic process requiring careful fraction handling.
In simple words: We find the inverse of the matrix by first calculating its determinant, which is \( \frac{7}{30} \). Then we swap the main diagonal elements and change the signs of the other two elements to get the adjoint matrix. Multiplying the adjoint by \( \frac{1}{\text{determinant}} \) gives us the inverse, which is option (d).

🎯 Exam Tip: Be careful with fraction calculations when finding the determinant. Also, remember to swap the main diagonal elements and change the signs of the off-diagonal elements for the adjoint of a 2x2 matrix.

Question 9. If \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) such that \( ad - bc \neq 0 \) then \( A^{-1} \) is:
(a) \( \frac{1}{ad-bc}\begin{pmatrix} d & b \\ -c & a \end{pmatrix} \)
(b) \( \frac{1}{ad-bc}\begin{pmatrix} d & b \\ c & a \end{pmatrix} \)
(c) \( \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \)
(d) \( \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ c & a \end{pmatrix} \)
Answer: (c) \( \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \)
For a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), its inverse \( A^{-1} \) exists only if the determinant \( |A| = ad - bc \) is not equal to zero. The formula for the inverse is given by \( A^{-1} = \frac{1}{|A|} \times \text{adj}(A) \). The adjoint of a \( 2 \times 2 \) matrix \( \text{adj}(A) \) is found by swapping the elements on the main diagonal (\( a \) and \( d \)) and changing the signs of the off-diagonal elements (\( b \) and \( c \)). So, \( \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \). Substituting these into the formula, we get \( A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \). This is a standard and very useful formula for finding the inverse of 2x2 matrices directly.
In simple words: To find the inverse of a 2x2 matrix, you take 1 divided by the determinant \( (ad-bc) \), and then multiply it by a modified matrix. This modified matrix is made by swapping the numbers on the main diagonal and changing the signs of the other two numbers. This matches option (c).

🎯 Exam Tip: The condition \( ad - bc \neq 0 \) (i.e., determinant is not zero) is essential for a matrix to have an inverse. Always check this first.

Question 10. The number of Hawkins-Simon conditions for the viability of input-output analysis is:
(a) 1
(b) 3
(c) 4
(d) 2
Answer: (d) 2
The Hawkins-Simon conditions are a set of criteria used in economics, specifically in input-output analysis, to ensure that the economic system is viable and produces positive outputs. There are two main conditions. These conditions state that all principal minors of the Leontief inverse matrix must be positive. More simply, it means (1) the diagonal elements of the Leontief inverse matrix \( (I - A)^{-1} \) must be positive, and (2) the determinant of \( (I - A) \) must be positive. These two criteria collectively ensure the system's ability to produce a net positive output.
In simple words: There are two main Hawkins-Simon conditions used in economics. These rules help to confirm that an economy's input-output system can actually produce goods and is working well.

🎯 Exam Tip: The Hawkins-Simon conditions are crucial for understanding the feasibility of an economic input-output model. Remember there are two main conditions involving the Leontief inverse.

Question 11. The inventor of input-output analysis is:
(a) Sir Francis Galton
(b) Fisher
(c) Prof. Wassily W. Leontief
(d) Arthur Cayley
Answer: (c) Prof. Wassily W. Leontief
Input-output analysis is a technique in economics developed to study the interdependence between different sectors of an economy. The primary developer and pioneer of this economic model was Professor Wassily W. Leontief. He was awarded the Nobel Memorial Prize in Economic Sciences in 1973 for his profound contributions to this field. His work provided a comprehensive framework for understanding the complex relationships within an economy.
In simple words: Professor Wassily W. Leontief invented input-output analysis. This economic tool helps us understand how different parts of an economy rely on each other.

🎯 Exam Tip: Wassily Leontief is a key figure in economics for developing input-output analysis. Remember his name when studying this topic.

Question 12. Which of the following matrix has no inverse?
(a) \( \begin{pmatrix} -1 & 1 \\ 1 & -4 \end{pmatrix} \)
(b) \( \begin{pmatrix} 2 & -1 \\ -4 & 2 \end{pmatrix} \)
(c) \( \begin{pmatrix} \cos a & \sin a \\ -\sin a & \cos a \end{pmatrix} \)
(d) \( \begin{pmatrix} \sin a & \sin a \\ -\cos a & \cos a \end{pmatrix} \)
Answer: (b) \( \begin{pmatrix} 2 & -1 \\ -4 & 2 \end{pmatrix} \)
A matrix has no inverse if and only if its determinant is zero. We need to calculate the determinant for each option to find which one is zero.
(a) For \( \begin{pmatrix} -1 & 1 \\ 1 & -4 \end{pmatrix} \), the determinant is \( (-1)(-4) - (1)(1) = 4 - 1 = 3 \). Since \( 3 \neq 0 \), an inverse exists.
(b) For \( \begin{pmatrix} 2 & -1 \\ -4 & 2 \end{pmatrix} \), the determinant is \( (2)(2) - (-1)(-4) = 4 - 4 = 0 \). Since the determinant is 0, this matrix has no inverse.
(c) For \( \begin{pmatrix} \cos a & \sin a \\ -\sin a & \cos a \end{pmatrix} \), the determinant is \( (\cos a)(\cos a) - (\sin a)(-\sin a) = \cos^2 a + \sin^2 a = 1 \). Since \( 1 \neq 0 \), an inverse exists.
(d) For \( \begin{pmatrix} \sin a & \sin a \\ -\cos a & \cos a \end{pmatrix} \), the determinant is \( (\sin a)(\cos a) - (\sin a)(-\cos a) = \sin a \cos a + \sin a \cos a = 2 \sin a \cos a \). This determinant is not generally zero unless \( \sin a = 0 \) or \( \cos a = 0 \).
Thus, matrix (b) is the only one with a determinant of zero, meaning it does not have an inverse. This is a crucial property in linear algebra.
In simple words: A matrix has no inverse if its determinant is zero. We checked each option: matrix (a) has determinant 3, matrix (b) has determinant 0, matrix (c) has determinant 1, and matrix (d) has determinant \( 2 \sin a \cos a \). Only matrix (b) has a determinant of zero, so it has no inverse.

🎯 Exam Tip: To check if a matrix has an inverse, always calculate its determinant. If the determinant is zero, the inverse does not exist. Remember this fundamental rule.

Question 13. Inverse of \( \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \) is:
(a) \( \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \)
(b) \( \begin{pmatrix} -2 & 5 \\ 1 & -3 \end{pmatrix} \)
(c) \( \begin{pmatrix} 3 & -1 \\ -5 & -3 \end{pmatrix} \)
(d) \( \begin{pmatrix} -3 & 5 \\ 1 & -2 \end{pmatrix} \)
Answer: (a) \( \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \)
Let the given matrix be \( A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \). To find its inverse, we first calculate the determinant \( |A| \). For a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), the determinant is \( ad - bc \). So, \( |A| = (3 \times 2) - (1 \times 5) = 6 - 5 = 1 \). Since the determinant is 1 (non-zero), the inverse exists. Next, we find the adjoint of A, \( \text{adj}(A) \). For a \( 2 \times 2 \) matrix, we swap the diagonal elements (\( a \) and \( d \)) and change the signs of the off-diagonal elements (\( b \) and \( c \)). So, \( \text{adj}(A) = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \). The inverse \( A^{-1} \) is \( \frac{1}{|A|} \times \text{adj}(A) \). Therefore, \( A^{-1} = \frac{1}{1} \times \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \). This process is a direct application of the 2x2 inverse formula.
In simple words: First, we find the determinant of the matrix, which is 1. Then, we swap the main diagonal numbers and change the signs of the other two numbers. Since the determinant is 1, the inverse is simply this new matrix, which matches option (a).

🎯 Exam Tip: When the determinant is 1, the inverse of a 2x2 matrix is simply its adjoint. This shortcut can save time in calculations.

Question 14. If \( A = \begin{pmatrix} -1 & 2 \\ 1 & -4 \end{pmatrix} \) then \( A (\text{adj} A) \) is:
(a) \( \begin{pmatrix} -4 & -2 \\ -1 & -1 \end{pmatrix} \)
(b) \( \begin{pmatrix} 4 & -2 \\ -1 & 1 \end{pmatrix} \)
(c) \( \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \)
(d) \( \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix} \)
Answer: (c) \( \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \)
We are given the matrix \( A = \begin{pmatrix} -1 & 2 \\ 1 & -4 \end{pmatrix} \). We need to find the product \( A (\text{adj} A) \). A very important property in matrix algebra states that for any square matrix A, the product of A and its adjoint \( (\text{adj} A) \) is equal to the determinant of A multiplied by the identity matrix I. That is, \( A (\text{adj} A) = |A| I \). First, calculate the determinant of A: \( |A| = (-1)(-4) - (2)(1) = 4 - 2 = 2 \). The identity matrix I for a \( 2 \times 2 \) matrix is \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \). Now, substitute these values into the property: \( A (\text{adj} A) = 2 \times \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).
\( \implies A (\text{adj} A) = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \). This result matches option (c). This property is often used to avoid direct calculation of the adjoint matrix.
In simple words: A special rule says that a matrix multiplied by its adjoint is equal to its determinant times the identity matrix. The determinant of A is 2. The identity matrix is \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \). So, the answer is \( 2 \times \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \), which is \( \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \).

🎯 Exam Tip: Memorize the fundamental property \( A (\text{adj} A) = (\text{adj} A) A = |A| I \). This relation frequently appears in questions and simplifies calculations significantly.

Question 15. If A and B are non-singular matrices, then which of the following is incorrect?
(a) \( A^2 = I \) implies \( A^{-1} = A \)
(b) \( I^{-1} = I \)
(c) If \( AX = B \) then \( X = B^{-1}A \)
(d) If A is square matrix of order 3 then \( |\text{adj} A| = |A|^2 \)
Answer: (c) If \( AX = B \) then \( X = B^{-1}A \)
We need to identify the incorrect statement among the given options for non-singular matrices.
(a) If \( A^2 = I \), multiplying both sides by \( A^{-1} \) (which exists because A is non-singular) gives \( A^{-1}A^2 = A^{-1}I \implies A = A^{-1} \). This statement is **correct**.
(b) The inverse of an identity matrix \( I \) is always itself, \( I^{-1} = I \). This statement is **correct**.
(c) To solve for \( X \) in the matrix equation \( AX = B \), we must pre-multiply (multiply from the left) both sides by \( A^{-1} \). So, \( A^{-1}(AX) = A^{-1}B \implies (A^{-1}A)X = A^{-1}B \implies IX = A^{-1}B \implies X = A^{-1}B \). The statement given, \( X = B^{-1}A \), is **incorrect** because matrix multiplication order matters.
(d) For a square matrix A of order \( n \), the property is \( |\text{adj} A| = |A|^{n-1} \). Since A is of order 3, \( n=3 \). So, \( |\text{adj} A| = |A|^{3-1} = |A|^2 \). This statement is **correct**.
Therefore, statement (c) is the incorrect one.
In simple words: We are looking for the false statement. Statements (a), (b), and (d) are true properties of matrices. Statement (c) is wrong because to solve \( AX=B \) for \( X \), you need to multiply by \( A^{-1} \) from the left, not \( B^{-1} \) from the left and \( A \) from the right. The correct answer should be \( X = A^{-1}B \).

🎯 Exam Tip: Matrix multiplication is not commutative, meaning \( AB \neq BA \) in general. Pay close attention to the order of matrix multiplication, especially when dealing with inverses and solving matrix equations.

Question 16. The value of \( \begin{vmatrix} 5 & 5 & 5 \\ 4x & 4y & 4z \\ -3x & -3y & -3z \end{vmatrix} \) is:
(a) 5
(b) 4
(c) 0
(d) -3
Answer: (c) 0
We need to evaluate the determinant \( \begin{vmatrix} 5 & 5 & 5 \\ 4x & 4y & 4z \\ -3x & -3y & -3z \end{vmatrix} \). We can simplify this by taking common factors out of each row. Take 5 as a common factor from Row 1 (\( R_1 \)). Take 4 as a common factor from Row 2 (\( R_2 \)). Take -3 as a common factor from Row 3 (\( R_3 \)). \[ 5 \times 4 \times (-3) \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x & y & z \end{vmatrix} \] Now, observe the new determinant: Row 2 is \( (x, y, z) \) and Row 3 is also \( (x, y, z) \). A fundamental property of determinants states that if any two rows (or columns) are identical, the value of the determinant is 0. Since Row 2 and Row 3 are identical, the determinant \( \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x & y & z \end{vmatrix} = 0 \). Therefore, the entire expression becomes \( 5 \times 4 \times (-3) \times 0 = 0 \). Recognizing identical rows simplifies the calculation immensely.
In simple words: We can take out common numbers from each row: 5 from the first, 4 from the second, and -3 from the third. This makes the second and third rows exactly the same. When a determinant has two identical rows, its value is always zero, so the final answer is 0.

🎯 Exam Tip: Always look for common factors in rows or columns, and check for identical or proportional rows/columns. These properties quickly simplify determinants to zero, saving complex expansion calculations.

Question 17. If A is an invertible matrix of order 2 then \( \text{det} (A^{-1}) \) be equal to:
(a) \( \text{det} (A) \)
(b) \( \frac{1}{\text{det}(A)} \)
(c) 1
(d) 0
Answer: (b) \( \frac{1}{\text{det}(A)} \)
We are given that A is an invertible matrix. By definition, an invertible matrix A has an inverse \( A^{-1} \) such that \( AA^{-1} = I \), where I is the identity matrix. We can take the determinant of both sides of the equation \( AA^{-1} = I \): \( \text{det}(AA^{-1}) = \text{det}(I) \). Using the property that \( \text{det}(AB) = \text{det}(A) \times \text{det}(B) \), we get \( \text{det}(A) \times \text{det}(A^{-1}) = \text{det}(I) \). The determinant of an identity matrix is always 1, regardless of its order. So, \( \text{det}(I) = 1 \). Therefore, \( \text{det}(A) \times \text{det}(A^{-1}) = 1 \). Since A is invertible, \( \text{det}(A) \neq 0 \), so we can divide by \( \text{det}(A) \) to find \( \text{det}(A^{-1}) = \frac{1}{\text{det}(A)} \). This relationship is fundamental for understanding matrix inverses.
In simple words: The determinant of an inverse matrix is always 1 divided by the determinant of the original matrix. This is because when you multiply a matrix by its inverse, the result is the identity matrix, whose determinant is always 1.

🎯 Exam Tip: Remember the key property \( \text{det}(A^{-1}) = \frac{1}{\text{det}(A)} \). This saves you from having to calculate the inverse matrix first before finding its determinant.

Question 18. If A is a \( 3 \times 3 \) matrix and \( |A| = 4 \) then \( |A^{-1}| \) is equal to:
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{16} \)
(d) 4
Answer: (a) \( \frac{1}{4} \)
We are given that A is a \( 3 \times 3 \) matrix and its determinant \( |A| = 4 \). We need to find the determinant of its inverse, \( |A^{-1}| \). From the properties of determinants, we know that for any invertible matrix A, the determinant of its inverse is the reciprocal of the determinant of the original matrix. That is, \( |A^{-1}| = \frac{1}{|A|} \). Substituting the given value \( |A| = 4 \), we get \( |A^{-1}| = \frac{1}{4} \). The order of the matrix (3x3 in this case) does not affect this specific relationship, making it a general rule.
In simple words: The determinant of an inverse matrix is always 1 divided by the determinant of the original matrix. Since the determinant of A is 4, the determinant of its inverse, \( |A^{-1}| \), will be \( \frac{1}{4} \).

🎯 Exam Tip: This relationship \( |A^{-1}| = \frac{1}{|A|} \) is fundamental and holds true for any invertible square matrix, regardless of its order.

Question 19. If A is a square matrix of order 3 and \( |A| = 3 \) then \( |\text{adj} A| \) is equal to:
(a) 81
(b) 27
(c) 3
(d) 9
Answer: (d) 9
We are given a square matrix A of order 3, and its determinant \( |A| = 3 \). We need to find the determinant of its adjoint, \( |\text{adj} A| \). There is a property that states for any square matrix A of order \( n \), the determinant of its adjoint is given by \( |\text{adj} A| = |A|^{n-1} \). In this problem, the order of the matrix \( n = 3 \). Substituting the values, we get \( |\text{adj} A| = |A|^{3-1} = |A|^2 \). Since \( |A| = 3 \), we have \( |\text{adj} A| = 3^2 = 9 \). This property is very useful for quickly determining the determinant of the adjoint without first calculating the adjoint matrix itself.
In simple words: To find the determinant of the adjoint of a matrix, we use a rule: it's the original determinant raised to the power of (the matrix's order minus one). Here, the matrix is order 3 and its determinant is 3. So, we calculate \( 3 \) raised to the power of \( (3-1) \), which is \( 3^2 = 9 \).

🎯 Exam Tip: Crucially remember the formula \( |\text{adj} A| = |A|^{n-1} \) for a square matrix of order \( n \). Misplacing the exponent is a common error.

Question 20. The value of \( \begin{vmatrix} x & x^{2}-yz & 1 \\ y & y^{2}-zx & 1 \\ z & z^{2}-xy & 1 \end{vmatrix} \) is:
(a) 1
(b) 0
(c) -1
(d) -xyz
Answer: (b) 0
Let the given determinant be \( D \). \[ D = \begin{vmatrix} x & x^{2}-yz & 1 \\ y & y^{2}-zx & 1 \\ z & z^{2}-xy & 1 \end{vmatrix} \] We can simplify this determinant using row operations. Perform the operations \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \). This will create zeros in the third column, simplifying expansion. \[ D = \begin{vmatrix} x & x^{2}-yz & 1 \\ y-x & (y^{2}-zx)-(x^{2}-yz) & 0 \\ z-x & (z^{2}-xy)-(x^{2}-yz) & 0 \end{vmatrix} \] Next, we simplify the elements in the second column:
\( (y^{2}-zx)-(x^{2}-yz) = y^2 - x^2 - zx + yz = (y-x)(y+x) + z(y-x) = (y-x)(y+x+z) \)
\( (z^{2}-xy)-(x^{2}-yz) = z^2 - x^2 - xy + yz = (z-x)(z+x) + y(z-x) = (z-x)(z+x+y) \)
Substitute these simplified expressions back into the determinant: \[ D = \begin{vmatrix} x & x^{2}-yz & 1 \\ y-x & (y-x)(x+y+z) & 0 \\ z-x & (z-x)(x+y+z) & 0 \end{vmatrix} \] Now, expand the determinant along the third column. Only the element '1' contributes:
\( D = 1 \times \begin{vmatrix} y-x & (y-x)(x+y+z) \\ z-x & (z-x)(x+y+z) \end{vmatrix} \) Take \( (y-x) \) common from the first row and \( (z-x) \) common from the second row of this \( 2 \times 2 \) determinant: \( D = (y-x)(z-x) \begin{vmatrix} 1 & x+y+z \\ 1 & x+y+z \end{vmatrix} \) Observe that the two rows in this \( 2 \times 2 \) determinant are identical. When any two rows (or columns) of a determinant are identical, its value is 0. So, \( D = (y-x)(z-x) \times 0 = 0 \). This type of problem often leads to a zero determinant when such patterns are revealed through row operations.
In simple words: We simplify the determinant by subtracting rows. This reveals common factors in the second column. After factoring, we see that two rows in the resulting smaller determinant are identical. When any two rows of a determinant are identical, its value is always zero.

🎯 Exam Tip: For determinants with complex algebraic terms, look for row/column operations that can reveal common factors or make rows/columns identical. This often leads to a quick solution of zero.

Question 21. If \( A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \), then \( |2A| \) is equal to:
(a) \( 4 \cos 2\theta \)
(b) 4
(c) 2
(d) 1
Answer: (b) 4
We are given the matrix \( A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \). We need to find \( |2A| \). First, recall the property for the determinant of a scalar multiple of a matrix: if A is a square matrix of order \( n \) and \( k \) is a scalar, then \( |kA| = k^n|A| \). In this case, the order of matrix A is \( n=2 \), and the scalar is \( k=2 \). So, \( |2A| = 2^2|A| = 4|A| \). Next, calculate the determinant of A: \( |A| = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) \implies |A| = \cos^2 \theta + \sin^2 \theta \). Using the fundamental trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \). So, \( |A| = 1 \). Finally, substitute \( |A|=1 \) back into the expression for \( |2A| \): \( |2A| = 4 \times 1 = 4 \). This problem effectively combines matrix properties with a basic trigonometric identity.
In simple words: The determinant of \( 2A \) is equal to \( 2^2 \) (which is 4) times the determinant of A. The determinant of A itself is \( \cos^2 \theta + \sin^2 \theta \), which always equals 1. So, \( |2A| \) is \( 4 \times 1 = 4 \).

🎯 Exam Tip: Don't forget the power \( n \) when using the property \( |kA| = k^n|A| \). Also, remember fundamental trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \), as they often appear in matrix problems.

Question 22. If \( \Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \) and \( A_{ij} \) is cofactor of \( a_{ij} \), then value of \( \Delta \) is given by:
(a) \( a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} \)
(b) \( a_{11}A_{11} + a_{12}A_{21} + a_{13}A_{31} \)
(c) \( a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13} \)
(d) \( a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31} \)
Answer: (d) \( a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31} \)
The value of a determinant can be found by expanding it along any row or any column. The rule for expansion is to multiply each element of a chosen row or column by its corresponding cofactor and then sum these products. Let's analyze the given options:
(a) \( a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} \) - This expression incorrectly combines elements of row 1 with cofactors of row 3. This sum is always 0, not the determinant's value.
(b) \( a_{11}A_{11} + a_{12}A_{21} + a_{13}A_{31} \) - This incorrectly mixes elements of row 1 with cofactors from different positions.
(c) \( a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13} \) - This incorrectly mixes elements of row 2 with cofactors of row 1.
(d) \( a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31} \) - This expression represents the sum of products of elements of column 1 with their *corresponding* cofactors (element \( a_{11} \) with cofactor \( A_{11} \), \( a_{21} \) with \( A_{21} \), and \( a_{31} \) with \( A_{31} \)). This is the correct expansion along Column 1. This formula is a key method for calculating determinants.
In simple words: To find the determinant, you multiply each number in a row or column by its own special 'cofactor' and then add them up. Option (d) correctly does this for the first column: \( a_{11} \) times \( A_{11} \), plus \( a_{21} \) times \( A_{21} \), plus \( a_{31} \) times \( A_{31} \).

🎯 Exam Tip: The determinant can be expanded along *any* row or *any* column. However, it's crucial to match each element \( a_{ij} \) with its *corresponding* cofactor \( A_{ij} \). If you use an element from one row/column with a cofactor from a different row/column, the sum will be zero.

Question 23. If \( \begin{vmatrix} x & 2 \\ 8 & 5 \end{vmatrix}=0 \) then the value of \( x \) is:
(a) \( \frac{-5}{6} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{-16}{5} \)
(d) \( \frac{16}{5} \)
Answer: (d) \( \frac{16}{5} \)
We are given a \( 2 \times 2 \) determinant that is equal to 0: \( \begin{vmatrix} x & 2 \\ 8 & 5 \end{vmatrix}=0 \). To evaluate a \( 2 \times 2 \) determinant \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} \), we calculate \( ad - bc \). Applying this formula to our given determinant, we get: \( (x \times 5) - (2 \times 8) = 0 \).
\( \implies 5x - 16 = 0 \). Now, we solve this simple linear equation for \( x \): \( 5x = 16 \implies x = \frac{16}{5} \). This value of \( x \) makes the determinant zero, which is a straightforward algebraic solution.
In simple words: To find \( x \), we calculate the determinant of the 2x2 matrix. This is done by multiplying the diagonal elements and subtracting. So, \( (x \times 5) - (2 \times 8) = 0 \). Solving \( 5x - 16 = 0 \) gives us \( x = \frac{16}{5} \).

🎯 Exam Tip: Remember the \( ad-bc \) formula for \( 2 \times 2 \) determinants. This is a fundamental calculation in matrix algebra and is often used in problem-solving.

Question 24. If \( \begin{vmatrix} 4 & 3 \\ 3 & 1 \end{vmatrix} = -5 \) then the value of \( \begin{vmatrix} 20 & 15 \\ 15 & 5 \end{vmatrix} \) is:
(a) 5
(b) -125
(c) -25
(d) 0
Answer: (b) -125
We are given that \( \begin{vmatrix} 4 & 3 \\ 3 & 1 \end{vmatrix} = -5 \). We need to find the value of \( \begin{vmatrix} 20 & 15 \\ 15 & 5 \end{vmatrix} \). We can observe the second determinant and factor out common multipliers from its rows. In the first row, we can factor out 5, since \( 20 = 5 \times 4 \) and \( 15 = 5 \times 3 \). In the second row, we can also factor out 5, since \( 15 = 5 \times 3 \) and \( 5 = 5 \times 1 \). A property of determinants states that if you take a common factor \( k \) from a row (or column), it multiplies the determinant by \( k \). If you take factors from multiple rows, they multiply together.
So, \( \begin{vmatrix} 20 & 15 \\ 15 & 5 \end{vmatrix} = 5 \times 5 \times \begin{vmatrix} 4 & 3 \\ 3 & 1 \end{vmatrix} \). Now, we can substitute the given value of \( \begin{vmatrix} 4 & 3 \\ 3 & 1 \end{vmatrix} = -5 \). Therefore, \( \begin{vmatrix} 20 & 15 \\ 15 & 5 \end{vmatrix} = 5 \times 5 \times (-5) \implies = 25 \times (-5) = -125 \). This showcases how factoring out scalars can simplify determinant calculations significantly.
In simple words: The second determinant has numbers that are 5 times larger than those in the given first determinant. We can take out a factor of 5 from the first row and another 5 from the second row. This leaves us with the first determinant, which has a value of -5. So, we calculate \( 5 \times 5 \times (-5) = -125 \).

🎯 Exam Tip: When evaluating determinants, always look for common factors in rows or columns that can be factored out. This can reduce complex determinants to simpler, known forms and prevent calculation errors.

Question 25. If any three rows or columns of a determinant are identical, then the value of the determinant is:
(a) 0
(b) 2
(c) 1
(d) 3
Answer: (a) 0
This question refers to a fundamental property of determinants. The property states that if any two rows or any two columns of a determinant are identical (meaning all corresponding elements are exactly the same), then the value of the determinant is zero. If three rows or columns are identical, it automatically means that any pair among those three rows or columns are also identical. For example, if Row 1, Row 2, and Row 3 are identical, then Row 1 and Row 2 are identical, and this is enough to make the determinant zero. Since the condition for "any two rows/columns being identical" is met, the determinant must be 0. This property significantly simplifies determinant evaluation.
In simple words: A basic rule for determinants says that if any two of its rows or columns are exactly the same, the determinant's value is zero. If three rows or columns are identical, it just means that at least two of them are the same, so the rule still applies, and the determinant's value will be zero.

🎯 Exam Tip: This property is a quick way to evaluate determinants without complex calculations. Always check for identical or proportional rows/columns, as they immediately imply a determinant value of zero.