RBSE Solutions Class 12 Maths Chapter 4 Determinants Exercise 4.1

Question 1. For which value of k, det \( \begin{vmatrix} k & 2 \\ 4 & -3 \end{vmatrix} \) will be zero?
Answer: To find the value of k for which the determinant is zero, we first calculate the determinant. The determinant of a 2x2 matrix \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} \) is \( ad - bc \).
So, for the given determinant:
\( \begin{vmatrix} k & 2 \\ 4 & -3 \end{vmatrix} = k \times (-3) - 2 \times 4 \)
\( = -3k - 8 \)
Now, we set this equal to zero to find k:
\( -3k - 8 = 0 \)
\( \implies -3k = 8 \)
\( \implies k = -\frac{8}{3} \)
This value of k makes the determinant equal to zero, which is a key property in linear algebra.
In simple words: We calculate the determinant by multiplying the numbers diagonally and subtracting. Then, we set this calculation equal to zero and solve for 'k'.

🎯 Exam Tip: Remember the formula for a 2x2 determinant: \( ad - bc \). This is crucial for solving such problems quickly and accurately.

Question 2. If \( \begin{vmatrix} x & y \\ 2 & 4 \end{vmatrix} = 0 \), find x : y.
Answer: We are given that the determinant of the matrix is zero. First, we calculate the determinant of the 2x2 matrix:
\( \begin{vmatrix} x & y \\ 2 & 4 \end{vmatrix} = (x \times 4) - (y \times 2) \)
\( = 4x - 2y \)
Since the determinant is equal to 0, we can write:
\( 4x - 2y = 0 \)
Now, we need to find the ratio x : y. We can move \( 2y \) to the other side of the equation:
\( 4x = 2y \)
To find the ratio \( x : y \), we divide both sides by \( y \) and by 4 (or simplify first by dividing by 2):
\( \frac{x}{y} = \frac{2}{4} \)
\( \implies \frac{x}{y} = \frac{1}{2} \)
This means the ratio \( x : y \) is \( 1 : 2 \). This type of problem highlights the relationship between elements that makes a matrix singular.
In simple words: Calculate the determinant and set it to zero. Then, rearrange the equation to find how 'x' and 'y' relate to each other as a simple ratio.

🎯 Exam Tip: Always simplify ratios to their lowest terms. Make sure to clearly show the steps for rearranging the equation.

Question 3. If \( \begin{vmatrix} 2 & 3 \\ y & x \end{vmatrix} = 4 \) and \( \begin{vmatrix} x & y \\ 4 & 2 \end{vmatrix} = 7 \), then find the value of x and y.
Answer: We have two determinant equations, and we need to solve them to find the values of x and y.
First equation:
\( \begin{vmatrix} 2 & 3 \\ y & x \end{vmatrix} = 4 \)
\( (2 \times x) - (3 \times y) = 4 \)
\( \implies 2x - 3y = 4 \) ...(i)
Second equation:
\( \begin{vmatrix} x & y \\ 4 & 2 \end{vmatrix} = 7 \)
\( (x \times 2) - (y \times 4) = 7 \)
\( \implies 2x - 4y = 7 \) ...(ii)
Now we have a system of two linear equations:
1. \( 2x - 3y = 4 \)
2. \( 2x - 4y = 7 \)
We can solve this system by subtracting equation (ii) from equation (i) to eliminate x:
\( (2x - 3y) - (2x - 4y) = 4 - 7 \)
\( 2x - 3y - 2x + 4y = -3 \)
\( \implies y = -3 \)
Now, substitute the value of \( y = -3 \) into equation (i) to find x:
\( 2x - 3(-3) = 4 \)
\( 2x + 9 = 4 \)
\( \implies 2x = 4 - 9 \)
\( \implies 2x = -5 \)
\( \implies x = -\frac{5}{2} \)
Thus, the values are \( x = -\frac{5}{2} \) and \( y = -3 \). This method uses the power of determinants to set up a system of equations.
In simple words: Convert each determinant into a simple equation. You will get two equations with 'x' and 'y'. Solve these two equations together to find the values for 'x' and 'y'.

🎯 Exam Tip: When solving a system of equations, be careful with signs, especially when subtracting one equation from another. Double-check your arithmetic.

Question 4. Find x, if \( \begin{vmatrix} x-1 & x-2 \\ x & x-3 \end{vmatrix} = 0 \)
Answer: We are given that the determinant of the matrix is zero. Let's calculate the determinant:
\( \begin{vmatrix} x-1 & x-2 \\ x & x-3 \end{vmatrix} = (x-1)(x-3) - (x-2)(x) \)
Now, expand the terms:
\( (x^2 - 3x - x + 3) - (x^2 - 2x) \)
\( = (x^2 - 4x + 3) - (x^2 - 2x) \)
\( = x^2 - 4x + 3 - x^2 + 2x \)
\( = -2x + 3 \)
Since the determinant is 0, we set this expression equal to zero:
\( -2x + 3 = 0 \)
Now, solve for x:
\( -2x = -3 \)
\( \implies x = \frac{-3}{-2} \)
\( \implies x = \frac{3}{2} \)
The value of x that makes the determinant zero is \( \frac{3}{2} \). Understanding how to set up and solve quadratic-like expressions from determinants is important.
In simple words: First, calculate the determinant by cross-multiplying and subtracting. You will get an equation with 'x'. Set this equation to zero and solve for 'x'.

🎯 Exam Tip: Be very careful when expanding and simplifying algebraic expressions, especially with negative signs. Distribute terms correctly.

Question 5. Write minors and co-factors of following determinants corresponding to first column, also find the value of determinants:
(i) \( \begin{vmatrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{vmatrix} \)
(ii) \( \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \)
Answer:
(i) Let \( A = \begin{vmatrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{vmatrix} \)
We need to find minors and co-factors corresponding to the first column (elements \( a_{11}, a_{21}, a_{31} \)).
Minor of \( a_{11} \) (element 1): \( M_{11} = \begin{vmatrix} -1 & 2 \\ 5 & 2 \end{vmatrix} = (-1 \times 2) - (2 \times 5) = -2 - 10 = -12 \)
Cofactor of \( a_{11} \) (element 1): \( F_{11} = (-1)^{1+1} M_{11} = (1) \times (-12) = -12 \)
Minor of \( a_{21} \) (element 4): \( M_{21} = \begin{vmatrix} -3 & 2 \\ 5 & 2 \end{vmatrix} = (-3 \times 2) - (2 \times 5) = -6 - 10 = -16 \)
Cofactor of \( a_{21} \) (element 4): \( F_{21} = (-1)^{2+1} M_{21} = (-1) \times (-16) = 16 \)
Minor of \( a_{31} \) (element 3): \( M_{31} = \begin{vmatrix} -3 & 2 \\ -1 & 2 \end{vmatrix} = (-3 \times 2) - (2 \times -1) = -6 - (-2) = -6 + 2 = -4 \)
Cofactor of \( a_{31} \) (element 3): \( F_{31} = (-1)^{3+1} M_{31} = (1) \times (-4) = -4 \)
Now, we find the value of the determinant \( |A| \) by expanding along the first column:
\( |A| = a_{11}F_{11} + a_{21}F_{21} + a_{31}F_{31} \)
\( |A| = 1 \times (-12) + 4 \times (16) + 3 \times (-4) \)
\( |A| = -12 + 64 - 12 \)
\( |A| = 40 \)

(ii) Let \( A = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \)
We need to find minors and co-factors corresponding to the first column (elements \( a_{11}, a_{21}, a_{31} \)).
Minor of \( a_{11} \) (element a): \( M_{11} = \begin{vmatrix} b & f \\ f & c \end{vmatrix} = (b \times c) - (f \times f) = bc - f^2 \)
Cofactor of \( a_{11} \) (element a): \( F_{11} = (-1)^{1+1} M_{11} = (1) \times (bc - f^2) = bc - f^2 \)
Minor of \( a_{21} \) (element h): \( M_{21} = \begin{vmatrix} h & g \\ f & c \end{vmatrix} = (h \times c) - (g \times f) = hc - fg \)
Cofactor of \( a_{21} \) (element h): \( F_{21} = (-1)^{2+1} M_{21} = (-1) \times (hc - fg) = fg - hc \)
Minor of \( a_{31} \) (element g): \( M_{31} = \begin{vmatrix} h & g \\ b & f \end{vmatrix} = (h \times f) - (g \times b) = hf - bg \)
Cofactor of \( a_{31} \) (element g): \( F_{31} = (-1)^{3+1} M_{31} = (1) \times (hf - bg) = hf - bg \)
Now, we find the value of the determinant \( |A| \) by expanding along the first column:
\( |A| = a_{11}F_{11} + a_{21}F_{21} + a_{31}F_{31} \)
\( |A| = a(bc - f^2) + h(fg - hc) + g(hf - bg) \)
\( |A| = abc - af^2 + hfg - h^2c + ghf - bg^2 \)
\( |A| = abc + 2fgh - af^2 - bg^2 - ch^2 \)
Calculating minors, cofactors, and the determinant is a fundamental skill in matrix algebra, showing how each element contributes to the overall value.
In simple words: For each element in the first column, find its minor by covering its row and column and calculating the smaller determinant. Then, find its cofactor by applying a sign rule (plus or minus). Finally, add up the products of each element and its cofactor to get the main determinant value.

🎯 Exam Tip: Carefully manage the signs when calculating cofactors: \( (-1)^{i+j} \) determines if the minor keeps its sign or changes it. Remember that \( (-1)^{odd} = -1 \) and \( (-1)^{even} = 1 \).

Question 6. Evaluate the following determinant: \( \begin{vmatrix} 3 & -11 & 1 \\ 5 & 0 & 0 \\ -10 & 3 & 0 \end{vmatrix} \)
Answer: We need to evaluate the determinant of the given 3x3 matrix. Since there are many zeros in the second and third columns, it's easiest to expand along the row or column that has the most zeros. In this case, Row 2 has two zeros, making it ideal.
Expanding along Row 2 (where elements are \( a_{21}=5, a_{22}=0, a_{23}=0 \)):
\( \begin{vmatrix} 3 & -11 & 1 \\ 5 & 0 & 0 \\ -10 & 3 & 0 \end{vmatrix} = a_{21}F_{21} + a_{22}F_{22} + a_{23}F_{23} \)
\( = 5 \times (-1)^{2+1} \begin{vmatrix} -11 & 1 \\ 3 & 0 \end{vmatrix} + 0 \times F_{22} + 0 \times F_{23} \)
(The terms with 0 will become 0, so we only need to calculate the first term.)
\( = 5 \times (-1) \times ((-11 \times 0) - (1 \times 3)) \)
\( = -5 \times (0 - 3) \)
\( = -5 \times (-3) \)
\( = 15 \)
The value of the determinant is 15. Expanding along the row or column with the most zeros simplifies calculations significantly.
In simple words: To find the value of the determinant, choose the row or column that has the most zeros. This makes the calculation much faster because you only need to work with the non-zero numbers.

🎯 Exam Tip: Always look for rows or columns with the most zeros before expanding a determinant. This strategy saves time and reduces the chance of calculation errors.

Question 7. Prove that: \( \begin{vmatrix} 1 & a & b \\ -a & 1 & c \\ -b & -c & 1 \end{vmatrix} = 1 + a^2 + b^2 + c^2 \)
Answer: To prove this, we will expand the determinant along the first row. The elements of the first row are \( a_{11}=1, a_{12}=a, a_{13}=b \).
\( \begin{vmatrix} 1 & a & b \\ -a & 1 & c \\ -b & -c & 1 \end{vmatrix} \)
\( = 1 \times \begin{vmatrix} 1 & c \\ -c & 1 \end{vmatrix} - a \times \begin{vmatrix} -a & c \\ -b & 1 \end{vmatrix} + b \times \begin{vmatrix} -a & 1 \\ -b & -c \end{vmatrix} \)
Now, calculate each 2x2 determinant:
\( = 1 \times ((1 \times 1) - (c \times -c)) - a \times ((-a \times 1) - (c \times -b)) + b \times ((-a \times -c) - (1 \times -b)) \)
\( = 1 \times (1 - (-c^2)) - a \times (-a - (-bc)) + b \times (ac - (-b)) \)
\( = 1 \times (1 + c^2) - a \times (-a + bc) + b \times (ac + b) \)
Now, distribute and simplify:
\( = 1 + c^2 + a^2 - abc + abc + b^2 \)
Notice that the \( -abc \) and \( +abc \) terms cancel each other out.
\( = 1 + a^2 + b^2 + c^2 \)
This matches the Right Hand Side (R.H.S.) of the equation. Hence, the identity is proven. This shows a property of skew-symmetric-like matrices.
In simple words: Expand the determinant carefully, multiplying each number in the first row by its smaller determinant (cofactor) and paying attention to the plus or minus signs. Then, simplify the whole expression; you will see that many terms cancel out, leaving the desired result.

🎯 Exam Tip: When proving identities with determinants, expand along any row or column, but be extremely careful with negative signs and algebraic simplification. One small error can prevent the final match.