JEE Mathematics Determinants and Cramers Rule MCQs Set 06

Answer: (a) 54

Explanation:
1. Apply C3 -> C3 - C2 to get third column (10, 3, 19).
2. Expand along the new column or evaluate directly: \( 2(29\cdot 87-32\cdot 68)-45(87-96)+55(68-87)=2(347)+405-1045=54 \).

Teacher's Note:
1. Column operations simplify before expanding.
2. Re-check with the sign pattern + - + along the first row.

Answer: (d) 0

Explanation:
1. Apply R2 -> R2 - R1 and R3 -> R3 - R2: both rows become (1, 1, 1).
2. Two identical rows give determinant 0.

Teacher's Note:
1. Entries in arithmetic progression along rows and columns always give a zero determinant for order 3.
2. Row operations of this type do not change the value.

Answer: (d) 2

Explanation:
1. Put \( a=b=c=1 \): the matrix is \( \begin{bmatrix} 4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4 \end{bmatrix} \) with determinant \( 4(15)-1(3)+1(-3)=54 \).
2. The right side is \( \lambda\cdot27 \), so \( \lambda=\frac{54}{27}=2 \).

Teacher's Note:
1. Both sides are homogeneous of degree 6, so testing a convenient value finds the constant.
2. The standard result of this determinant is \( 2abc(a+b+c)^3 \).

Answer: (c) 0

Explanation:
1. Row 2 is \( \omega \) times Row 1, because \( \omega\cdot\omega^2=\omega^3=1 \).
2. Proportional rows give determinant 0.

Teacher's Note:
1. Look for a row that is a multiple of another before expanding.
2. Powers of omega repeat with period 3.

Answer: (b) 0

Explanation:
1. Add all three rows: R1+R2+R3 \( = \) (0, 0, 0).
2. A determinant whose rows add to the zero row is 0.

Teacher's Note:
1. Whenever each column sums to zero, the rows are linearly dependent.
2. Row addition without replacement shows the dependence.

Answer: (c) \( 2(x+y+z)^3 \)

Explanation:
1. Apply C1 -> C1 + C2 + C3 to get first column 2(x+y+z) in every row.
2. Take 2(x+y+z) out: remaining determinant has first column (1,1,1).
3. Subtract R1 from R2 and R3, then expand to get \( (x+y+z)^2 \), so the value is \( 2(x+y+z)^3 \).

Teacher's Note:
1. Check with \( x=y=z=1 \): matrix \( \begin{bmatrix} 4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4 \end{bmatrix} \) has determinant 54 \( = 2\cdot 27 \).
2. Common factors are the first thing to look for.

Answer: (c) 0

Explanation:
1. Test an equilateral triangle: every entry equals sin 60 degrees \( = \) sqrt3/2 (sin120 \( = \) sin60).
2. All rows are identical, so the determinant is 0.
3. The general identity gives 0 for every triangle (verified numerically for scalene triangles too).

Teacher's Note:
1. A special triangle suggests the constant value.
2. Numerical checks with scalene triangles also give 0, so the value is 0 for every triangle.

Answer: (b) -8

Explanation:
1. Apply C1 -> C1 - C2 and C2 -> C2 - C3, then R1 -> R1 - R2 and R2 -> R2 - R3; the x terms cancel and the determinant reduces to a constant.
2. Evaluate the constant by putting \( x=0 \): the matrix is \( \begin{bmatrix} 4 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 4 \end{bmatrix} \).
3. Its determinant is \( 4(0-1)-1(4-0)+0=-8 \).

Teacher's Note:
1. The result is independent of x, so test \( x=0 \): matrix \( \begin{bmatrix} 4 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 4 \end{bmatrix} \) has determinant \( 4(-1)-1(4)+0=-8 \).
2. Consecutive-square determinants always reduce to a constant.

Answer: (a) -3

Explanation:
1. Singular means determinant 0.
2. Expand: \( 1(6-28)-2(-24-14)+x(16+2)=-22+76+18x=54+18x \).
3. Setting \( 54+18x=0 \) gives \( x=-3 \).

Teacher's Note:
1. A singular matrix has zero determinant.
2. Substitute \( x=-3 \) to confirm.

Answer: (a) -6

Explanation:
1. Add all columns into C1: each entry of C1 becomes 6+x.
2. Take 6+x out: (6+x) times the determinant with first column (1,1,1); subtracting rows gives \( x^2 \).
3. So the determinant is \( x^2(x+6)=0 \) and the non-zero root is \( x=-6 \).

Teacher's Note:
1. Column sums are equal to 6+x in every row, which makes the factorisation quick.
2. The other root \( x=0 \) is a double root and is excluded by the wording.

Answer: (d) 0

Explanation:
1. (log_b a)(log_a \( b)=1 \).
2. So the determinant is \( 1\cdot 1-1=0 \).

Teacher's Note:
1. The two logarithms are reciprocals of each other.
2. Change of base gives log_b a \( = 1/log_a \) b.

Answer: (b) -1, -2

Explanation:
1. Expanding the determinant (after row operations such as R2 -> R2 - R1) gives \( -3(x+1)^2(x+2) \).
2. Setting it to zero gives \( x=-1 \) or \( x=-2 \).

Teacher's Note:
1. Reduce using row operations first; direct expansion is error-prone.
2. Substitute \( x=-1 \): rows become (1,1,1),(1,1,1),(2,3,7): two equal rows give 0.

Answer: (c) \( (3abc-a^3-b^3-c^3)^2 \)

Explanation:
1. For this type of determinant, \( \Delta=(r^2-s^2)^2(r^2+2s^2) \).
2. \( r^2-s^2=a^2+b^2+c^2-ab-bc-ca \) and \( r^2+2s^2=(a+b+c)^2 \).
3. So \( \Delta=[(a+b+c)(a^2+b^2+c^2-ab-bc-ca)]^2=(a^3+b^3+c^3-3abc)^2 \).

Teacher's Note:
1. Add all columns to extract the common factor, then subtract rows.
2. The square of a factor appears because the factor r squared minus s squared occurs twice.

Answer: (c) 0

Explanation:
1. \( A^n \) is a rotation by \( n\theta \), so \( |A^n|=\cos^2n\theta+\sin^2n\theta=1 \).
2. The limit of \( \frac1n\cdot1 \) as \( n\to\infty \) is 0.

Teacher's Note:
1. A rotation matrix always has determinant 1, whatever the power.
2. Dividing a bounded quantity by n gives a limit of 0.

Answer: (a) \( (2015)^2 \)

Explanation:
1. \( |A_k|=k^2-(k-1)^2=2k-1 \).
2. The sum of the first 2015 odd numbers is \( 2015^2 \).

Teacher's Note:
1. The sum of the first n odd numbers is n squared.
2. Check small cases: the first two values are 1 and 3, and 1 + 3 \( = \) 4 \( = \) 2 squared.

Answer: (b) \( (-11)2^{2015} \)

Explanation:
1. \( A^{2015}-6A^{2014}=A^{2014}(A-6I) \).
2. \( |A|=24-22=2 \) and \( |A-6I|=\begin{vmatrix}0&11\\2&-2\end{vmatrix}=-22 \).
3. So the determinant is \( 2^{2014}\times(-22)=-11\times2^{2015} \).

Teacher's Note:
1. Factor out the lowest power of A before taking the determinant.
2. The determinant of a product is the product of the determinants.

Answer: (b) -1

Explanation:
1. Take out \( x^ny^nz^n \): the remaining determinant has columns \( 1, x^2, x^3 \) etc.
2. That determinant equals \( (x-y)(y-z)(z-x)(xy+yz+zx) \).
3. So the value is \( x^ny^nz^n(xy+yz+zx)\cdot(\ldots) \), and \( xy+yz+zx=xyz(\frac1x+\frac1y+\frac1z) \).
4. Matching requires \( x^ny^nz^n\cdot xyz=1 \), so \( n=-1 \).

Teacher's Note:
1. Factor out the common powers from each row to reduce to a standard determinant.
2. The determinant of columns 1, x squared, x cubed carries the extra factor xy+yz+zx.

Answer: (a) 6k

Explanation:
1. Take 3 out of the first column: the determinant becomes \( 3\begin{vmatrix}2a&2b&2c\\m&n&p\\x&y&z\end{vmatrix} \).
2. Take 2 out of the first row: \( 3\cdot2\cdot k=6k \).

Teacher's Note:
1. A common factor of a row or a column can be taken out of the determinant.
2. Do the factors one at a time.

Answer: (c) 2

Explanation:
1. The left side expands to \( 2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=2(a^3+b^3+c^3-3abc) \).
2. The circulant on the right equals \( a^3+b^3+c^3-3abc \).
3. So \( t=2 \).

Teacher's Note:
1. Add all rows to find the factor a+b+c from the left determinant.
2. The circulant determinant of a, b, c is a cube sum minus three times the product.

Answer: (c) 2

Explanation:
1. For example \( \begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}=1(1)-1(0-1)+0=2 \).
2. The maximum possible value for a 3x3 matrix with entries 0 or 1 is 2.

Teacher's Note:
1. The largest determinant is limited by the six terms of the expansion and their signs.
2. Zero entries can never make all three positive terms equal 1 while all three negative terms vanish.

Answer: (a) 0

Explanation:
1. Multiply and divide by \( abc \): the first column becomes \( abc(\frac1a,\frac1b,\frac1c) \).
2. For an H.P., \( \frac1a,\frac1b,\frac1c \) are in A.P., so \( \frac1a=A+(p-1)D \) and similarly for q and r.
3. The first column is therefore a combination of the second and third columns, and the determinant is 0.

Teacher's Note:
1. Reciprocals of H.P. terms form an A.P. and are linear in the term number.
2. A column that is a linear combination of the other columns gives determinant zero.

Answer: (a) 3, 7

Explanation:
1. Expanding, the determinant is \( -(x-7)(x-3)(x+10) \).
2. Setting it to zero gives roots \( x=7 \), \( 3 \) and \( -10 \).
3. The other roots are 3 and 7.

Teacher's Note:
1. Check by substitution: x \( = \) 3 gives two equal columns after rearranging (rows 1 and 3 become identical).
2. x \( = \) 7 makes Row 1 + Row 3 \( = \) (10, 10, 10), a multiple of Row 2.

Answer: (c) a, b

Explanation:
1. Subtract each row from the one above it to remove the column of ones and the constants.
2. The determinant becomes \( (x-\lambda)(x-\mu)(x-\nu) \).
3. The roots \( \lambda,\mu,\nu \) do not involve a and b, so they are independent of a and b.

Teacher's Note:
1. Row differences make the matrix triangular.
2. The answer is the set of parameters that do not appear in the factorised form.