JEE Mathematics Matrices and Determinants MCQs Set 10

Answer: (b) \( \frac12 \)

Explanation:
1. A homogeneous system has a non-trivial solution only if the coefficient determinant is 0.
2. \( \begin{vmatrix}1&-c&-c\\c&-1&c\\c&c&-1\end{vmatrix}=-2c^3-3c^2+1=-(c+1)^2(2c-1)=0 \).
3. So \( c=-1 \) or \( c=\frac12 \), and the greatest value is \( \frac12 \). Answer: (b)

Teacher's Note:
1. Set the determinant of the coefficient matrix to zero.
2. Factorise the cubic fully before comparing the roots.

Answer: (c) \( [4,6] \)

Explanation:
1. Let \( b=2+d \), \( c=2+2d \). Then \( \det A=(b-2)(c-2)(c-b)=d\cdot2d\cdot d=2d^3 \).
2. \( 2\le2d^3\le16\Rightarrow1\le d\le2 \).
3. So \( c=2+2d\in[4,6] \). Answer: (c)

Teacher's Note:
1. The determinant is a Vandermonde type in \( 2,b,c \).
2. Express \( b \) and \( c \) through the common difference.

Answer: (b) \( \begin{bmatrix}1&-13\\0&1\end{bmatrix} \)

Explanation:
1. Multiplying such matrices adds the upper-right entries: the product has entry \( 1+2+\cdots+(n-1)=\frac{n(n-1)}2 \).
2. \( \frac{n(n-1)}2=78\Rightarrow n(n-1)=156=13\times12\Rightarrow n=13 \).
3. \( \begin{bmatrix}1&13\\0&1\end{bmatrix}^{-1}=\begin{bmatrix}1&-13\\0&1\end{bmatrix} \). Answer: (b)

Teacher's Note:
1. Multiply the first two matrices to see the pattern.
2. The inverse of a unit upper triangular matrix of this form just changes the sign of the corner entry.

Answer: (d) \( y^3 \)

Explanation:
1. Here \( \alpha+\beta=-1 \) and \( \alpha\beta=1 \). Apply \( R_1\to R_1+R_2+R_3 \): every entry of row 1 becomes \( y \).
2. Take \( y \) common and apply \( C_2\to C_2-C_1 \), \( C_3\to C_3-C_1 \) to get zeros in row 1 apart from the first entry.
3. Expanding gives \( y\left[y^2-(\alpha+\beta)^2+4\alpha\beta-(\alpha+\beta)-1\right]\ldots=y^3 \). Answer: (d) (Checked by direct expansion with \( \alpha=\omega,\ \beta=\omega^2 \).)

Teacher's Note:
1. Add all rows into the first row and use \( \alpha+\beta=-1 \).
2. Then create zeros using column operations.

Answer: (b) \( \frac12 \)

Explanation:
1. Non-trivial solution needs \( \begin{vmatrix}2&3&-1\\1&k&-2\\2&-1&1\end{vmatrix}=0\Rightarrow4k-18=0 \), so \( k=\frac92 \).
2. From equations 1 and 3: \( 4x+2y=0\Rightarrow\frac xy=-\frac12 \). Adding them to eliminate \( x \) with equation 2 gives \( \frac yz=\frac12 \), hence \( \frac zx=\frac zy\cdot\frac yx=2\times(-2)=-4 \).
3. Sum \( =-\frac12+\frac12-4+\frac92=\frac12 \). Answer: (b) (Example solution: \( (x,y,z)=(-\tfrac12,1,2) \).)

Teacher's Note:
1. First find \( k \) from the zero determinant.
2. Then find the ratios \( x:y:z \) from two of the equations.

Answer: (b) 4

Explanation:
1. \( A^TA \) has diagonal entries \( 8x^2,\ 6y^2,\ 3 \) and all other entries are 0 for every \( x,y \).
2. So \( 8x^2=3 \) and \( 6y^2=3 \), giving \( x=\pm\sqrt{3/8} \) and \( y=\pm\frac1{\sqrt2} \).
3. The two values have different magnitudes so \( x\ne y \) always, and the four sign choices give 4 matrices. Answer: (b)

Teacher's Note:
1. Multiply \( A^T \) by \( A \) and compare entries with \( 3I_3 \).
2. Count the independent sign choices for \( x \) and \( y \).

Answer: (c) \( \lambda^2-\lambda-6=0 \)

Explanation:
1. Infinitely many solutions need \( \begin{vmatrix}1&1&1\\4&\lambda&-\lambda\\3&2&-4\end{vmatrix}=0 \), which gives \( -8\lambda+24=0 \), so \( \lambda=3 \).
2. For \( \lambda=3 \) the system is consistent (the coefficient matrix and augmented matrix both have rank 2), so there are infinitely many solutions.
3. Test \( \lambda=3 \): \( 9-3-6=0 \) holds for option (c) only. Answer: (c)

Teacher's Note:
1. Set the coefficient determinant to zero.
2. Substitute the value of \( \lambda \) into each quadratic.

Answer: (a) 0

Explanation:
1. Expanding the determinant gives \( -5(x^3-7x+6)=0 \), i.e. \( x^3-7x+6=0 \).
2. It factorises as \( (x-1)(x-2)(x+3)=0 \), so all three roots are real.
3. Sum of roots \( =1+2-3=0 \). Answer: (a)

Teacher's Note:
1. Expand along the first row carefully.
2. The coefficient of \( x^2 \) in the cubic is zero.

Answer: (b) \( \begin{bmatrix}4&-2\\-1&-4\end{bmatrix} \)

Explanation:
1. Symmetric part: \( A=\frac{M+M^T}2=\begin{bmatrix}2&4\\4&-1\end{bmatrix} \); skew part: \( B=\frac{M-M^T}2=\begin{bmatrix}0&-1\\1&0\end{bmatrix} \).
2. \( AB=\begin{bmatrix}2\cdot0+4\cdot1&2(-1)+4\cdot0\\4\cdot0-1\cdot1&4(-1)+(-1)\cdot0\end{bmatrix}=\begin{bmatrix}4&-2\\-1&-4\end{bmatrix} \). Answer: (b)

Teacher's Note:
1. Every square matrix splits uniquely into a symmetric and a skew-symmetric part.
2. Use \( A=\frac{M+M^T}{2} \) and \( B=\frac{M-M^T}{2} \).

Answer: (c) 1

Explanation:
1. \( \det A=\frac1{\det B} \), so \( \det A+1=0 \) means \( \det B=-1 \).
2. \( \det B=5(-2-3)-2\alpha(0-\alpha)+1(0-2\alpha)=2\alpha^2-2\alpha-25 \).
3. \( 2\alpha^2-2\alpha-25=-1\Rightarrow\alpha^2-\alpha-12=0 \), so \( \alpha=4,-3 \) with sum 1. Answer: (c)

Teacher's Note:
1. Use \( \det(A^{-1})=\frac1{\det A} \).
2. Sum of roots of the quadratic equals minus the coefficient of \( \alpha \).

Answer: (a) \( \frac\pi9 \)

Explanation:
1. Apply \( C_1\to C_1+C_2 \), then \( R_1\to R_1-2R_3 \), \( R_2\to R_2-2R_3 \); the determinant reduces to \( (2+4\cos6\theta)(\sin^2\theta+\cos^2\theta) \), i.e. \( 2+4\cos6\theta \).
2. So \( \cos6\theta=-\frac12 \), giving \( 6\theta=\frac{2\pi}3 \) in the range \( 0\lt6\theta\lt2\pi \) (the other value \( \frac{4\pi}{3} \) gives \( \theta=\frac{2\pi}{9} \), which is not among the options).
3. \( \theta=\frac\pi9 \). Answer: (a) (Numerical check: the determinant equals \( 2+4\cos6\theta \), which is 0 at \( \theta=\frac\pi9 \) and is nonzero for the other options.)

Teacher's Note:
1. Use column and row operations to make the determinant factorise.
2. Solve \( \cos6\theta=-\tfrac12 \) within the given interval.

Answer: (b) invertible for every \( t\in R \)

Explanation:
1. Take \( e^t \) from column 1, \( e^{-t} \) from each of columns 2 and 3 to get \( e^{t}e^{-2t}=e^{-t} \) times a determinant with column 1 all ones.
2. Using \( R_2\to R_2-R_1 \), \( R_3\to R_3-R_1 \) and expanding gives \( e^{-t}(2\cos t+\sin t)^2+e^{-t}(2\sin t-\cos t)^2=5e^{-t} \).
3. \( |A|=5e^{-t}\ne0 \) for all real \( t \), so \( A \) is always invertible. Answer: (b)

Teacher's Note:
1. Factor the exponentials out of the columns.
2. A determinant that never vanishes means the matrix is always invertible.

Answer: (a) \( 2g+h+k=0 \)

Explanation:
1. The coefficient determinant is 0, so the system is consistent only when the right-hand sides satisfy the same dependency as the rows.
2. Row 3 + 2 (Row 1) \( =(0,5-8,-9+14)=(0,-3,5)=-\,\text{Row 2} \).
3. So \( k+2g=-h \), i.e. \( 2g+h+k=0 \). Answer: (a)

Teacher's Note:
1. Find the linear relation between the three left-hand sides.
2. Apply the same relation to \( g,h,k \).

Answer: (a) 8

Explanation:
1. Subtract equation 1 from 2: \( y+2z=4 \). Subtract equation 1 from 3: \( 2y+(\alpha-1)z=\beta-5 \).
2. The last must be twice the previous one: \( \alpha-1=4 \) and \( \beta-5=8 \), so \( \alpha=5 \), \( \beta=13 \).
3. \( \beta-\alpha=8 \). Answer: (a)

Teacher's Note:
1. Eliminate \( x \) first.
2. Make the two reduced equations proportional.

Answer: (a) \( -5 \)

Explanation:
1. Apply \( R_3\to R_3-2R_2+R_1 \); the determinant factorises as \( \det A=(d+2)^2-\sin^2\theta \).
2. It is least when \( \sin^2\theta=1 \), so \( \det A_{\min}=(d+2)^2-1=8 \).
3. \( (d+2)^2=9\Rightarrow d=1 \) or \( -5 \). Only \( -5 \) is among the options. Answer: (a)

Teacher's Note:
1. Use a row operation to produce zeros in the third row.
2. Minimise using \( \sin^2\theta\le1 \).

Answer: (c) \( 2\sqrt3 \)

Explanation:
1. Expanding, \( \det A=2(2b^2+4-b^2)-b(2b-b)+1(b^2-b^2-1)=b^2+3 \).
2. \( \frac{\det A}b=b+\frac3b \). By AM-GM, for \( b\gt0 \), \( b+\frac3b\ge2\sqrt3 \).
3. The minimum is \( 2\sqrt3 \) (at \( b=\sqrt3 \)). Answer: (c)

Teacher's Note:
1. Expand the determinant to get a simple expression in \( b \).
2. Use AM-GM for positive \( b \).

Answer: (a) \( b-c-a=0 \)

Explanation:
1. The coefficient determinant is 0 because \( R_2=R_1+R_3 \), so more than one solution requires the system to be consistent: \( D_1=0 \).
2. \( D_1=\begin{vmatrix}a&2&3\\b&-1&5\\c&-3&2\end{vmatrix}=13a-13b+13c \).
3. So \( a-b+c=0 \), i.e. \( b-c-a=0 \). Answer: (a)

Teacher's Note:
1. Check that the coefficient determinant vanishes first.
2. Then impose \( D_1=0 \) for consistency.

Answer: (b) \( \frac1{\sqrt2} \)

Explanation:
1. Rows are orthonormal: row 1 gives \( 4q^2+r^2=1 \); row 2 gives \( p^2+q^2+r^2=1 \); rows 1 and 2 give \( 2q^2-r^2=0 \); rows 2 and 3 give \( p^2-q^2-r^2=0 \).
2. From \( r^2=2q^2 \) and \( p^2=q^2+r^2=3q^2 \); then \( 4q^2+2q^2=1\Rightarrow q^2=\frac16 \).
3. \( p^2=3q^2=\frac12 \), so \( |p|=\frac1{\sqrt2} \). Answer: (b)

Teacher's Note:
1. \( AA^T=I \) means the rows are mutually perpendicular unit vectors.
2. Solve the resulting equations for \( p^2,q^2,r^2 \).

Answer: (c) \( \frac1{16} \)

Explanation:
1. \( |A|^2|B|=8 \) and \( \frac{|A|}{|B|}=8 \). Multiplying: \( |A|^3=64 \), so \( |A|=4 \) and \( |B|=\frac12 \).
2. \( |BA^{-1}B^T|=|B|\cdot\frac1{|A|}\cdot|B|=\frac12\cdot\frac14\cdot\frac12=\frac1{16} \). Answer: (c)

Teacher's Note:
1. Use \( |XY|=|X||Y| \), \( |A^T|=|A| \) and \( |A^{-1}|=\frac1{|A|} \).
2. Solve for \( |A| \) and \( |B| \) first.

Answer: (b) \( -2(a+b+c) \)

Explanation:
1. Apply \( R_1\to R_1+R_2+R_3 \), take \( (a+b+c) \) out, then \( C_2\to C_2-C_1 \), \( C_3\to C_3-C_1 \).
2. The determinant becomes \( (a+b+c)^3 \).
3. \( (x+a+b+c)^2=(a+b+c)^2\Rightarrow x=0 \) or \( x=-2(a+b+c) \); since \( x\ne0 \), \( x=-2(a+b+c) \). Answer: (b)

Teacher's Note:
1. Add all rows into the first row.
2. A perfect-square equality gives two cases; reject \( x=0 \).

Answer: (a) \( (2,4) \)

Explanation:
1. A unique solution needs a non-zero coefficient determinant.
2. Expanding, \( D=\alpha+\beta+2 \).
3. Option (a): \( D=8\ne0 \). Options (b), (c), (d) all give \( D=0 \). Answer: (a)

Teacher's Note:
1. Compute the determinant of the coefficient matrix.
2. Test each option in the simplified expression.

Answer: (a) \( \left(\frac32,3\right] \)

Explanation:
1. Expanding, \( \det A=2(1+\sin^2\theta) \).
2. For \( \theta\in\left(\frac{3\pi}4,\frac{5\pi}4\right) \), \( \sin\theta\in\left(-\frac1{\sqrt2},\frac1{\sqrt2}\right) \), so \( \sin^2\theta\in\left[0,\frac12\right) \).
3. Hence \( \det A\in[2,3) \), which lies inside \( \left(\frac32,3\right] \) and no other option. Answer: (a)

Teacher's Note:
1. Expand along the first row.
2. Find the range of \( \sin^2\theta \) on the given interval.

Answer: (c) \( (-4,5) \)

Explanation:
1. Apply \( C_1\to C_1+C_2+C_3 \) and take out \( (5x-4) \); then use \( R_2\to R_2-R_1 \), \( R_3\to R_3-R_1 \).
2. The determinant equals \( (5x-4)(x+4)^2 \).
3. Compare with \( (A+Bx)(x-A)^2 \): \( A=-4 \), \( B=5 \). Answer: (c)

Teacher's Note:
1. Add all columns into the first column.
2. Compare the factorised form with the given expression.

Answer: (b) 10

Explanation:
1. \( \begin{vmatrix}1&k&3\\3&k&-2\\2&4&-3\end{vmatrix}=44-4k=0\Rightarrow k=11 \).
2. Put \( z=t \). From the first two equations: \( x=\frac{5t}2 \), \( y=-\frac t2 \).
3. \( \frac{xz}{y^2}=\frac{\frac{5t}2\cdot t}{\frac{t^2}4}=10 \). Answer: (b)

Teacher's Note:
1. Find \( k \) from the zero determinant.
2. Express \( x,y \) in terms of \( z \).

Answer: (c) a singleton set

Explanation:
1. The coefficient determinant is \( -(a-1)^2 \). If \( a\ne1 \) it is non-zero, so the solution is unique.
2. If \( a=1 \) the system is \( x+y+z=1 \), \( x+y+z=1 \), \( x+by+z=0 \). For \( b\ne1 \), subtracting gives \( (b-1)y=-1 \), which has a solution.
3. For \( b=1 \) we get \( x+y+z=1 \) and \( x+y+z=0 \) together, which is impossible. So \( S=\{1\} \). Answer: (c)

Teacher's Note:
1. Separate the cases \( a=1 \) and \( a\ne1 \).
2. Check whether the equations contradict one another.

Answer: (d) \( \left(2,\frac12\right) \)

Explanation:
1. Area condition: \( \frac12\left|\begin{vmatrix}k&-3k&1\\5&k&1\\-k&2&1\end{vmatrix}\right|=28 \) gives \( 5k^2+13k+10=\pm56 \), i.e. \( 5k^2+13k-46=0 \) or \( 5k^2+13k+66=0 \). Only \( k=2 \) is an integer.
2. The vertices are \( A(2,-6),\ B(5,2),\ C(-2,2) \). \( BC \) is horizontal, so the altitude from \( A \) is \( x=2 \).
3. Slope of \( AB \) is \( \frac83 \), so the altitude from \( C \) has slope \( -\frac38 \): \( y-2=-\frac38(x+2) \). At \( x=2 \), \( y=\frac12 \). Orthocentre \( \left(2,\frac12\right) \). Answer: (d)

Teacher's Note:
1. Use the area formula to find the integer \( k \).
2. Intersect two altitudes to find the orthocentre.

Answer: (d) exactly three values of \( \lambda \)

Explanation:
1. A non-trivial solution needs \( \begin{vmatrix}1&\lambda&-1\\ \lambda&-1&-1\\1&1&-\lambda\end{vmatrix}=0 \).
2. Expanding gives \( \lambda^3-\lambda=0\Rightarrow\lambda(\lambda-1)(\lambda+1)=0 \).
3. So \( \lambda=0,1,-1 \): exactly three values. Answer: (d)

Teacher's Note:
1. Set the coefficient determinant to zero.
2. Count the distinct roots of the resulting cubic.

Answer: (d) \( (-2,-1) \)

Explanation:
1. Rows must be mutually perpendicular with squared length 9.
2. Row 1 . Row 3: \( a+4+2b=0 \); Row 2 . Row 3: \( 2a+2-2b=0 \). So \( a+2b=-4 \) and \( a-b=-1 \).
3. Subtracting: \( 3b=-3\Rightarrow b=-1 \), \( a=-2 \). Check: \( a^2+4+b^2=9 \). Answer: (d)

Teacher's Note:
1. \( AA^T=9I \) is a condition on the dot products of the rows.
2. Solve the two linear equations and verify the length condition.