JEE Mathematics Matrices MCQs Set 13

Answer: (a) \( \begin{bmatrix} 1 & 3 \\ -1 & 0 \\ 2 & 4 \end{bmatrix} \)

Explanation:
1. Transpose the second equation: \( 2B^T-3A=\begin{bmatrix}-1&4&-5\\18&-6&-7\end{bmatrix} \).
2. Add it to the first equation \( 3A+4B^T=\begin{bmatrix}7&-10&17\\0&6&31\end{bmatrix} \): the A terms cancel and \( 6B^T=\begin{bmatrix}6&-6&12\\18&0&24\end{bmatrix} \).
3. So \( B^T=\begin{bmatrix}1&-1&2\\3&0&4\end{bmatrix} \) and \( B=\begin{bmatrix}1&3\\-1&0\\2&4\end{bmatrix} \).

Teacher's Note:
1. Transposing one equation turns the system into two equations in A and B transpose.
2. Verify: 3A + 4 B transpose reproduces the given matrix.

Answer: (d) 6

Explanation:
1. Symmetric means \( a_{12}=a_{21} \): \( x-3=3 \), so \( x=6 \).
2. Also \( a_{13}=a_{31} \): \( x-2=4 \), so \( x=6 \) again.

Teacher's Note:
1. Equate each pair of mirror-image entries.
2. The two conditions agree, so the matrix exists.

Answer: (a) 14641

Explanation:
1. The determinant of the cofactor matrix equals \( |A|^{n-1}=11^2=121 \).
2. The square of this is \( 121^2=14641=11^4 \).

Teacher's Note:
1. The cofactor matrix is the transpose of the adjoint, so it has the same determinant.
2. Square after applying the adjoint determinant formula.

Answer: (b) 4

Explanation:
1. \( A^TA=I \) gives \( |A|=\pm1 \), and \( |A|=3abc-(a^3+b^3+c^3) \).
2. So \( a^3+b^3+c^3=3\mp1 \), i.e. 2 or 4.
3. By AM-GM, \( a^3+b^3+c^3\ge3abc=3 \) for positive numbers, so the sum is 4.

Teacher's Note:
1. An orthogonal matrix has determinant plus or minus 1.
2. The AM-GM inequality eliminates the value 2.

Answer: (c) 64

Explanation:
1. Compute \( |A|=\begin{vmatrix}1&4&9\\4&9&16\\9&16&25\end{vmatrix}=-8 \).
2. \( |\operatorname{adj}A|=|A|^2=64 \).

Teacher's Note:
1. Row differences make this determinant easy: it is -8.
2. The adjoint determinant is the square for a 3x3 matrix.

Answer: (a) 16

Explanation:
1. \( A(\operatorname{adj}A)=|A|I \) gives \( |A|=4 \).
2. \( |\operatorname{adj}A|=|A|^2=16 \).

Teacher's Note:
1. The diagonal entry of A times its adjoint is the determinant of A.
2. The adjoint determinant is the square of the determinant for order 3.

Answer: (a) [4 1]

Explanation:
1. Cofactor \( C_{21} \) of the matrix (delete row 2, column 1): \( -\begin{vmatrix}0&2\\2&1\end{vmatrix}=4 \), and the adjoint's (1,2) entry is \( C_{21}=4 \), so \( a=4 \).
2. The (3,3) entry of the adjoint is \( C_{33}=\begin{vmatrix}1&0\\-1&1\end{vmatrix}=1 \), so \( b=1 \).
3. So \( [a\ b]=[4\ \ 1] \).

Teacher's Note:
1. The (i, j) entry of the adjoint is the cofactor C_ji.
2. Check the other entries of the adjoint against the same formula to be safe.

Answer: (a) \( \pm4 \)

Explanation:
1. \( |\operatorname{adj}A|=|A|^2=16 \).
2. So \( |A|=\pm4 \).

Teacher's Note:
1. Squaring loses the sign, so both signs are possible.
2. Both 4 and -4 give the same adjoint determinant.

Answer: (d) \( 3^2\times2^8 \)

Explanation:
1. \( |2A|=2^3\cdot6=48 \).
2. \( |\operatorname{adj}(2A)|=|2A|^2=48^2=2304=2^8\times3^2 \).

Teacher's Note:
1. Scale the matrix first, then apply the adjoint formula.
2. 48 squared is 2304.

Answer: (d) 1

Explanation:
1. No inverse means determinant 0.
2. The determinant equals \( -(x-1)(x^2+x+2) \).
3. \( x^2+x+2 \) has negative discriminant, so the only real root is \( x=1 \).

Teacher's Note:
1. Expand and factor the determinant.
2. Check the quadratic factor for real roots using the discriminant.

Answer: (b) 1

Explanation:
1. The inverse of the given matrix has second row \( (0,\frac1x,0) \) and first row \( (1,-\frac ax,-b) \).
2. Comparing with the given inverse gives \( \frac1x=1 \), so \( x=1 \).

Teacher's Note:
1. Multiply the two matrices; the product should be the identity.
2. The (2,2) entry of the product is x, which must equal 1.

Answer: (c) \( \begin{bmatrix} a-ib & -c-id \\ c-id & a+ib \end{bmatrix} \)

Explanation:
1. \( |A|=(a+ib)(a-ib)-(c+id)(-c+id)=a^2+b^2+c^2+d^2=1 \).
2. For a 2x2 matrix with determinant 1, \( A^{-1}=\begin{bmatrix}a-ib&-(c+id)\\-(-c+id)&a+ib\end{bmatrix} \).
3. So \( A^{-1}=\begin{bmatrix}a-ib&-c-id\\c-id&a+ib\end{bmatrix} \).

Teacher's Note:
1. Swap the diagonal entries and change the sign of the off-diagonal entries, then divide by the determinant.
2. The condition a squared + b squared + c squared + d squared \( = \) 1 makes the determinant 1.

Answer: (c) \( I-A+A^2 \)

Explanation:
1. \( (I+A)(I-A+A^2)=I-A+A^2+A-A^2+A^3=I+A^3=I \).
2. So \( (I+A)^{-1}=I-A+A^2 \).

Teacher's Note:
1. This is the sum of cubes factorisation with A cubed equal to zero.
2. Multiply out to verify the product is I.

Answer: (a) \( \frac{5I-A}6 \)

Explanation:
1. Expanding gives \( A^2-5A+6I=O \).
2. Multiply by \( A^{-1} \): \( A-5I+6A^{-1}=O \).
3. So \( A^{-1}=\frac{5I-A}{6} \).

Teacher's Note:
1. Multiply the polynomial identity by the inverse of A.
2. The constant term 6 becomes the coefficient of A inverse.

Answer: (a) 1

Explanation:
1. Row 2 is 2 times Row 1, so the rows are dependent.
2. Only one independent non-zero row remains, so the rank is 1.

Teacher's Note:
1. Check whether one row is a multiple of the other.
2. Every 2x2 minor is zero.

Answer: (a) 9

Explanation:
1. \( |I|=1\ne0 \), so the largest non-zero minor has order 9.
2. Hence the rank is 9.

Teacher's Note:
1. The rank of the identity matrix equals its order.
2. A non-singular square matrix has full rank.

Answer: (b) A, C, B

Explanation:
1. \( |A|=1(6)-4(4)-1(2)=-12\ne0 \), so rank A = 3.
2. B has all rows equal, so rank B \( = \) 1.
3. \( |C|=0 \) because \( 2R_1=R_2+R_3 \), but no two rows are proportional, so rank C = 2.
4. Descending order: A (3), C (2), B (1).

Teacher's Note:
1. Compute determinants for the 3x3 matrices to find which have full rank.
2. A matrix with all equal rows has rank 1.

Answer: (d) 5

Explanation:
1. For rank 3, \( |A|=0 \).
2. The determinant equals \( -12(\alpha-5) \).
3. So \( \alpha=5 \) (and a 3x3 minor of A is non-zero, so the rank is exactly 3).

Teacher's Note:
1. A 4x4 matrix has rank less than 4 only if its determinant is zero.
2. Check that some 3x3 minor is non-zero to confirm the rank is exactly 3.

Answer: (d) (1, -3, 2)

Explanation:
1. Substitute option (c) \( (1,-3,2) \): \( 2-3+2=1 \), \( 1+6-6=1 \), \( 3-6+8=5 \).
2. All three equations are satisfied, and the coefficient determinant is non-zero, so the solution is unique.
3. The solution is \( (1,-3,2) \).

Teacher's Note:
1. Substituting the options is the fastest check in a multiple-choice question.
2. A non-zero coefficient determinant guarantees uniqueness.

Answer: (b) is inconsistent

Explanation:
1. Written as \( 2x+6y=-11 \), \( 6y-18z=-1 \), \( 6x+20y-6z=-3 \), the coefficient matrix has determinant 0, so its rank is 2.
2. The augmented matrix has rank 3.
3. Rank of the coefficient matrix is less than the rank of the augmented matrix, so the system is inconsistent.

Teacher's Note:
1. Compare the two ranks.
2. Rank A less than rank \( [A|B] \) means no solution.

Answer: (d) \( -\frac72 \)

Explanation:
1. Failing to have a unique solution means the coefficient determinant is 0.
2. The determinant equals \( -2a-7 \).
3. So \( a=-\frac72 \).

Teacher's Note:
1. Set the coefficient determinant equal to zero.
2. A zero determinant means no solution or infinitely many solutions.

Answer: (b) 36

Explanation:
1. Note that \( \text{Row 3}=3\times\text{Row 2}-\text{Row 1} \) on the left side.
2. For consistency the right side must satisfy \( a=3(14)-6=36 \).
3. Then the system has infinitely many solutions.

Teacher's Note:
1. Find the combination of the first two equations that reproduces the left side of the third.
2. Apply the same combination to the constants.

Answer: (a) infinite

Explanation:
1. Subtract the third equation from the first: \( x+y=2 \).
2. Then \( z=1 \) from the second equation, and the third gives \( 2(2)-1=3 \), which holds.
3. The system is consistent with only two independent equations in three unknowns, so there are infinitely many solutions.

Teacher's Note:
1. The coefficient determinant is 0 and the system is consistent.
2. One free parameter remains because x and y appear only as x + y.

Answer: (a) 0

Explanation:
1. The coefficient determinant is \( \begin{vmatrix}1&-1&1\\1&2&-1\\2&1&3\end{vmatrix}=9\ne0 \).
2. So only the trivial solution exists and there are 0 non-trivial solutions.

Teacher's Note:
1. A homogeneous system has non-trivial solutions only if the determinant is zero.
2. Here the determinant is 9.

Answer: (a) \( a^2+b^2+c^2+2abc=1 \)

Explanation:
1. Write as a homogeneous system: \( x-cy-bz=0 \), \( -cx+y-az=0 \), \( -bx-ay+z=0 \).
2. Non-trivial solutions need \( \begin{vmatrix}1&-c&-b\\-c&1&-a\\-b&-a&1\end{vmatrix}=0 \).
3. The determinant is \( 1-a^2-b^2-c^2-2abc \), so \( a^2+b^2+c^2+2abc=1 \).

Teacher's Note:
1. Bring all terms to one side to form a homogeneous system.
2. The determinant condition gives the relation.

Answer: (d) \( -5\pm2\sqrt2 \)

Explanation:
1. The coefficient determinant must vanish.
2. It simplifies to \( -(a^2+10a+17)=0 \).
3. So \( a=\frac{-10\pm\sqrt{100-68}}2=-5\pm2\sqrt2 \).

Teacher's Note:
1. Set the determinant of coefficients to zero and solve the quadratic.
2. Use the quadratic formula: discriminant 32.