JEE Mathematics Matrices and Determinants MCQs Set 05

Answer: (b) 4

Explanation:
1. Write a = [x], b = [y], c = [z]. The matrix equals \( I+\mathbf 1\,(a\ b\ c) \), so its determinant is \( 1+a+b+c \).
2. Possible values: a ∈ {0, 1}, b ∈ {−1, 0}, c ∈ {1, 2}. Maximum \( 1+1+0+2=4 \).

Teacher's Note:
1. Subtract/add rows, or use det(I + uvᵀ) = 1 + vᵀu.
2. Find the integer values each greatest integer can take.

Answer: (b) λ = −1, 4

Explanation:
1. Consistency requires \( \begin{vmatrix}\lambda&-1&2\\2&-3&-\lambda\\3&-2&-1\end{vmatrix}=0 \).
2. Expanding: \( -2\lambda^2+6\lambda+8=0\Rightarrow\lambda^2-3\lambda-4=0 \), so λ = 4 or −1.
3. Check: λ = 4 gives x = 1, y = 2; λ = −1 gives x = −1, y = −1 — both satisfy all three equations.

Teacher's Note:
1. Three equations in two unknowns: the augmented determinant must vanish.
2. Verify by solving two equations and checking the third.

Answer: (a) \( \begin{bmatrix}1&2005\\0&1\end{bmatrix} \)

Explanation:
1. P is orthogonal: \( P^TP=I \). So \( Q^n=PA^nP^T \).
2. Then \( P^TQ^{2005}P=P^TPA^{2005}P^TP=A^{2005} \).
3. \( A^n=\begin{bmatrix}1&n\\0&1\end{bmatrix} \), so the answer is \( \begin{bmatrix}1&2005\\0&1\end{bmatrix} \).

Teacher's Note:
1. Check that P is orthogonal (rotation matrix).
2. Powers of PAPᵀ telescope.

Answer: (d) none of these

Explanation:
1. Since \( (I-A)(I+A+A^2+A^3+A^4)=I-A^5=I \), we get \( (I-A)^{-1}=I+A+A^2+A^3+A^4 \).
2. This is not equal to A⁴, A³ or I + A (those would need A² + A³ + A⁴ = O, which is not given).

Teacher's Note:
1. Use the geometric-series factorisation of I − A⁵.
2. The answer has five terms.

Answer: (d) 0

Explanation:
1. Replace x by −x: every entry of the first column (x, sin x, tan x) changes sign, while the other columns are unchanged (cos x, x², e^(x²), sec x, 1, 2 are even).
2. So \( f(-x)=-f(x) \): f is odd.
3. The integral over the symmetric interval is 0.

Teacher's Note:
1. Look at the parity of each column.
2. An odd integrand over [−a, a] integrates to zero.

Answer: (b) \( x=\begin{bmatrix}2&-1&1\\1&2&0\end{bmatrix} \)

Explanation:
1. Let the two given matrices be P and Q. Then \( x=\frac{2P+Q}{5} \) and \( y=\frac{2Q-P}{5} \).
2. \( 2P+Q=\begin{bmatrix}10&-5&5\\5&10&0\end{bmatrix} \), so \( x=\begin{bmatrix}2&-1&1\\1&2&0\end{bmatrix} \).
3. Also \( y=\begin{bmatrix}1&1&2\\-1&1&-2\end{bmatrix} \), so the other options fail (the (1,3) entry of y is +2, not −2).

Teacher's Note:
1. Solve the two matrix equations like simultaneous linear equations.
2. Check each option against the computed x and y.

Answer: (d) \( (2+i\sqrt3)\begin{bmatrix}1&0\\0&1\end{bmatrix} \)

Explanation:
1. Simplify the entries using \( \frac1i=-i \): \( A=\begin{pmatrix}\frac{\sqrt3}2+\frac i2&-\frac{\sqrt3}2+\frac i2\\ \frac{\sqrt3}2-\frac i2&-\frac{\sqrt3}2-\frac i2\end{pmatrix} \).
2. Multiplying out, \( A^2=i\sqrt3\,I \) (the off-diagonal entries cancel and each diagonal entry equals \( i\sqrt3 \)).
3. So \( f(A)=A^2+2I=(2+i\sqrt3)I \).

Teacher's Note:
1. Rewrite each entry in the form x + iy before multiplying.
2. A² turns out to be a scalar matrix.

Answer: (a) \( A^{-1}+B^{-1} \)

Explanation:
1. Use \( (XYZ)^{-1}=Z^{-1}Y^{-1}X^{-1} \): \( B^{-1}(A+B)A^{-1} \).
2. Expand: \( B^{-1}AA^{-1}+B^{-1}BA^{-1}=B^{-1}+A^{-1} \).

Teacher's Note:
1. The inverse of a product reverses the order.
2. Distribute (A + B) and cancel.

Answer: (b) 2007

Explanation:
1. The matrix is \( \operatorname{diag}(x,y,z)+\mathbf 1(a\ b\ c) \), so \( \Delta=xyz\left(1+\frac ax+\frac by+\frac cz\right)=xyz+ayz+bzx+cxy \).
2. Δ = 0 gives \( ayz+bzx+cxy=-xyz=2007 \).

Teacher's Note:
1. Use the matrix determinant lemma, or subtract rows to expand.
2. Δ = xyz + ayz + bzx + cxy.

Answer: (d) −2n

Explanation:
1. Only the first row depends on r, so \( \sum\Delta_r \) is the determinant with first row replaced by its sum: \( \left(n,\ \frac{n(n+1)}2,\ 2^{n+1}-2\right) \).
2. The third row is \( \left(n,\frac{n(n+1)}2,2^{n+1}\right) \). Subtracting row 1 from row 3 gives \( (0,0,2) \).
3. Expanding along the new third row: \( 2\left(n\cdot n-2\cdot\frac{n(n+1)}2\right)=2(n^2-n^2-n)=-2n \).

Teacher's Note:
1. Sum the first row entries over r, since the other rows are independent of r.
2. Row operation R₃ − R₁ simplifies the determinant.

Answer: (b) \( A' \)

Explanation:
1. A is orthogonal if \( AA'=A'A=I \).
2. By the definition of the inverse, \( A^{-1}=A' \).

Teacher's Note:
1. Use the definition of an orthogonal matrix.
2. The transpose equals the inverse.

Answer: (c) null matrix

Explanation:
1. The cofactor of an entry of Aᵀ is the cofactor of the transposed entry of A, so \( \operatorname{adj}(A^T)=(\operatorname{adj}A)^T \).
2. The difference is the null matrix.

Teacher's Note:
1. Recall adj(Aᵀ) = (adj A)ᵀ.
2. Both matrices are identical.

Answer: (d) \( 2007^2 \)

Explanation:
1. \( \det M_r=r^2-(r-1)^2=2r-1 \).
2. The sum is \( \sum_{r=1}^{2007}(2r-1)=2007^2 \) (sum of the first 2007 odd numbers).

Teacher's Note:
1. Compute det M_r.
2. The sum of the first n odd numbers is n².

Answer: (d) x = y

Explanation:
1. The matrix sends (x, y) to (y, x): it swaps the coordinates.
2. This is the reflection in the line y = x.

Teacher's Note:
1. Apply the matrix to a general point.
2. Swapping x and y is reflection in y = x.

Answer: (a) 3

Explanation:
1. \( \det A=8(7\lambda-16)+6(-6\lambda+8)+2(24-14)=20\lambda-60 \).
2. Singular means det A = 0, so λ = 3.

Teacher's Note:
1. Expand the determinant along the first row.
2. Set it to zero.

Answer: (a) 128B

Explanation:
1. A = iB, and \( B^2=2B \), so \( B^n=2^{n-1}B \).
2. \( A^8=i^8B^8=2^7B=128B \).

Teacher's Note:
1. Notice that A = iB.
2. B² = 2B gives Bⁿ = 2ⁿ⁻¹B.