Determinants — JEE MCQs with answers

Answer: B — $\operatorname{adj} A$ is the transpose of $[A_{ij}]$, so its (1, 3) entry is the cofactor $A_{31} = (-1)^{4}M_{31} = -11$. So the answer is option B. Option A ($11$) is wrong: it uses the opposite sign for $(-1)^{i+j}$. Option C ($9$) is wrong: it takes the cofactor $A_{13}$ instead of $A_{31}$, forgetting the transpose. Option D ($-9$) is wrong: it works the minor $M_{31}$ as $ad + bc$.

Answer: C — When $i + j$ is even the sign is $+$, when odd it is $-$. So position $(1, 1)$ is $+$, $(1, 2)$ is $-$, and the signs alternate along every row and every column. That is option C. Option A uses only the row number, $(-1)^{i+1}$. Option B is the reversed pattern, as if $(-1)^{i+j+1}$ were used. Option D uses only the column number, $(-1)^{j+1}$.

Answer: B — $|A| = 1 \cdot 3 - 2 \cdot 2 = -1 \neq 0$, so Case I applies: $A^{-1}$ exists and there is a unique solution ($x = 0$, $y = 1$). The answer key prints "Consistent". Option B. The book computes $(\operatorname{adj} A)B$ only "In this case", that is, when $|A| = 0$; option A applies that test to a nonsingular matrix. Option C adds the two products ($3 + 4 = 7$) instead of subtracting them, and still misuses the test. Option D quotes the Case II wording for $(\operatorname{adj} A)B = O$, which does not apply here on either count.

Answer: A — Expanding along $R_1$: $2(12 - (-2)) - k(4 - 0) + 1(2 - 0) = 28 - 4k + 2 = 30 - 4k$. So $30 - 4k = 10$ and $k = 5$. So the answer is option A. Option B ($3$) is wrong: it works out the first minor as $12 - 2 = 10$, losing the sign of $-1$: $22 - 4k = 10$. Option C ($4$) is wrong: it gives the last term a $-$ sign: $26 - 4k = 10$. Option D ($-5$) is wrong: it gives the middle term a $+$ sign: $30 + 4k = 10$.

Answer: D — $P$ is chosen on $AB$, so $A$, $B$, $P$ are collinear, and Remark (iii) says the area of the triangle formed by three collinear points is zero. That is what makes the determinant equal to $0$ and gives the line $y = 3x$. So the answer is option D. Option A is false: the area with vertex $(0, 0)$ is $\frac{1}{2}|x_1 y_2 - x_2 y_1|$, which need not be $0$. Option B is true but explains only the sign, not why the value is zero. Option C is false: Example 6's determinant has that column and equals $61$.

Answer: B — $a_1 b_2 - a_2 b_1 = 2k - 3(-5) = 2k + 15 = 19$, so $k = 2$. So the answer is option B. Option A ($\frac{13}{5}$) is wrong: it multiplies down the columns: $a_1 a_2 - b_1 b_2 = 6 + 5k = 19$. Option C ($17$) is wrong: it drops the sign of $b_1 = -5$: $2k - 15 = 19$. Option D ($-17$) is wrong: it uses the reversed order $a_2 b_1 - a_1 b_2 = -15 - 2k = 19$.

Answer: D — $A \cdot A^2 = A^2 \cdot A = A^3 = I$, so by the Summary's rule (if $AB = BA = I$ then $B = A^{-1}$), $A$ is invertible with $A^{-1} = A^2$: option D. Also $|A|^3 = |I| = 1$ by Theorem 3, so the real number $|A|$ is $1$. That rules out option C. Option A need not hold: $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ has cube $I$ but is not $I$. Option B would need $A^2 = I$, which that same matrix fails.

Answer: A — $\Delta$ = sum of the products of the elements of $C_2$ with their cofactors. The elements are $7, 2, 5$ and their cofactors are $22, 13, -5$, so $\Delta = (7)(22) + (2)(13) + (5)(-5) = 155$. So the answer is option A. Option B ($-103$) is wrong: it multiplies each element by its minor, leaving out the signs $(-1)^{i+j}$. Option C ($-155$) is wrong: it uses each cofactor with its sign reversed. Option D ($205$) is wrong: it adds the sizes of the three products, ignoring every sign.

Answer: B — $\operatorname{adj} A = \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$, so $X = -\frac{1}{11}\begin{bmatrix} 2 - 35 \\ -3 + 14 \end{bmatrix} = -\frac{1}{11}\begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$: option B. Option A stops at $(\operatorname{adj} A)B$ without dividing by $|A|$. Option C uses the cofactor matrix without transposing it. Option D divides by $+11$, dropping the sign of $|A|$.

Answer: B — $\begin{vmatrix} k & 2 & 1 \\ 1 & 4 & 1 \\ 3 & 8 & 1 \end{vmatrix} = k(4 - 8) - 2(1 - 3) + (8 - 12) = -4k + 4 - 4 = -4k = 0$, so $k = 0$. The triangle $(0, 0)$, $(0, 3)$, $(2, 0)$ has area $\frac{1}{2}\left|\begin{vmatrix} 0 & 0 & 1 \\ 0 & 3 & 1 \\ 2 & 0 & 1 \end{vmatrix}\right| = \frac{1}{2}|-6| = 3$. So the answer is option B. Option A, $6$, reverses the sign of the middle term, $-4k - 4 - 4 = 0$, so $k = -2$ and the triangle $(-2, 0)$, $(0, 3)$, $(2, 0)$ has area $6$. Option C, $0$, computes the last minor as $3 \cdot 4 - 1 \cdot 8 = 4$, so $-4k + 8 = 0$, $k = 2$, and two vertices coincide. Option D, $12$, makes the sign slip ($k = -2$) and also drops the $\frac{1}{2}$.