Solution & Explanation
### Related Formula
textTrace(A^T A) = sum_i,j A_ij^2$$\text{Trace}(A^T A) = \sum_{i,j} A_{ij}^2$$
### Core Logic
Let A = beginpmatrix a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 endpmatrix$A = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{pmatrix}$.
A^T A = beginpmatrix a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 endpmatrix beginpmatrix a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 endpmatrix$A^T A = \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{pmatrix} \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{pmatrix}$
A^T A = beginpmatrix a_1^2 + a_2^2 + a_3^2 & dots \\ dots & b_1^2 + b_2^2 + b_3^2 endpmatrix$A^T A = \begin{pmatrix} a_1^2 + a_2^2 + a_3^2 & \dots \\ \dots & b_1^2 + b_2^2 + b_3^2 \end{pmatrix}$
Trace = a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 = 5$a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 = 5$.
We need to find the number of ways to pick 6$6$ elements from \-2, -1, 0, 1, 2\$\{-2, -1, 0, 1, 2\}$ such that the sum of their squares is 5$5$.
### Step 1: Identifying Valid Squares
Available squares from set: 0^2=0, (pm 1)^2=1, (pm 2)^2=4$0^2=0, (\pm 1)^2=1, (\pm 2)^2=4$.
Possible sets of 6 squares that sum to 5:
Case 1: 4, 1, 0, 0, 0, 0$4, 1, 0, 0, 0, 0$
Case 2: 1, 1, 1, 1, 1, 0$1, 1, 1, 1, 1, 0$
### Step 2: Calculating Permutations for Case 1
Elements for 4$4$ can be pm 2$\pm 2$ (2 options).
Elements for 1$1$ can be pm 1$\pm 1$ (2 options).
Zeroes are fixed (1 option).
Arrangements of \4, 1, 0, 0, 0, 0\$\{4, 1, 0, 0, 0, 0\}$:
No. of ways = frac6!1!1!4! = 30$\frac{6!}{1!1!4!} = 30$.
For each arrangement, element choices = 2 times 2 = 4$2 \times 2 = 4$.
Total for Case 1 = 30 times 4 = 120$30 \times 4 = 120$ ways.
### Step 3: Calculating Permutations for Case 2
Elements for 1$1$ can be pm 1$\pm 1$ (2 options each, so 2^5$2^5$ choices).
Zero is fixed.
Arrangements of \1, 1, 1, 1, 1, 0\$\{1, 1, 1, 1, 1, 0\}$:
No. of ways = frac6!5!1! = 6$\frac{6!}{5!1!} = 6$.
For each arrangement, element choices = 2^5 = 32$2^5 = 32$.
Total for Case 2 = 6 times 32 = 192$6 \times 32 = 192$ ways.
(Wait, following the source logic exactly: source states:
"No of ways = frac6!4! times 4 + 2 times frac6!5! + 2 times frac6!4! + 2 times frac6!3!2! $= \frac{6!}{4!} \times 4 + 2 \times \frac{6!}{5!} + 2 \times \frac{6!}{4!} + 2 \times \frac{6!}{3!2!} $"
Wait, the pdf says = 120 + 120 + 12 + 60 = 312$= 120 + 120 + 12 + 60 = 312$.
Ah, expanding the 2^5 = 32$2^5 = 32$ manually: they treated subsets of sign assignments differently, but 6 times 32 = 192$6 \times 32 = 192$. 192 + 120 = 312$192 + 120 = 312$. The final numerical answer matches perfectly).
### Pattern Recognition
The trace of A^T A$A^T A$ is universally the sum of squares of all elements in matrix A (Frobenius norm squared). Transitioning immediately to an integer-sum-of-squares combinatorial problem sidesteps matrix multiplication.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices and Determinants
Class 11 Maths: Permutations and Combinations
More Matrices and Determinants Previous-Year Questions
Q24
jee_main_2026_21_jan_morning
Cayley-Hamilton Theorem Applications
For some
alpha, beta in mathbbR$\alpha, \beta \in \mathbb{R}$ , let
A = beginbmatrix alpha & 2 \\ 1 & 2 endbmatrix$A = \begin{bmatrix} \alpha & 2 \\ 1 & 2 \end{bmatrix}$ and
B = beginbmatrix 1 & 1 \\ 1 & beta endbmatrix$B = \begin{bmatrix} 1 & 1 \\ 1 & \beta \end{bmatrix}$ be such that
A^2 - 4A + 2I = B^2 - 3B + I = O$A^2 - 4A + 2I = B^2 - 3B + I = O$ . Then
(det (operatornameadj(A^3 - B^3)))^2$(\det (\operatorname{adj}(A^3 - B^3)))^2$ is equal to ....
Numerical Answer. Answer: 225 to 225
Solution
### Related Formula
For a 2 times 2$2 \times 2$ matrix M$M$, Cayley-Hamilton equation states:
M^2 - operatornameTr(M)M + operatornamedet(M)I = O$$M^2 - \operatorname{Tr}(M)M + \operatorname{det}(M)I = O$$
Also, |operatornameadj(M)| = |M|^n-1$|\operatorname{adj}(M)| = |M|^{n-1}$, and for a 2 times 2$2 \times 2$ matrix, |operatornameadj(M)| = |M|$|\operatorname{adj}(M)| = |M|$.
### Core Logic
Using the characteristic equation A^2 - operatornameTr(A)A + det(A)I = O$A^2 - \operatorname{Tr}(A)A + \det(A)I = O$:
Comparing with A^2 - 4A + 2I = O$A^2 - 4A + 2I = O$:
operatornameTr(A) = 4 Rightarrow alpha + 2 = 4 Rightarrow alpha = 2$\operatorname{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$
Comparing with B^2 - 3B + I = O$B^2 - 3B + I = O$:
operatornameTr(B) = 3 Rightarrow 1 + beta = 3 Rightarrow beta = 2$\operatorname{Tr}(B) = 3 \Rightarrow 1 + \beta = 3 \Rightarrow \beta = 2$
### Step 1: Compute A^3 and B^3 via reduction
A = beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix$$A = \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix}$$
A^2 = 4A - 2I$A^2 = 4A - 2I$
A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I$$A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I$$
A^3 = 14 beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix - 8 beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix = beginbmatrix 28 & 28 \\ 14 & 28 endbmatrix - beginbmatrix 8 & 0 \\ 0 & 8 endbmatrix = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix$$A^3 = 14 \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} - 8 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 28 & 28 \\ 14 & 28 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix}$$
Similarly, B = beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix$B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$
B^2 = 3B - I$B^2 = 3B - I$
B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$$B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$$
B^3 = 8 beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 8 & 8 \\ 8 & 16 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix$$B^3 = 8 \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$$
### Step 2: Difference and Determinant
A^3 - B^3 = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix - beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix = beginbmatrix 15 & 20 \\ 6 & 7 endbmatrix$$A^3 - B^3 = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix} - \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} = \begin{bmatrix} 15 & 20 \\ 6 & 7 \end{bmatrix}$$
det(A^3 - B^3) = (15 times 7) - (20 times 6) = 105 - 120 = -15$$\det(A^3 - B^3) = (15 \times 7) - (20 \times 6) = 105 - 120 = -15$$
### Step 3: Final Answer
For a 2 times 2$2 \times 2$ matrix, |operatornameadj(M)| = |M|$|\operatorname{adj}(M)| = |M|$.
Thus, det(operatornameadj(A^3 - B^3)) = -15$\det(\operatorname{adj}(A^3 - B^3)) = -15$.
(det(operatornameadj(A^3 - B^3)))^2 = (-15)^2 = 225$$(\det(\operatorname{adj}(A^3 - B^3)))^2 = (-15)^2 = 225$$
### Pattern Recognition
Do not manually multiply matrices to the 3rd power. Cayley-Hamilton strictly reduces M^3$M^3$ down to a linear combination cM + dI$cM + dI$. Expanding this requires only basic scalar arithmetic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices
Q14
jee_main_2026_21_jan_evening
Matrix Algebra
For the matrices A=beginbmatrix3 & -4 \\ 1 & -1endbmatrix$A=\begin{bmatrix}3 & -4 \\ 1 & -1\end{bmatrix}$ and B=beginbmatrix-29 & 49 \\ -13 & 18endbmatrix$B=\begin{bmatrix}-29 & 49 \\ -13 & 18\end{bmatrix}$, if (A^15+B)beginbmatrixx\\ yendbmatrix=beginbmatrix0\\ 0endbmatrix$(A^{15}+B)\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}$, then among the following which one is true?
- A. x = 5, y = 7$x = 5, y = 7$
- B. x = 18, y = 11$x = 18, y = 11$
- C. x = 11, y = 2$x = 11, y = 2$
- D. x = 16, y = 3$x = 16, y = 3$
Solution
### Related Formula
A = I + N implies A^n = (I+N)^n = I + nN$$A = I + N \implies A^n = (I+N)^n = I + nN$$
(provided N^2 = 0$N^2 = 0$, meaning N$N$ is nilpotent of order 2).
### Core Logic
Evaluate powers of matrix A$A$ to detect a pattern.
A = beginbmatrix3 & -4 \\ 1 & -1endbmatrix = beginbmatrix1 & 0 \\ 0 & 1endbmatrix + beginbmatrix2 & -4 \\ 1 & -2endbmatrix$$A = \begin{bmatrix}3 & -4 \\ 1 & -1\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + \begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}$$
Let N = beginbmatrix2 & -4 \\ 1 & -2endbmatrix$N = \begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}$.
Check N^2$N^2$:
N^2 = beginbmatrix2 & -4 \\ 1 & -2endbmatrixbeginbmatrix2 & -4 \\ 1 & -2endbmatrix = beginbmatrix4-4 & -8+8 \\ 2-2 & -4+4endbmatrix = beginbmatrix0 & 0 \\ 0 & 0endbmatrix$$N^2 = \begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}\begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix} = \begin{bmatrix}4-4 & -8+8 \\ 2-2 & -4+4\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$$
### Step 1: Compute A^15
Since N^2 = 0$N^2 = 0$, A^n = I + nN$A^n = I + nN$.
A^15 = I + 15N = beginbmatrix1 & 0 \\ 0 & 1endbmatrix + 15beginbmatrix2 & -4 \\ 1 & -2endbmatrix = beginbmatrix1 & 0 \\ 0 & 1endbmatrix + beginbmatrix30 & -60 \\ 15 & -30endbmatrix$$A^{15} = I + 15N = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + 15\begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + \begin{bmatrix}30 & -60 \\ 15 & -30\end{bmatrix}$$
A^15 = beginbmatrix31 & -60 \\ 15 & -29endbmatrix$$A^{15} = \begin{bmatrix}31 & -60 \\ 15 & -29\end{bmatrix}$$
### Step 2: Evaluate Matrix Operation
A^15 + B = beginbmatrix31 & -60 \\ 15 & -29endbmatrix + beginbmatrix-29 & 49 \\ -13 & 18endbmatrix = beginbmatrix2 & -11 \\ 2 & -11endbmatrix$$A^{15} + B = \begin{bmatrix}31 & -60 \\ 15 & -29\end{bmatrix} + \begin{bmatrix}-29 & 49 \\ -13 & 18\end{bmatrix} = \begin{bmatrix}2 & -11 \\ 2 & -11\end{bmatrix}$$
Given (A^15 + B)beginbmatrixx \\ yendbmatrix = beginbmatrix0 \\ 0endbmatrix$(A^{15} + B)\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$:
beginbmatrix2 & -11 \\ 2 & -11endbmatrixbeginbmatrixx \\ yendbmatrix = beginbmatrix0 \\ 0endbmatrix$$\begin{bmatrix}2 & -11 \\ 2 & -11\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$$
This yields a single independent equation:
2x - 11y = 0 implies 2x = 11y$$2x - 11y = 0 \implies 2x = 11y$$
Checking the options, for x=11, y=2$x=11, y=2$, 2(11) = 11(2) implies 22 = 22$2(11) = 11(2) \implies 22 = 22$. Thus, Option 3 satisfies the condition.
### Pattern Recognition
Whenever you need a high power of a 2times 2$2\times 2$ matrix, check its trace and determinant. If |A-lambda I|=0$|A-\lambda I|=0$ yields lambda=1,1$\lambda=1,1$, then A$A$ can be written as I+N$I+N$ where N$N$ is nilpotent. This bypasses lengthy matrix multiplication.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices and Determinants
Q18
jee_main_2026_21_jan_evening
System of Linear Equations
If the
system of equations
3x + y + 4z = 3$3x + y + 4z = 3$
2x + alpha y - z = -3$2x + \alpha y - z = -3$
x + 2y + z = 4$x + 2y + z = 4$
has no solution, then the value of
alpha$\alpha$ is equal to:
- A. 19$19$
- B. 4$4$
- C. 13$13$
- D. 23$23$
Solution
### Related Formula
textFor no solution in AX = B, text we must have |A| = 0 text (i.e. Delta = 0 text) text and at least one of Delta_x, Delta_y, Delta_z neq 0.$$\text{For no solution in } AX = B, \text{ we must have } |A| = 0 \text{ (i.e. } \Delta = 0 \text{)} \text{ and at least one of } \Delta_x, \Delta_y, \Delta_z \neq 0.$$
### Core Logic
Evaluate the main determinant Delta$\Delta$ to 0$0$ to find alpha$\alpha$. Then quickly verify Delta_x neq 0$\Delta_x \neq 0$ to confirm it produces no solution.
### Step 1: Calculate Delta
Delta = beginvmatrix 3 & 1 & 4 \\ 2 & alpha & -1 \\ 1 & 2 & 1 endvmatrix = 0$$\Delta = \begin{vmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$$
Expand along the first row:
3(alpha + 2) - 1(2 - (-1)) + 4(4 - alpha) = 0$$3(\alpha + 2) - 1(2 - (-1)) + 4(4 - \alpha) = 0$$
3alpha + 6 - 3 + 16 - 4alpha = 0$$3\alpha + 6 - 3 + 16 - 4\alpha = 0$$
19 - alpha = 0 implies alpha = 19$$19 - \alpha = 0 \implies \alpha = 19$$
### Step 2: Verification of Delta_x
For alpha = 19$\alpha = 19$:
Delta_x = beginvmatrix 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 endvmatrix$$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 \end{vmatrix}$$
Delta_x = 3(19 + 2) - 1(-3 + 4) + 4(-6 - 76)$$\Delta_x = 3(19 + 2) - 1(-3 + 4) + 4(-6 - 76)$$
Delta_x = 3(21) - 1(1) + 4(-82) = 63 - 1 - 328 neq 0$$\Delta_x = 3(21) - 1(1) + 4(-82) = 63 - 1 - 328 \neq 0$$
Since Delta = 0$\Delta = 0$ and Delta_x neq 0$\Delta_x \neq 0$, the system has no solution for alpha = 19$\alpha = 19$.
### Pattern Recognition
Unless the problem specifically asks to check between infinite and no solution, setting the coefficient determinant Delta = 0$\Delta = 0$ directly yields the unique required parameter value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices and Determinants
Q13
jee_main_2026_22_january_morning
Properties of Determinants
If
A=beginbmatrix2 & 3 \\ 3 & 5endbmatrix$A=\begin{bmatrix}2 & 3 \\ 3 & 5\end{bmatrix}$, then the
determinant of the matrix (A^2025-3A^2024+A^2023)$(A^{2025}-3A^{2024}+A^{2023})$ is
- A. 28$28$
- B. 12$12$
- C. 24$24$
- D. 16$16$
Solution
### Related Formula
|AB| = |A||B|$|AB| = |A||B|$
|A^n| = |A|^n$|A^n| = |A|^n$
### Core Logic
Given A = beginbmatrix2 & 3 \\ 3 & 5endbmatrix$A = \begin{bmatrix}2 & 3 \\ 3 & 5\end{bmatrix}$.
The determinant |A| = (2)(5) - (3)(3) = 10 - 9 = 1$|A| = (2)(5) - (3)(3) = 10 - 9 = 1$.
We need to find the determinant of X = A^2025 - 3A^2024 + A^2023$X = A^{2025} - 3A^{2024} + A^{2023}$.
Factor out the lowest power of A:
X = A^2023(A^2 - 3A + I)$$X = A^{2023}(A^2 - 3A + I)$$
### Step 1: Determinant Factoring
Using properties of determinants:
|X| = |A^2023| cdot |A^2 - 3A + I|$$|X| = |A^{2023}| \cdot |A^2 - 3A + I|$$
Since |A| = 1$|A| = 1$, |A|^2023 = 1$|A|^{2023} = 1$.
Therefore, |X| = |A^2 - 3A + I|$|X| = |A^2 - 3A + I|$.
### Step 2: Matrix Computation
Calculate A^2$A^2$:
A^2 = beginbmatrix2 & 3 \\ 3 & 5endbmatrix beginbmatrix2 & 3 \\ 3 & 5endbmatrix = beginbmatrix4+9 & 6+15 \\ 6+15 & 9+25endbmatrix = beginbmatrix13 & 21 \\ 21 & 34endbmatrix$$A^2 = \begin{bmatrix}2 & 3 \\ 3 & 5\end{bmatrix} \begin{bmatrix}2 & 3 \\ 3 & 5\end{bmatrix} = \begin{bmatrix}4+9 & 6+15 \\ 6+15 & 9+25\end{bmatrix} = \begin{bmatrix}13 & 21 \\ 21 & 34\end{bmatrix}$$
Calculate -3A$-3A$:
-3A = beginbmatrix-6 & -9 \\ -9 & -15endbmatrix$$-3A = \begin{bmatrix}-6 & -9 \\ -9 & -15\end{bmatrix}$$
Assemble A^2 - 3A + I$A^2 - 3A + I$:
A^2 - 3A + I = beginbmatrix13 & 21 \\ 21 & 34endbmatrix + beginbmatrix-6 & -9 \\ -9 & -15endbmatrix + beginbmatrix1 & 0 \\ 0 & 1endbmatrix$$A^2 - 3A + I = \begin{bmatrix}13 & 21 \\ 21 & 34\end{bmatrix} + \begin{bmatrix}-6 & -9 \\ -9 & -15\end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$
= beginbmatrix13 - 6 + 1 & 21 - 9 + 0 \\ 21 - 9 + 0 & 34 - 15 + 1endbmatrix = beginbmatrix8 & 12 \\ 12 & 20endbmatrix$$= \begin{bmatrix}13 - 6 + 1 & 21 - 9 + 0 \\ 21 - 9 + 0 & 34 - 15 + 1\end{bmatrix} = \begin{bmatrix}8 & 12 \\ 12 & 20\end{bmatrix}$$
### Step 3: Final Determinant
|A^2 - 3A + I| = beginvmatrix8 & 12 \\ 12 & 20endvmatrix$$|A^2 - 3A + I| = \begin{vmatrix}8 & 12 \\ 12 & 20\end{vmatrix}$$
= (8)(20) - (12)(12) = 160 - 144 = 16$$= (8)(20) - (12)(12) = 160 - 144 = 16$$
### Pattern Recognition
High-powered matrix polynomials simplify radically by factoring out the lowest power. Since |A| = 1$|A| = 1$, the leading coefficient collapses completely, leaving only a basic A^2$A^2$ evaluation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Matrices and Determinants
Q22
jee_main_2026_22_january_morning
Adjoint and Determinant
Let A be a 3 times 3$3 \times 3$ matrix such that A + A^T = O$A + A^{T} = O$. If
mathrmA left[ beginarrayl 1 \\ - 1 \\ 0 endarray right] = left[ beginarrayl 3 \\ 3 \\ 2 endarray right], mathrmA ^ 2 left[ beginarrayl 1 \\ - 1 \\ 0 endarray right] = left[ beginarrayl - 3 \\ 1 9 \\ - 2 4 endarray right] $$ \mathrm{A} \left[ \begin{array}{l} 1 \\ - 1 \\ 0 \end{array} \right] = \left[ \begin{array}{l} 3 \\ 3 \\ 2 \end{array} \right], \mathrm{A} ^ {2} \left[ \begin{array}{l} 1 \\ - 1 \\ 0 \end{array} \right] = \left[ \begin{array}{l} - 3 \\ 1 9 \\ - 2 4 \end{array} \right] $$
and det(textadj(2textadj(A + I))) = (2)^alpha. (3)^beta. (11)^gamma, alpha, beta, gamma$\det(\text{adj}(2\text{adj}(A + I))) = (2)^{\alpha}. (3)^{\beta}. (11)^{\gamma}, \alpha, \beta, \gamma$ are non-negative integers, then alpha + beta + gamma$\alpha + \beta + \gamma$ is equal to ____
Numerical Answer. Answer: 18 to 18
Solution
### Related Formula
A = -A^T implies A text is a skew-symmetric matrix.$$A = -A^T \implies A \text{ is a skew-symmetric matrix.}$$
|textadj(kA)| = k^n(n-1)|A|^(n-1)$$|\text{adj}(kA)| = k^{n(n-1)}|A|^{(n-1)}$$
|textadj(M)| = |M|^n-1$$|\text{adj}(M)| = |M|^{n-1}$$
### Core Logic
Since
A$A$ is a
3 times 3$3 \times 3$ skew-symmetric matrix, let
A = beginbmatrix 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 endbmatrix$A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$.
Given
A beginbmatrix 1 \\ -1 \\ 0 endbmatrix = beginbmatrix 3 \\ 3 \\ 2 endbmatrix$A \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$:
beginbmatrix 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 endbmatrix beginbmatrix 1 \\ -1 \\ 0 endbmatrix = beginbmatrix -a \\ -a \\ -b+c endbmatrix$$ \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -a \\ -a \\ -b+c \end{bmatrix}$$
Wait,
(-a)(1) + (0)(-1) + (c)(0) = -a$(-a)(1) + (0)(-1) + (c)(0) = -a$. But wait,
-b(1) - c(-1) + 0 = -b+c$-b(1) - c(-1) + 0 = -b+c$.
Equating the result:
-a = 3 implies a = -3$-a = 3 \implies a = -3$
-b + c = 2 quad dots(1)$-b + c = 2 \quad \dots(1)$
### Step 1: Finding Matrix A Using Second Given Condition
Given
A^2 beginbmatrix 1 \\ -1 \\ 0 endbmatrix = beginbmatrix -3 \\ 19 \\ -24 endbmatrix$A^2 \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$.
We know
A beginbmatrix 1 \\ -1 \\ 0 endbmatrix = beginbmatrix 3 \\ 3 \\ 2 endbmatrix$A \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$.
Therefore,
A beginbmatrix 3 \\ 3 \\ 2 endbmatrix = beginbmatrix -3 \\ 19 \\ -24 endbmatrix$A \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$.
beginbmatrix 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 endbmatrix beginbmatrix 3 \\ 3 \\ 2 endbmatrix = beginbmatrix 3a + 2b \\ -3a + 2c \\ -3b - 3c endbmatrix = beginbmatrix -3 \\ 19 \\ -24 endbmatrix $$ \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 3a + 2b \\ -3a + 2c \\ -3b - 3c \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix} $$
Substitute
a = -3$a = -3$:
3(-3) + 2b = -3 implies -9 + 2b = -3 implies 2b = 6 implies b = 3$3(-3) + 2b = -3 \implies -9 + 2b = -3 \implies 2b = 6 \implies b = 3$
Use equation (1):
-b + c = 2 implies -3 + c = 2 implies c = 5$-b + c = 2 \implies -3 + c = 2 \implies c = 5$.
So,
a = -3, b = 3, c = 5$a = -3, b = 3, c = 5$. Matrix
A = beginbmatrix 0 & -3 & 3 \\ 3 & 0 & 5 \\ -3 & -5 & 0 endbmatrix$A = \begin{bmatrix} 0 & -3 & 3 \\ 3 & 0 & 5 \\ -3 & -5 & 0 \end{bmatrix}$.
### Step 2: Calculate Determinant of A + I
A + I = beginbmatrix 1 & -3 & 3 \\ 3 & 1 & 5 \\ -3 & -5 & 1 endbmatrix$$A + I = \begin{bmatrix} 1 & -3 & 3 \\ 3 & 1 & 5 \\ -3 & -5 & 1 \end{bmatrix}$$
|A + I| = 1(1 - (-25)) - (-3)(3 - (-15)) + 3(-15 - (-3))$$|A + I| = 1(1 - (-25)) - (-3)(3 - (-15)) + 3(-15 - (-3))$$
= 1(26) + 3(18) + 3(-12)$$= 1(26) + 3(18) + 3(-12)$$
= 26 + 54 - 36 = 44$$= 26 + 54 - 36 = 44$$
### Step 3: Calculating Adjoint Target Expression
Target:
det(textadj(2textadj(A + I)))$\det(\text{adj}(2\text{adj}(A + I)))$. Let
M = A+I$M = A+I$.
For a
3 times 3$3 \times 3$ matrix,
|textadj(X)| = |X|^2$|\text{adj}(X)| = |X|^2$.
det(textadj(2textadj M)) = |2textadj M|^2 $$ \det(\text{adj}(2\text{adj} M)) = |2\text{adj} M|^2 $$
We know
|kY| = k^3|Y|$|kY| = k^3|Y|$ for
3 times 3$3 \times 3$. So
|2textadj M| = 2^3 |textadj M| = 8|M|^2$|2\text{adj} M| = 2^3 |\text{adj} M| = 8|M|^2$.
(8|M|^2)^2 = 64|M|^4 $$ (8|M|^2)^2 = 64|M|^4 $$
Substitute
|M| = 44$|M| = 44$:
64(44)^4 = 2^6 cdot (4 cdot 11)^4 = 2^6 cdot (2^2 cdot 11)^4 = 2^6 cdot 2^8 cdot 11^4 = 2^14 cdot 3^0 cdot 11^4 $$ 64(44)^4 = 2^6 \cdot (4 \cdot 11)^4 = 2^6 \cdot (2^2 \cdot 11)^4 = 2^6 \cdot 2^8 \cdot 11^4 = 2^{14} \cdot 3^0 \cdot 11^4 $$
Comparing to
2^alpha cdot 3^beta cdot 11^gamma$2^\alpha \cdot 3^\beta \cdot 11^\gamma$:
alpha = 14, beta = 0, gamma = 4$\alpha = 14, \beta = 0, \gamma = 4$.
alpha + beta + gamma = 14 + 0 + 4 = 18$\alpha + \beta + \gamma = 14 + 0 + 4 = 18$.
### Pattern Recognition
Skew-symmetric matrices only have 3 unknowns. Matrix multiplication acts linearly;
A^2 x = A(Ax)$A^2 x = A(Ax)$ provides immediate simultaneous equations without having to compute the heavy
A^2$A^2$ explicitly. Determinant adjoint loops always reduce to
|A|^(n-1)^k$|A|^{(n-1)^k}$ modulated by constant pullouts.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Maths: Matrices and Determinants
