CBSE Class 12 Mathematics Matrices And Determinants Worksheet

CBSE Class 12 Mathematics Matrices & Determinants (1). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

ONE MARKS QUESTIONS

Question. Show by means of an example that the product of two non-zero matrices can be a zero matrix.
Answer: Let \( A \) and \( B \) be two non-zero matrices defined as:
\( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \)
Now, let's compute their product \( AB \):
\( AB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0\cdot 1 + 1\cdot 0 & 0\cdot 0 + 1\cdot 0 \\ 0\cdot 1 + 0\cdot 0 & 0\cdot 0 + 0\cdot 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
Since \( AB \) is a zero matrix while \( A \neq O \) and \( B \neq O \), it is shown that the product of two non-zero matrices can be a zero matrix.

Question. Construct a 3 × 2 matrix whose elements are given by \( a_{ij} = e^{ix}\sin jx \). (Exemplar).
Answer: A \( 3 \times 2 \) matrix \( A \) is represented as:
\( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} \)
Given \( a_{ij} = e^{ix}\sin(jx) \), we compute each element by substituting the corresponding values of \( i \) and \( j \):
For \( i = 1, j = 1 \implies a_{11} = e^x\sin x \)
For \( i = 1, j = 2 \implies a_{12} = e^x\sin 2x \)
For \( i = 2, j = 1 \implies a_{21} = e^{2x}\sin x \)
For \( i = 2, j = 2 \implies a_{22} = e^{2x}\sin 2x \)
For \( i = 3, j = 1 \implies a_{31} = e^{3x}\sin x \)
For \( i = 3, j = 2 \implies a_{32} = e^{3x}\sin 2x \)
Thus, the required matrix is:
\( A = \begin{bmatrix} e^x\sin x & e^x\sin 2x \\ e^{2x}\sin x & e^{2x}\sin 2x \\ e^{3x}\sin x & e^{3x}\sin 2x \end{bmatrix} \)

Question. Solve for x and y for \( x \begin{bmatrix} 2 \\ 1 \end{bmatrix} + y \begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \) (Exemplar).
Answer: Simplifying the matrix equation yields:
\( \begin{bmatrix} 2x \\ x \end{bmatrix} + \begin{bmatrix} 3y \\ 5y \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2x + 3y - 8 \\ x + 5y - 11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)
Equating corresponding elements, we obtain a system of two linear equations:
\( 2x + 3y = 8 \) --- (1)
\( x + 5y = 11 \) --- (2)
From equation (2), we get \( x = 11 - 5y \). Substituting this value into equation (1):
\( 2(11 - 5y) + 3y = 8 \)
\( 22 - 10y + 3y = 8 \)
\( -7y = -14 \implies y = 2 \)
Substituting \( y = 2 \) into \( x = 11 - 5y \):
\( x = 11 - 5(2) = 1 \)
Thus, \( x = 1, y = 2 \).

Question. Give an example of matrices A, B and C such that AB = AC, Where A is non-zero matrix, but B≠C.
Answer: Let \( A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), and \( C = \begin{bmatrix} 1 & 2 \\ 5 & 6 \end{bmatrix} \).
Here, \( A \) is a non-zero matrix and \( B \neq C \). Let's compute \( AB \) and \( AC \):
\( AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1(1)+0(3) & 1(2)+0(4) \\ 0(1)+0(3) & 0(2)+0(4) \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \)
\( AC = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 1(1)+0(5) & 1(2)+0(6) \\ 0(1)+0(5) & 0(2)+0(6) \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \)
Since \( AB = AC \) and \( B \neq C \), this serves as a valid example.

Question. Show that \( A^T A \) and \( AA^T \) are both symmetric matrices for any matrix A. (Exemplar).
Answer: A matrix \( X \) is symmetric if and only if \( X^T = X \).
Let \( X = A^T A \). Taking the transpose of both sides:
\( X^T = (A^T A)^T = A^T (A^T)^T \)
Since \( (A^T)^T = A \), we have:
\( X^T = A^T A = X \)
Therefore, \( A^T A \) is a symmetric matrix.

Similarly, let \( Y = AA^T \). Taking the transpose of both sides:
\( Y^T = (AA^T)^T = (A^T)^T A^T \)
Since \( (A^T)^T = A \), we have:
\( Y^T = AA^T = Y \)
Therefore, \( AA^T \) is also a symmetric matrix.

FOUR MARKS QUESTIONS

Question. If \( A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \) prove that \( A^2 - 4A - 5I = O \) Hence find \( A^{-1} \).
Answer: First, let's compute \( A^2 \):
\( A^2 = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} \)
Now, let's evaluate \( A^2 - 4A - 5I \):
\( A^2 - 4A - 5I = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - 4 \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \)
\( = \begin{bmatrix} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O \).
Hence, \( A^2 - 4A - 5I = O \).

To find \( A^{-1} \), we pre-multiply the equation by \( A^{-1} \):
\( A^{-1}(A^2 - 4A - 5I) = A^{-1}O \)
\( A - 4I - 5A^{-1} = O \)
\( \Rightarrow 5A^{-1} = A - 4I \)
\( \Rightarrow A^{-1} = \frac{1}{5} (A - 4I) \)
Let's find \( A - 4I \):
\( A - 4I = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix} \)
Thus,
\( A^{-1} = \frac{1}{5} \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix} \).

Question. Given \( A = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \) show by induction that \( A^n = \begin{pmatrix} 1+2n & -4n \\ n & 1-2n \end{pmatrix} \).
Answer: Let \( P(n) \) be the statement:
\( A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix} \)
Step 1: For \( n=1 \)
\( L.H.S = A^1 = A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \)
\( R.H.S = \begin{bmatrix} 1+2(1) & -4(1) \\ 1 & 1-2(1) \end{bmatrix} = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \)
Since \( L.H.S = R.H.S \), \( P(1) \) is true.

Step 2: Assume \( P(k) \) is true for some \( k \in \mathbb{N} \)
\( A^k = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix} \) --- (1)

Step 3: Prove that \( P(k+1) \) is true
We need to prove that:
\( A^{k+1} = \begin{bmatrix} 1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1) \end{bmatrix} = \begin{bmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{bmatrix} \)
Now, using equation (1):
\( A^{k+1} = A^k \cdot A = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \)
\( = \begin{bmatrix} (1+2k)(3) + (-4k)(1) & (1+2k)(-4) + (-4k)(-1) \\ k(3) + (1-2k)(1) & k(-4) + (1-2k)(-1) \end{bmatrix} \)
\( = \begin{bmatrix} 3 + 6k - 4k & -4 - 8k + 4k \\ 3k + 1 - 2k & -4k - 1 + 2k \end{bmatrix} \)
\( = \begin{bmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{bmatrix} \)
Therefore, \( P(k+1) \) is true. By the principle of mathematical induction, the statement is true for all \( n \in \mathbb{N} \).

Question. If \( X = \begin{bmatrix} 3 & 1 & -1 \\ 5 & -2 & -3 \end{bmatrix} \) and \( Y = \begin{bmatrix} 2 & 1 & -1 \\ 7 & 2 & 4 \end{bmatrix} \), Find a matrix Z such that X+Y+Z is a zero matrix. (Exemplar).
Answer: Given \( X+Y+Z = O \), we have:
\( Z = -(X+Y) \)
First, compute \( X+Y \):
\( X+Y = \begin{bmatrix} 3 & 1 & -1 \\ 5 & -2 & -3 \end{bmatrix} + \begin{bmatrix} 2 & 1 & -1 \\ 7 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 3+2 & 1+1 & -1-1 \\ 5+7 & -2+2 & -3+4 \end{bmatrix} = \begin{bmatrix} 5 & 2 & -2 \\ 12 & 0 & 1 \end{bmatrix} \)
Now, find \( Z \):
\( Z = -\begin{bmatrix} 5 & 2 & -2 \\ 12 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -5 & -2 & 2 \\ -12 & 0 & -1 \end{bmatrix} \)

Question. Find the matrix A satisfying the matrix equation : \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). (Exemplar).
Answer: Let the given equation be \( P A Q = I \), where:
\( P = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \) and \( Q = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} \)
To find \( A \), we pre-multiply by \( P^{-1} \) and post-multiply by \( Q^{-1} \):
\( P^{-1}(PAQ)Q^{-1} = P^{-1} I Q^{-1} \implies A = P^{-1} Q^{-1} \)
First, let's find \( P^{-1} \):
\( |P| = 2(2) - 1(3) = 4 - 3 = 1 \)
\( P^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \)
Next, find \( Q^{-1} \):
\( |Q| = (-3)(-3) - 2(5) = 9 - 10 = -1 \)
\( Q^{-1} = \frac{1}{-1} \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} \)
Now, compute \( A \):
\( A = P^{-1} Q^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} 2(3)+(-1)(5) & 2(2)+(-1)(3) \\ -3(3)+2(5) & -3(2)+2(3) \end{bmatrix} \)
\( A = \begin{bmatrix} 6-5 & 4-3 \\ -9+10 & -6+6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \)

Question. Prove by mathematical induction that \( (A^T)^n = (A^n)^T \), where \( n \in \mathbb{N} \) for any square matrix A. (Exemplar).
Answer: Let the statement be \( P(n) \): \( (A^T)^n = (A^n)^T \).
Step 1: For \( n=1 \)
\( L.H.S = (A^T)^1 = A^T \)
\( R.H.S = (A^1)^T = A^T \)
Since \( L.H.S = R.H.S \), the statement \( P(1) \) is true.

Step 2: Assume \( P(k) \) is true for some \( k \in \mathbb{N} \)
\( (A^T)^k = (A^k)^T \) --- (1)

Step 3: Prove that \( P(k+1) \) is true
We need to prove that \( (A^T)^{k+1} = (A^{k+1})^T \).
\( L.H.S = (A^T)^{k+1} = (A^T)^k \cdot A^T \)
Using equation (1):
\( (A^T)^{k+1} = (A^k)^T \cdot A^T \)
Using the transpose property \( Y^T X^T = (XY)^T \):
\( (A^k)^T A^T = (A \cdot A^k)^T = (A^{k+1})^T = R.H.S \)
Therefore, \( P(k+1) \) is true. By the principle of mathematical induction, the statement is true for all \( n \in \mathbb{N} \).

Question. If \( F(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \) show that \( F(\theta)F(\phi) = F(\theta + \phi) \).
Answer: Let's multiply \( F(\theta) \) and \( F(\phi) \):
\( F(\theta) F(\phi) = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos\phi & -\sin\phi & 0 \\ \sin\phi & \cos\phi & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} \cos\theta\cos\phi - \sin\theta\sin\phi + 0 & \cos\theta(-\sin\phi) - \sin\theta\cos\phi + 0 & 0 \\ \sin\theta\cos\phi + \cos\theta\sin\phi + 0 & \sin\theta(-\sin\phi) + \cos\theta\cos\phi + 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} \cos\theta\cos\phi - \sin\theta\sin\phi & -(\sin\theta\cos\phi + \cos\theta\sin\phi) & 0 \\ \sin\theta\cos\phi + \cos\theta\sin\phi & \cos\theta\cos\phi - \sin\theta\sin\phi & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Using the trigonometric sum formulas:
\( \cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi \)
\( \sin(\theta+\phi) = \sin\theta\cos\phi + \cos\theta\sin\phi \)
We obtain:
\( F(\theta) F(\phi) = \begin{bmatrix} \cos(\theta+\phi) & -\sin(\theta+\phi) & 0 \\ \sin(\theta+\phi) & \cos(\theta+\phi) & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(\theta + \phi) \).

Question. Find the inverse by elementary Operations \( \begin{bmatrix} 2 & -1 & 3 \\ 1 & 3 & -1 \\ 3 & 2 & 1 \end{bmatrix} \).
Answer: Let \( A = \begin{bmatrix} 2 & -1 & 3 \\ 1 & 3 & -1 \\ 3 & 2 & 1 \end{bmatrix} \). We write \( A = IA \):
\( \begin{bmatrix} 2 & -1 & 3 \\ 1 & 3 & -1 \\ 3 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \)
Applying \( R_1 \leftrightarrow R_2 \):
\( \begin{bmatrix} 1 & 3 & -1 \\ 2 & -1 & 3 \\ 3 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \) and \( R_3 \rightarrow R_3 - 3R_1 \):
\( \begin{bmatrix} 1 & 3 & -1 \\ 0 & -7 & 5 \\ 0 & -7 & 4 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 0 & -3 & 1 \end{bmatrix} A \)
Applying \( R_2 \rightarrow R_2 - R_3 \):
\( \begin{bmatrix} 1 & 3 & -1 \\ 0 & 0 & 1 \\ 0 & -7 & 4 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & -1 \\ 0 & -3 & 1 \end{bmatrix} A \)
Applying \( R_2 \leftrightarrow R_3 \):
\( \begin{bmatrix} 1 & 3 & -1 \\ 0 & -7 & 4 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & -3 & 1 \\ 1 & 1 & -1 \end{bmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 4R_3 \):
\( \begin{bmatrix} 1 & 3 & -1 \\ 0 & -7 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ -4 & -7 & 5 \\ 1 & 1 & -1 \end{bmatrix} A \)
Applying \( R_2 \rightarrow -\frac{1}{7}R_2 \):
\( \begin{bmatrix} 1 & 3 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ \frac{4}{7} & 1 & -\frac{5}{7} \\ 1 & 1 & -1 \end{bmatrix} A \)
Applying \( R_1 \rightarrow R_1 + R_3 \):
\( \begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & -1 \\ \frac{4}{7} & 1 & -\frac{5}{7} \\ 1 & 1 & -1 \end{bmatrix} A \)
Applying \( R_1 \rightarrow R_1 - 3R_2 \):
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{5}{7} & -1 & \frac{8}{7} \\ \frac{4}{7} & 1 & -\frac{5}{7} \\ 1 & 1 & -1 \end{bmatrix} A \)
Thus, the inverse matrix is:
\( A^{-1} = \frac{1}{7} \begin{bmatrix} -5 & -7 & 8 \\ 4 & 7 & -5 \\ 7 & 7 & -7 \end{bmatrix} \)

Question. Express the matrix \( \begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{bmatrix} \) as the sum of a symmetric and skew symmetric matrix. (Exemplar).
Answer: Any square matrix \( A \) can be expressed as \( A = P + Q \), where \( P = \frac{1}{2}(A + A^T) \) is a symmetric matrix and \( Q = \frac{1}{2}(A - A^T) \) is a skew-symmetric matrix.
Given:
\( A = \begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{bmatrix} \implies A^T = \begin{bmatrix} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{bmatrix} \)
Let's find \( P \):
\( P = \frac{1}{2}(A + A^T) = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{bmatrix} + \begin{bmatrix} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix} 4 & 4 & 5 \\ 4 & -2 & 3 \\ 5 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{bmatrix} \)
Here, we can verify that \( P^T = P \), making \( P \) symmetric.

Let's find \( Q \):
\( Q = \frac{1}{2}(A - A^T) = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -\frac{3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} & 0 \end{bmatrix} \)
Here, we can verify that \( Q^T = -Q \), making \( Q \) skew-symmetric.

Thus, the matrix \( A \) expressed as the sum of a symmetric and skew-symmetric matrix is:
\( \begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{bmatrix} + \begin{bmatrix} 0 & 1 & -\frac{3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} & 0 \end{bmatrix} \)

Question. Find the value of x, if \( \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O \).
Answer: Let's multiply the first two matrices first:
\( \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 1(1) + 2x + 15(1) & 1(3) + 5x + 3(1) & 1(2) + x + 2(1) \end{bmatrix} \)
\( = \begin{bmatrix} 16 + 2x & 6 + 5x & 4 + x \end{bmatrix} \)
Now multiply this result by the third matrix:
\( \begin{bmatrix} 16 + 2x & 6 + 5x & 4 + x \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = [0] \)
\( \Rightarrow (16 + 2x)\cdot 1 + (6 + 5x)\cdot 2 + (4 + x)\cdot x = 0 \)
\( \Rightarrow 16 + 2x + 12 + 10x + 4x + x^2 = 0 \)
\( \Rightarrow x^2 + 16x + 28 = 0 \)
Factoring the quadratic equation:
\( x^2 + 14x + 2x + 28 = 0 \)
\( x(x + 14) + 2(x + 14) = 0 \)
\( (x + 2)(x + 14) = 0 \)
Thus, \( x = -2 \) or \( x = -14 \).