Misc Exercise Practice
Matrices & Determinants: Systems of Equations, Adjoint, and Properties
Q1
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Evaluate the determinant: \( \Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix} \)
Step 1: Row Operations
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \).
\( \Delta = \begin{vmatrix} 1 & x & yz \\ 0 & y-x & zx-yz \\ 0 & z-x & xy-yz \end{vmatrix} \).
\( \Delta = \begin{vmatrix} 1 & x & yz \\ 0 & y-x & zx-yz \\ 0 & z-x & xy-yz \end{vmatrix} \).
Step 2: Factorize
Row 2 becomes \( 0, -(x-y), -z(x-y) \). Row 3 becomes \( 0, -(x-z), -y(x-z) \).
Take out \( (x-y) \) and \( (z-x) \).
Take out \( (x-y) \) and \( (z-x) \).
Step 3: Expand
\( (x-y)(z-x) \begin{vmatrix} 1 & x & yz \\ 0 & -1 & -z \\ 0 & 1 & -y \end{vmatrix} \).
Expand along C1: \( 1(y - (-z)) = y+z \).
Expand along C1: \( 1(y - (-z)) = y+z \).
Value = \( (x-y)(y-z)(z-x) \)
Q2
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If \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \), show that \( A^3 - 6A^2 + 5A +
11I = O \). Hence find \( A^{-1} \).
Part 1: Verification
Calculate \( A^2 \) and \( A^3 \). Substitute into the equation.
Result will be the Zero Matrix \( O \).
Result will be the Zero Matrix \( O \).
Part 2: Find Inverse
Multiply equation by \( A^{-1} \):
\( A^2 - 6A + 5I + 11A^{-1} = O \).
\( 11A^{-1} = 6A - A^2 - 5I \).
\( A^{-1} = \frac{1}{11} (6A - A^2 - 5I) \).
\( A^2 - 6A + 5I + 11A^{-1} = O \).
\( 11A^{-1} = 6A - A^2 - 5I \).
\( A^{-1} = \frac{1}{11} (6A - A^2 - 5I) \).
Q3
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Solve the system of equations using matrix method:
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)
Step 1: Form Matrices
\( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y
\\ z \end{bmatrix} \), \( B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \).
Step 2: Determinant of A
\( |A| = 2(0) - (-3)(-2) + 5(1) = -6 + 5 = -1 \).
Since \( |A| \ne 0 \), inverse exists.
Since \( |A| \ne 0 \), inverse exists.
Step 3: Adjoint and Inverse
Calculate Cofactors.
\( A^{-1} = \frac{1}{|A|} \text{adj } A = -1 \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \).
\( A^{-1} = \frac{1}{|A|} \text{adj } A = -1 \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \).
Step 4: Solve X = A⁻¹B
Multiply \( A^{-1} \) with \( B \).
\( x=1, y=2, z=3 \)
Q4
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If \( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \), find \( A^{-1} \). Using \(
A^{-1} \), solve the system:
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)
Note
This matrix A is identical to the coefficient matrix in Q3. The solution follows the exact same steps.
\( A^{-1} \) is calculated above. Solution: \( x=1, y=2, z=3 \).
Q5
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Use the product \( \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 0 &
1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \) to solve the system of equations:
\( x - y + 2z = 1 \)
\( 2y - 3z = 1 \)
\( 3x - 2y + 4z = 2 \)
\( x - y + 2z = 1 \)
\( 2y - 3z = 1 \)
\( 3x - 2y + 4z = 2 \)
Step 1: Calculate Product
The product of the two matrices results in the Identity Matrix \( I \).
Thus, the second matrix is the inverse of the first.
Thus, the second matrix is the inverse of the first.
Step 2: Solve
The system's coefficient matrix matches the first matrix.
\( X = A^{-1}B \). Use the second matrix as \( A^{-1} \).
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \).
\( X = A^{-1}B \). Use the second matrix as \( A^{-1} \).
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \).
\( x=0, y=5, z=3 \)
Q6
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If \( x, y, z \) are non-zero real numbers, then find the inverse of matrix \( A = \begin{bmatrix} x & 0 & 0 \\
0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \).
Diagonal Matrix Property
The inverse of a diagonal matrix is a diagonal matrix with reciprocal elements.
\( A^{-1} = \begin{bmatrix} 1/x & 0 & 0 \\ 0 & 1/y & 0 \\ 0 & 0 & 1/z \end{bmatrix} \)
Q7
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Given \( A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix} \), show that \( A^T A^{-1} =
\begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix} \).
Step 1: Find Inverse
\( |A| = 1 + \tan^2 x = \sec^2 x \).
\( A^{-1} = \frac{1}{\sec^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \).
\( A^{-1} = \frac{1}{\sec^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \).
Step 2: Multiply
Multiply \( A^T \) with \( A^{-1} \) and simplify using trig identities like \( \frac{1-\tan^2 x}{1+\tan^2 x}
= \cos 2x \).
Result verified.
Q8
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Let \( A \) be a non-singular square matrix of order \( 3 \times 3 \). Then \( | \text{adj } A | \) is equal
to:
(A) \( |A| \)
(B) \( |A|^2 \)
(C) \( |A|^3 \)
(D) \( 3|A| \)
(A) \( |A| \)
(B) \( |A|^2 \)
(C) \( |A|^3 \)
(D) \( 3|A| \)
Formula
For a matrix of order \( n \), \( |\text{adj } A| = |A|^{n-1} \).
Here \( n=3 \), so \( |A|^{3-1} = |A|^2 \).
Here \( n=3 \), so \( |A|^{3-1} = |A|^2 \).
Correct Option: (B)
Q9
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If \( A \) is an invertible matrix of order 2, then \( \text{det}(A^{-1}) \) is equal to:
(A) \( \text{det}(A) \)
(B) \( 1/\text{det}(A) \)
(C) 1
(D) 0
(A) \( \text{det}(A) \)
(B) \( 1/\text{det}(A) \)
(C) 1
(D) 0
Property
\( A A^{-1} = I \Rightarrow |A||A^{-1}| = |I| = 1 \).
\( |A^{-1}| = 1/|A| \).
\( |A^{-1}| = 1/|A| \).
Correct Option: (B)