Exercise 3.4 Practice
Invertible Matrices: Concepts and Properties
Q1
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Two matrices \( X \) and \( Y \) of the same order are said to be invertible matrices of each other if and only
if:
(A) \( XY = O, YX = I \)
(B) \( XY = YX = I \)
(C) \( XY = YX = O \)
(D) \( XY = I, YX = O \)
(A) \( XY = O, YX = I \)
(B) \( XY = YX = I \)
(C) \( XY = YX = O \)
(D) \( XY = I, YX = O \)
Definition
By definition of invertible matrices, a square matrix \( Y \) is called the inverse of square matrix \( X \)
if \( XY = YX = I \), where \( I \) is the identity matrix.
Correct Option: (B) \( XY = YX = I \)
Q2
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Let \( A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & -5 \\ -1 & 2
\end{bmatrix} \). Check whether \( B \) is the inverse of \( A \).
Calculate AB
\( AB = \begin{bmatrix} 2(3)+5(-1) & 2(-5)+5(2) \\ 1(3)+3(-1) & 1(-5)+3(2) \end{bmatrix} \)
\( = \begin{bmatrix} 6-5 & -10+10 \\ 3-3 & -5+6 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \)
\( = \begin{bmatrix} 6-5 & -10+10 \\ 3-3 & -5+6 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \)
Calculate BA
\( BA = \begin{bmatrix} 3(2)+(-5)(1) & 3(5)+(-5)(3) \\ -1(2)+2(1) & -1(5)+2(3) \end{bmatrix} \)
\( = \begin{bmatrix} 6-5 & 15-15 \\ -2+2 & -5+6 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \)
\( = \begin{bmatrix} 6-5 & 15-15 \\ -2+2 & -5+6 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \)
Since \( AB = BA = I \), \( B \) is the inverse of \( A \).
Q3
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If \( A \) and \( B \) are invertible matrices of the same order, then \( (AB)^{-1} \) is equal to:
(A) \( A^{-1}B^{-1} \)
(B) \( B^{-1}A^{-1} \)
(C) \( A^{-1} + B^{-1} \)
(D) \( B^{-1} - A^{-1} \)
(A) \( A^{-1}B^{-1} \)
(B) \( B^{-1}A^{-1} \)
(C) \( A^{-1} + B^{-1} \)
(D) \( B^{-1} - A^{-1} \)
Property
This is a standard property of matrices known as the Reversal Law of Inverses: The inverse of a product is the
product of the inverses in reverse order.
Correct Option: (B) \( B^{-1}A^{-1} \)
Q4
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Given \( A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \). If \( A^2 - 4A + I = O \), use this equation to
find \( A^{-1} \).
Multiply by Inverse
Given \( A^2 - 4A + I = O \).
Multiply entire equation by \( A^{-1} \):
\( A^{-1}(A^2) - 4A^{-1}(A) + A^{-1}(I) = A^{-1}(O) \)
\( A - 4I + A^{-1} = O \).
Multiply entire equation by \( A^{-1} \):
\( A^{-1}(A^2) - 4A^{-1}(A) + A^{-1}(I) = A^{-1}(O) \)
\( A - 4I + A^{-1} = O \).
Solve for A Inverse
\( A^{-1} = 4I - A \)
\( = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} \).
\( = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} \).
\( A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} \)
Q5
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Verify that for \( A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -2 \\ -1 &
3 \end{bmatrix} \), the result \( (AB)^{-1} = B^{-1}A^{-1} \) holds true. (Assume inverses exist and calculate
values).
Step 1: Calculate AB
\( AB = \begin{bmatrix} 2-1 & -4+3 \\ 3-4 & -6+12 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 6
\end{bmatrix} \).
Step 2: Find (AB)⁻¹
Det(AB) = \( 6 - 1 = 5 \).
\( (AB)^{-1} = \frac{1}{5} \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \).
\( (AB)^{-1} = \frac{1}{5} \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \).
Step 3: Find A⁻¹ and B⁻¹
Det(A) = 5. \( A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix} \).
Det(B) = 1. \( B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \).
Det(B) = 1. \( B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \).
Step 4: Calculate B⁻¹A⁻¹
\( B^{-1}A^{-1} = \frac{1}{5} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 4 & -1 \\ -3 & 2
\end{bmatrix} \)
\( = \frac{1}{5} \begin{bmatrix} 12-6 & -3+4 \\ 4-3 & -1+2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \).
\( = \frac{1}{5} \begin{bmatrix} 12-6 & -3+4 \\ 4-3 & -1+2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} \).
Verified: LHS = RHS.