# Exercise 3.4 Solutions

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Exercise 3.4 Solutions
Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17.
1. {\left[\begin{array}{cc} 1 & -1 \$5pt] 2 & 3 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 1 & -1 \\[5pt] 2 & 3 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 1 & -1 \\[5pt] 2 & 3 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 1 & -1 \\[5pt] 2 & 3 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - 2\text{R}_1} {\left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 5 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] -2 & 1 \end{array}\right]}A Applying {\text{R}_2 → \dfrac15\text{R}_2} {\left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] \dfrac{-2}{5} & \dfrac15 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 + \text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} \dfrac35 & \dfrac15 \\[10pt] \dfrac{-2}{5} & \dfrac15 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} \dfrac35 & \dfrac15 \\[10pt] \dfrac{-2}{5} & \dfrac15 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 1 & -1 \\[5pt] 2 & 3 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 1 & -1 \\[5pt] 2 & 3 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 1 & -1 \\[5pt] 2 & 3 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 + \text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 2 & 5 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 1 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_2 → \dfrac15\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 2 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & \dfrac15 \\[10pt] 0 & \dfrac15 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - 2\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} \dfrac35 & \dfrac15 \\[10pt] \dfrac{-2}{5} & \dfrac15 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} \dfrac35 & \dfrac15 \\[10pt] \dfrac{-2}{5} & \dfrac15 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 2. {\left[\begin{array}{cc} 2 & 1 \\[5pt] 1 & 1 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 1 \\[5pt] 1 & 1 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 2 & 1 \\[5pt] 1 & 1 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 2 & 1 \\[5pt] 1 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - \text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 1 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - \text{R}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] -1 & 2 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} 1 & -1 \\[5pt] -1 & 2 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 1 \\[5pt] 1 & 1 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 2 & 1 \\[5pt] 1 & 1 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 2 & 1 \\[5pt] 1 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - \text{C}_2} {\left[\begin{array}{cc} 1 & 1 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] -1 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 - \text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & -1 \\[5pt] -1 & 2 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} 1 & -1 \\[5pt] -1 & 2 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 3. {\left[\begin{array}{cc} 1 & 3 \\[5pt] 2 & 7 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 1 & 3 \\[5pt] 2 & 7 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 1 & 3 \\[5pt] 2 & 7 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 1 & 3 \\[5pt] 2 & 7 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - 2\text{R}_1} {\left[\begin{array}{cc} 1 & 3 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] -2 & 1 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - 3\text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 7 & -3 \\[5pt] -2 & 1 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} 7 & -3 \\[5pt] -2 & 1 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 1 & 3 \\[5pt] 2 & 7 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 1 & 3 \\[5pt] 2 & 7 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 1 & 3 \\[5pt] 2 & 7 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 - 3\text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 2 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & -3 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - 2\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 7 & -3 \\[5pt] -2 & 1 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} 7 & -3 \\[5pt] -2 & 1 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 4. {\left[\begin{array}{cc} 2 & 3 \\[5pt] 5 & 7 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 3 \\[5pt] 5 & 7 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 2 & 3 \\[5pt] 5 & 7 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 2 & 3 \\[5pt] 5 & 7 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → 3\text{R}_1 - \text{R}_2} {\left[\begin{array}{cc} 1 & 2 \\[5pt] 5 & 7 \end{array}\right]} {= \left[\begin{array}{cc} 3 & -1 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - 5\text{R}_1} {\left[\begin{array}{cc} 1 & 2 \\[5pt] 0 & -3 \end{array}\right]} {= \left[\begin{array}{cc} 3 & -1 \\[5pt] -15 & 6 \end{array}\right]}A Applying {\text{R}_2 → -\dfrac13\text{R}_2} {\left[\begin{array}{cc} 1 & 2 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 3 & -1 \\[5pt] 5 & -2 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - 2\text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} -7 & 3 \\[5pt] 5 & -2 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} -7 & 3 \\[5pt] 5 & -2 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 3 \\[5pt] 5 & 7 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 2 & 3 \\[5pt] 5 & 7 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 2 & 3 \\[5pt] 5 & 7 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → 2\text{C}_1 - \text{C}_2} {\left[\begin{array}{cc} 1 & 3 \\[5pt] 3 & 7 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & 0 \\[5pt] -1 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 - 3\text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 3 & -2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & -6 \\[5pt] -1 & 4 \end{array}\right]} Applying {\text{C}_2 → -\dfrac12\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 3 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & 3 \\[5pt] -1 & -2 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - 3\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} -7 & 3 \\[5pt] 5 & -2 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} -7 & 3 \\[5pt] 5 & -2 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 5. {\left[\begin{array}{cc} 2 & 1 \\[5pt] 7 & 4 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 1 \\[5pt] 7 & 4 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 2 & 1 \\[5pt] 7 & 4 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 2 & 1 \\[5pt] 7 & 4 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → 4\text{R}_1 - \text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 7 & 4 \end{array}\right]} {= \left[\begin{array}{cc} 4 & -1 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - 7\text{R}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 4 \end{array}\right]} {= \left[\begin{array}{cc} 4 & -1 \\[5pt] -28 & 8 \end{array}\right]}A Applying {\text{R}_2 → \dfrac14\text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 4 & -1 \\[5pt] -7 & 2 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} 4 & -1 \\[5pt] -7 & 2 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 1 \\[5pt] 7 & 4 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 2 & 1 \\[5pt] 7 & 4 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 2 & 1 \\[5pt] 7 & 4 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - \text{C}_2} {\left[\begin{array}{cc} 1 & 1 \\[5pt] 3 & 4 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] -1 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 - \text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 3 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & -1 \\[5pt] -1 & 2 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - 3\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 4 & -1 \\[5pt] -7 & 2 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} 4 & -1 \\[5pt] -7 & 2 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 6. {\left[\begin{array}{cc} 2 & 5 \\[5pt] 1 & 3 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 5 \\[5pt] 1 & 3 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 2 & 5 \\[5pt] 1 & 3 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 2 & 5 \\[5pt] 1 & 3 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - \text{R}_2} {\left[\begin{array}{cc} 1 & 2 \\[5pt] 1 & 3 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - \text{R}_1} {\left[\begin{array}{cc} 1 & 2 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] -1 & 2 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - 2\text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 3 & -5 \\[5pt] -1 & 2 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} 3 & -5 \\[5pt] -1 & 2 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 5 \\[5pt] 1 & 3 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 2 & 5 \\[5pt] 1 & 3 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 2 & 5 \\[5pt] 1 & 3 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → 3\text{C}_1 - \text{C}_1} {\left[\begin{array}{cc} 1 & 5 \\[5pt] 0 & 3 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 3 & 0 \\[5pt] -1 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2} - 5\text{C}_1 {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 3 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 3 & -15 \\[5pt] -1 & 6 \end{array}\right]} Applying {\text{C}_2 → \dfrac13\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 3 & -5 \\[5pt] -1 & 2 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} 3 & -5 \\[5pt] -1 & 2 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 7. {\left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & 2 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & 2 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & 2 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → 2\text{R}_1 - \text{R}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 5 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 2 & -1 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - 5\text{R}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 2 & -1 \\[5pt] -10 & 6 \end{array}\right]}A Applying {\text{R}_2 → \dfrac12\text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 2 & -1 \\[5pt] -5 & 3 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} 2 & -1 \\[5pt] -5 & 3 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & 2 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & 2 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & 2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - 2\text{C}_2} {\left[\begin{array}{cc} 1 & 1 \\[5pt] 1 & 2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] -2 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 - \text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 1 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & -1 \\[5pt] -2 & 3 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - \text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & -1 \\[5pt] -5 & 3 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} 2 & -1 \\[5pt] -5 & 3 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 8. {\left[\begin{array}{cc} 4 & 5 \\[5pt] 3 & 4 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 4 & 5 \\[5pt] 3 & 4 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 4 & 5 \\[5pt] 3 & 4 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 4 & 5 \\[5pt] 3 & 4 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - \text{R}_2} {\left[\begin{array}{cc} 1 & 1 \\[5pt] 3 & 4 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - 3\text{R}_1} {\left[\begin{array}{cc} 1 & 1 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] -3 & 4 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - \text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 4 & -5 \\[5pt] -3 & 4 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} 4 & -5 \\[5pt] -3 & 4 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 4 & 5 \\[5pt] 3 & 4 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 4 & 5 \\[5pt] 3 & 4 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 4 & 5 \\[5pt] 3 & 4 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → 4\text{C}_1 - 3\text{C}_2} {\left[\begin{array}{cc} 1 & 5 \\[5pt] 0 & 4 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 4 & 0 \\[5pt] -3 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 - 5\text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 4 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 4 & -20 \\[5pt] -3 & 16 \end{array}\right]} Applying {\text{C}_1 → \dfrac14\text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 4 & -5 \\[5pt] -3 & 4 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} 4 & -5 \\[5pt] -3 & 4 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 9. {\left[\begin{array}{cc} 3 & 10 \\[5pt] 2 & 7 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 3 & 10 \\[5pt] 2 & 7 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 3 & 10 \\[5pt] 2 & 7 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 3 & 10 \\[5pt] 2 & 7 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - \text{R}_2} {\left[\begin{array}{cc} 1 & 3 \\[5pt] 2 & 7 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - 2\text{R}_1} {\left[\begin{array}{cc} 1 & 3 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] -2 & 3 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - 3\text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 7 & -10 \\[5pt] -2 & 3 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} 7 & -10 \\[5pt] -2 & 3 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 3 & 10 \\[5pt] 2 & 7 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 3 & 10 \\[5pt] 2 & 7 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 3 & 10 \\[5pt] 2 & 7 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → 7\text{C}_1 - 2\text{C}_2} {\left[\begin{array}{cc} 1 & 10 \\[5pt] 0 & 7 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 7 & 0 \\[5pt] -2 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 - 10\text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 7 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 7 & -70 \\[5pt] -2 & 21 \end{array}\right]} Applying {\text{C}_2 → \dfrac17\text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 7 & -10 \\[5pt] -2 & 3 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} 7 & -10 \\[5pt] -2 & 3 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 10. {\left[\begin{array}{cc} 3 & -1 \\[5pt] -4 & 2 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 3 & -1 \\[5pt] -4 & 2 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 3 & -1 \\[5pt] -4 & 2 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 3 & -1 \\[5pt] -4 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → 3\text{R}_2 + 2\text{R}_2} {\left[\begin{array}{cc} 1 & 1 \\[5pt] -4 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 3 & 2 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 + 4\text{R}_2} {\left[\begin{array}{cc} 1 & 1 \\[5pt] 0 & 6 \end{array}\right]} {= \left[\begin{array}{cc} 3 & 2 \\[5pt] 12 & 9 \end{array}\right]}A Applying {\text{R}_2 → \dfrac16\text{R}_2} {\left[\begin{array}{cc} 1 & 1 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 3 & 2 \\[5pt] 2 & \dfrac32 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - \text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & \dfrac12 \\[10pt] 2 & \dfrac32 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} 1 & \dfrac12 \\[10pt] 2 & \dfrac32 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 3 & -1 \\[5pt] -4 & 2 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 3 & -1 \\[5pt] -4 & 2 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 3 & -1 \\[5pt] -4 & 2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 + 2\text{C}_2} {\left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 2 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 + \text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 3 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 1 \\[5pt] 2 & 3 \end{array}\right]} Applying {\text{C}_2 → \dfrac12\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & \dfrac12 \\[10pt] 2 & \dfrac32 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} 1 & \dfrac12 \\[10pt] 2 & \dfrac32 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 11. {\left[\begin{array}{cc} 2 & -6 \\[5pt] 1 & -2 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & -6 \\[5pt] 1 & -2 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 2 & -6 \\[5pt] 1 & -2 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 2 & -6 \\[5pt] 1 & -2 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 - \text{R}_2} {\left[\begin{array}{cc} 1 & -4 \\[5pt] 1 & -2 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - \text{R}_1} {\left[\begin{array}{cc} 1 & -4 \\[5pt] 0 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 1 & -1 \\[5pt] -1 & 2 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 + 2\text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 2 \end{array}\right]} {= \left[\begin{array}{cc} -1 & 3 \\[5pt] -1 & 2 \end{array}\right]}A Applying {\text{R}_2 → \dfrac12\text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} -1 & 3 \\[5pt] -\dfrac12 & 1 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} -1 & 3 \\[5pt] -\dfrac12 & 1 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & -6 \\[5pt] 1 & -2 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 2 & -6 \\[5pt] 1 & -2 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 2 & -6 \\[5pt] 1 & -2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → 2\text{C}_1 + \dfrac12\text{C}_2} {\left[\begin{array}{cc} 1 & -6 \\[5pt] 1 & -2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & 0 \\[5pt] \dfrac12 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 + 6\text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 1 & 4 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & 12 \\[5pt] \dfrac12 & 4 \end{array}\right]} Applying {\text{C}_2 → \dfrac14\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 1 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & 3 \\[5pt] \dfrac12 & 1 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - \text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} -1 & 3 \\[5pt] -\dfrac12 & 1 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} -1 & 3 \\[5pt] -\dfrac12 & 1 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 12. {\left[\begin{array}{cc} 6 & -3 \\[5pt] -2 & 1 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 6 & -3 \\[5pt] -2 & 1 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 6 & -3 \\[5pt] -2 & 1 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 6 & -3 \\[5pt] -2 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → \dfrac12\text{R}_1 + 3\text{R}_2} {\left[\begin{array}{cc} 0 & 0 \\[5pt] -2 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 3 \\[5pt] 0 & 1 \end{array}\right]}A As the row transformation resulted in one of the rows with all 0s, the Inverse does not exist. Info: In general, if any row is multiple of another row, then we can say that the inverse does not exist. Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 6 & -3 \\[5pt] -2 & 1 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 6 & -3 \\[5pt] -2 & 1 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 6 & -3 \\[5pt] -2 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 + 2\text{C}_2} {\left[\begin{array}{cc} 0 & -3 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 2 & 1 \end{array}\right]} As the column transformation resulted in one of the columns with all 0s, the Inverse does not exist. Info: In general, if any column is multiple of another column, then we can say that the inverse does not exist. Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 13. {\left[\begin{array}{cc} 2 & -3 \\[5pt] -1 & 2 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & -3 \\[5pt] -1 & 2 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 2 & -3 \\[5pt] -1 & 2 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 2 & -3 \\[5pt] -1 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 + \text{R}_2} {\left[\begin{array}{cc} 1 & -1 \\[5pt] -1 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 1 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 + \text{R}_1} {\left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 1 \\[5pt] 1 & 2 \end{array}\right]}A Applying {\text{R}_1 → \text{R}_1 + \text{R}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \left[\begin{array}{cc} 2 & 3 \\[5pt] 1 & 2 \end{array}\right]}A Thus {\text{A}^{-1} = \left[\begin{array}{cc} 2 & 3 \\[5pt] 1 & 2 \end{array}\right]} Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & -3 \\[5pt] -1 & 2 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 2 & -3 \\[5pt] -1 & 2 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 2 & -3 \\[5pt] -1 & 2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → 2\text{C}_1 + \text{C}_2} {\left[\begin{array}{cc} 1 & -3 \\[5pt] 0 & 2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & 0 \\[5pt] 1 & 1 \end{array}\right]} Applying {\text{C}_2 → \text{C}_2 + 3\text{C}_1} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & 6 \\[5pt] 1 & 4 \end{array}\right]} Applying {\text{C}_2 → \dfrac12\text{C}_2} {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 2 & 3 \\[5pt] 1 & 2 \end{array}\right]} Thus {\text{A}^{-1} = \left[\begin{array}{cc} 2 & 3 \\[5pt] 1 & 2 \end{array}\right]} Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 14. {\left[\begin{array}{cc} 2 & 1 \\[5pt] 4 & 2 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 1 \\[5pt] 4 & 2 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{cc} 2 & 1 \\[5pt] 4 & 2 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{cc} 2 & 1 \\[5pt] 4 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - 2\text{R}_1} {\left[\begin{array}{cc} 2 & 1 \\[5pt] 0 & 0 \end{array}\right]} {= \left[\begin{array}{cc} 1 & 0 \\[5pt] -2 & 1 \end{array}\right]}A As the row transformation resulted in one of the rows with all 0s, the Inverse does not exist. Info: In general, if any row is multiple of another row, then we can say that the inverse does not exist. Method 2: To find the inverse of the matrix {\left[\begin{array}{cc} 2 & 1 \\[5pt] 4 & 2 \end{array}\right]} using the elementary column transformations: Let A = {\left[\begin{array}{cc} 2 & 1 \\[5pt] 4 & 2 \end{array}\right]} Using the elementary row operations, we may write A = AI {⇒ \left[\begin{array}{cc} 2 & 1 \\[5pt] 4 & 2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]} Applying {\text{C}_1 → \text{C}_1 - 2\text{C}_2} {\left[\begin{array}{cc} 0 & 1 \\[5pt] 0 & 2 \end{array}\right]} {= \text{A} \left[\begin{array}{cc} 1 & 0 \\[5pt] -2 & 1 \end{array}\right]} As the column transformation resulted in one of the columns with all 0s, the Inverse does not exist. Info: In general, if any column is multiple of another column, then we can say that the inverse does not exist. Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. 15. {\left[\begin{array}{ccc} 2 & -3 & 3 \\[5pt] 2 & 2 & 3 \\[5pt] 3 & -2 & 2 \end{array}\right]} Method 1: To find the inverse of the matrix {\left[\begin{array}{ccc} 2 & -3 & 3 \\[5pt] 2 & 2 & 3 \\[5pt] 3 & -2 & 2 \end{array}\right]} using the elementary row transformations: Let A = {\left[\begin{array}{ccc} 2 & -3 & 3 \\[5pt] 2 & 2 & 3 \\[5pt] 3 & -2 & 2 \end{array}\right]} Using the elementary row operations, we may write A = IA {⇒ \left[\begin{array}{ccc} 2 & -3 & 3 \\[5pt] 2 & 2 & 3 \\[5pt] 3 & -2 & 2 \end{array}\right]} {= \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}A Applying {\text{R}_1 → 2\text{R}_1 - \text{R}_3} {\left[\begin{array}{ccc} 1 & -4 & 4 \\[5pt] 2 & 2 & 3 \\[5pt] 3 & -2 & 2 \end{array}\right]} {= \left[\begin{array}{cc} 2 & 0 & -1 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}A Applying {\text{R}_2 → \text{R}_2 - 2\text{R}_1} {\left[\begin{array}{ccc} 1 & -4 & 4 \\[5pt] 0 & 10 & -5 \\[5pt] 3 & -2 & 2 \end{array}\right]} {= \left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] -4 & 1 & 2 \\[5pt] 0 & 0 & 1 \end{array}\right]}A Applying {\text{R}_3 → \text{R}_3 - 3\text{R}_1} {\left[\begin{array}{ccc} 1 & -4 & 4 \\[5pt] 0 & 10 & -5 \\[5pt] 0 & 10 & -10 \end{array}\right]} {= \left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] -4 & 1 & 2 \\[5pt] -6 & 0 & 4 \end{array}\right]}A Applying {\text{R}_2 → \dfrac15\text{R}_2} and {\text{R}_3 → \dfrac{1}{10}\text{R}_3} {\left[\begin{array}{ccc} 1 & -4 & 4 \$ 0 & 2 & -1 \\[5pt] 0 & 1 & -1 \end{array}\right]}
{= \left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] -\dfrac{4}{5} & \dfrac15 & \dfrac25 \\[10pt] -\dfrac35 & 0 & \dfrac25 \end{array}\right]}A
Applying {\text{R}_1 → \text{R}_1 + 2\text{R}_2}
{\left[\begin{array}{ccc} 1 & 0 & 2 \\[5pt] 0 & 2 & -1 \\[5pt] 0 & 1 & -1 \end{array}\right]}
{= \left[\begin{array}{ccc} \dfrac25 & \dfrac25 & -\dfrac15 \\[10pt] -\dfrac{4}{5} & \dfrac15 & \dfrac25 \\[10pt] -\dfrac35 & 0 & \dfrac25 \end{array}\right]}A
Applying {\text{R}_2 → \text{R}_2 - \text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 2 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 1 & -1 \end{array}\right]}
{= \left[\begin{array}{ccc} \dfrac25 & \dfrac25 & -\dfrac15 \\[10pt] -\dfrac15 & \dfrac15 & 0 \\[10pt] -\dfrac35 & 0 & \dfrac25 \end{array}\right]}A
Applying {\text{R}_3 → \text{R}_3 - \text{R}_2}
{\left[\begin{array}{ccc} 1 & 0 & 2 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & -1 \end{array}\right]}
{= \left[\begin{array}{ccc} \dfrac25 & \dfrac25 & -\dfrac15 \\[10pt] -\dfrac15 & \dfrac15 & 0 \\[10pt] -\dfrac25 & -\dfrac15 & \dfrac25 \end{array}\right]}A
Applying {\text{R}_1 → \text{R}_1 + 2\text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & -1 \end{array}\right]}
{= \left[\begin{array}{ccc} -\dfrac25 & 0 & \dfrac35 \\[10pt] -\dfrac15 & \dfrac15 & 0 \\[10pt] -\dfrac25 & -\dfrac15 & \dfrac25 \end{array}\right]}A
Applying {\text{R}_3 → -\text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{ccc} -\dfrac25 & 0 & \dfrac35 \\[10pt] -\dfrac15 & \dfrac15 & 0 \\[10pt] \dfrac25 & \dfrac15 & -\dfrac25 \end{array}\right]}A
Thus {\text{A}^{-1} = \left[\begin{array}{ccc} -\dfrac25 & 0 & \dfrac35 \\[10pt] -\dfrac15 & \dfrac15 & 0 \\[10pt] \dfrac25 & \dfrac15 & -\dfrac25 \end{array}\right]}
Method 2: To find the inverse of the matrix {\left[\begin{array}{ccc} 2 & -3 & 3 \\[5pt] 2 & 2 & 3 \\[5pt] 3 & -2 & 2 \end{array}\right]} using the elementary column transformations:
Let A = {\left[\begin{array}{ccc} 2 & -3 & 3 \\[5pt] 2 & 2 & 3 \\[5pt] 3 & -2 & 2 \end{array}\right]}
Using the elementary row operations, we may write
A = AI
{⇒ \left[\begin{array}{ccc} 2 & -3 & 3 \\[5pt] 2 & 2 & 3 \\[5pt] 3 & -2 & 2 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
Applying {\text{C}_1 → 2\text{C}_1 + \text{C}_2}
{\left[\begin{array}{ccc} 1 & -3 & 3 \\[5pt] 6 & 2 & 3 \\[5pt] 4 & -2 & 2 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 2 & 0 & 0 \\[5pt] 1 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
Applying {\text{C}_2 → \text{C}_2 + \text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 3 \\[5pt] 6 & 5 & 3 \\[5pt] 4 & 0 & 2 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 2 & 0 & 0 \\[5pt] 1 & 1 & 0 \\[5pt] 0 & 1 & 1 \end{array}\right]}
Applying {\text{C}_3 → \text{C}_3 - 3\text{C}_1}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 6 & 5 & -15 \\[5pt] 4 & 0 & -10 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 2 & 0 & -6 \\[5pt] 1 & 1 & -3 \\[5pt] 0 & 1 & 1 \end{array}\right]}
Applying {\text{C}_3 → \dfrac15\text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 6 & 5 & -3 \\[5pt] 4 & 0 & -2 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 2 & 0 & -\dfrac65 \\[10pt] 1 & 1 & -\dfrac35 \\[10pt] 0 & 1 & \dfrac15 \end{array}\right]}
Applying {\text{C}_1 → \text{C}_1 + 2\text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 5 & -3 \\[5pt] 0 & 0 & -2 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} -\dfrac25 & 0 & -\dfrac65 \\[10pt] -\dfrac15 & 1 & -\dfrac35 \\[10pt] \dfrac25 & 1 & \dfrac15 \end{array}\right]}
Applying {\text{C}_2 → \dfrac15\text{C}_2}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & -3 \\[5pt] 0 & 0 & -2 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} -\dfrac25 & 0 & -\dfrac65 \\[10pt] -\dfrac15 & \dfrac15 & -\dfrac35 \\[10pt] \dfrac25 & \dfrac15 & \dfrac15 \end{array}\right]}
Applying {\text{C}_3 → \text{C}_3 + 3\text{C}_2}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & -2 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} -\dfrac25 & 0 & -\dfrac65 \\[10pt] -\dfrac15 & \dfrac15 & 0 \\[10pt] \dfrac25 & \dfrac15 & \dfrac45 \end{array}\right]}
Applying {\text{C}_3 → -\dfrac12\text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} -\dfrac25 & 0 & \dfrac35 \\[10pt] -\dfrac15 & \dfrac15 & 0 \\[10pt] \dfrac25 & \dfrac15 & -\dfrac25 \end{array}\right]}
Thus {\text{A}^{-1} = \left[\begin{array}{ccc} -\dfrac25 & 0 & \dfrac35 \\[10pt] -\dfrac15 & \dfrac15 & 0 \\[10pt] \dfrac25 & \dfrac15 & -\dfrac25 \end{array}\right]}
Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17.
16. {\left[\begin{array}{ccc} 1 & 3 & -2 \\[5pt] -3 & 0 & -5 \\[5pt] 2 & 5 & 0 \end{array}\right]}
Method 1: To find the inverse of the matrix {\left[\begin{array}{ccc} 1 & 3 & -2 \\[5pt] -3 & 0 & -5 \\[5pt] 2 & 5 & 0 \end{array}\right]} using the elementary row transformations:
Let A = {\left[\begin{array}{ccc} 1 & 3 & -2 \\[5pt] -3 & 0 & -5 \\[5pt] 2 & 5 & 0 \end{array}\right]}
Using the elementary row operations, we may write
A = IA
{⇒ \left[\begin{array}{ccc} 1 & 3 & -2 \\[5pt] -3 & 0 & -5 \\[5pt] 2 & 5 & 0 \end{array}\right]}
{= \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_2 → \text{R}_2 + 3\text{R}_1}
{\left[\begin{array}{ccc} 1 & 3 & -2 \\[5pt] 0 & 9 & -11 \\[5pt] 2 & 5 & 0 \end{array}\right]}
{= \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 3 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_3 → \text{R}_3 - 2\text{R}_1}
{\left[\begin{array}{ccc} 1 & 3 & -2 \\[5pt] 0 & 9 & -11 \\[5pt] 0 & -1 & 4 \end{array}\right]}
{= \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 3 & 1 & 0 \\[5pt] -2 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_1 → \text{R}_1 + 3\text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 10 \\[5pt] 0 & 9 & -11 \\[5pt] 0 & -1 & 4 \end{array}\right]}
{= \left[\begin{array}{ccc} -5 & 0 & 3 \\[5pt] 3 & 1 & 0 \\[5pt] -2 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_2 → \text{R}_2 + 8\text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 10 \\[5pt] 0 & 1 & 21 \\[5pt] 0 & -1 & 4 \end{array}\right]}
{= \left[\begin{array}{ccc} -5 & 0 & 3 \\[5pt] -13 & 1 & 8 \\[5pt] -2 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_3 → \text{R}_3 + \text{R}_2}
{\left[\begin{array}{ccc} 1 & 0 & 10 \\[5pt] 0 & 1 & 21 \\[5pt] 0 & 0 & 25 \end{array}\right]}
{= \left[\begin{array}{ccc} -5 & 0 & 3 \\[5pt] -13 & 1 & 8 \\[5pt] -15 & 1 & 9 \end{array}\right]}A
Applying {\text{R}_3 → \dfrac{1}{25}\text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 10 \\[5pt] 0 & 1 & 21 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{ccc} -5 & 0 & 3 \\[5pt] -13 & 1 & 8 \\[5pt] -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \end{array}\right]}A
Applying {\text{R}_1 → \text{R}_1 - 10\text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 21 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{ccc} 1 & -\dfrac{2}{5} & -\dfrac{3}{5} \\[5pt] -13 & 1 & 8 \\[5pt] -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \end{array}\right]}A
Applying {\text{R}_2 → \text{R}_2 - 21\text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{ccc} 1 & -\dfrac{2}{5} & -\dfrac{3}{5} \\[10pt] -\dfrac25 & \dfrac{4}{25} & \dfrac{11}{25} \\[10pt] -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \end{array}\right]}A
Thus {\text{A}^{-1} = \left[\begin{array}{ccc} 1 & -\dfrac{2}{5} & -\dfrac{3}{5} \\[10pt] -\dfrac25 & \dfrac{4}{25} & \dfrac{11}{25} \\[10pt] -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \end{array}\right]}
Method 2: To find the inverse of the matrix {\left[\begin{array}{ccc} 1 & 3 & -2 \\[5pt] -3 & 0 & -5 \\[5pt] 2 & 5 & 0 \end{array}\right]} using the elementary column transformations:
Let A = {\left[\begin{array}{ccc} 1 & 3 & -2 \\[5pt] -3 & 0 & -5 \\[5pt] 2 & 5 & 0 \end{array}\right]}
Using the elementary row operations, we may write
A = AI
{⇒ \left[\begin{array}{ccc} 1 & 3 & -2 \\[5pt] -3 & 0 & -5 \\[5pt] 2 & 5 & 0 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
Applying {\text{C}_2 → \text{C}_2 - 3\text{C}_1}
{\left[\begin{array}{ccc} 1 & 0 & -2 \\[5pt] -3 & 9 & -5 \\[5pt] 2 & -1 & 0 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 1 & -3 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
Applying {\text{C}_3 → \text{C}_3 + 2\text{C}_1}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] -3 & 9 & -11 \\[5pt] 2 & -1 & 4 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 1 & -3 & 2 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
Applying {\text{C}_1 → \text{C}_1 + 4\text{C}_2 + 3\text{C}_1}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 9 & -11 \\[5pt] 10 & -1 & 4 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} -5 & -3 & 2 \\[5pt] 4 & 1 & 0 \\[5pt] 3 & 0 & 1 \end{array}\right]}
Applying {\text{C}_2 → 5\text{C}_2 + 4\text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & -11 \\[5pt] 10 & 11 & 4 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} -5 & -7 & 2 \\[5pt] 4 & 5 & 0 \\[5pt] 3 & 4 & 1 \end{array}\right]}
Applying {\text{C}_3 → \text{C}_3 + 11\text{C}_2}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 10 & 11 & 125 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} -5 & -7 & -75 \\[5pt] 4 & 5 & 55 \\[5pt] 3 & 4 & 45 \end{array}\right]}
Applying {\text{C}_3 → \dfrac{1}{125}\text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 10 & 11 & 1 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} -5 & -7 & -\dfrac{3}{5} \\[10pt] 4 & 5 & \dfrac{11}{25} \\[10pt] 3 & 4 & \dfrac{9}{25} \end{array}\right]}
Applying {\text{C}_1 → \text{C}_1 - 10\text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 11 & 1 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 1 & -7 & -\dfrac{3}{5} \\[10pt] -\dfrac25 & 5 & \dfrac{11}{25} \\[10pt] -\dfrac35 & 4 & \dfrac{9}{25} \end{array}\right]}
Applying {\text{C}_2 → \text{C}_2 - 11\text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 1 & -\dfrac{2}{5} & -\dfrac{3}{5} \\[10pt] -\dfrac25 & \dfrac{4}{25} & \dfrac{11}{25} \\[10pt] -\dfrac35 & \dfrac{1}{25} & \dfrac{9}{25} \end{array}\right]}
Thus {\text{A}^{-1} = \left[\begin{array}{ccc} 1 & -\dfrac{2}{5} & -\dfrac{3}{5} \\[10pt] -\dfrac25 & \dfrac{4}{25} & \dfrac{11}{25} \\[10pt] -\dfrac35 & \dfrac{1}{25} & \dfrac{9}{25} \end{array}\right]}
Using the elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17.
17. {\left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] 5 & 1 & 0 \\[5pt] 0 & 1 & 3 \end{array}\right]}
Method 1: To find the inverse of the matrix {\left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] 5 & 1 & 0 \\[5pt] 0 & 1 & 3 \end{array}\right]} using the elementary row transformations:
Let A = {\left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] 5 & 1 & 0 \\[5pt] 0 & 1 & 3 \end{array}\right]}
Using the elementary row operations, we may write
A = IA
{⇒ \left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] 5 & 1 & 0 \\[5pt] 0 & 1 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_1 → 3\text{R}_1 - \text{R}_2}
{\left[\begin{array}{ccc} 1 & -1 & -3 \\[5pt] 5 & 1 & 0 \\[5pt] 0 & 1 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 & -1 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_2 → \text{R}_2 - 5\text{R}_1}
{\left[\begin{array}{ccc} 1 & -1 & -3 \\[5pt] 0 & 6 & 15 \\[5pt] 0 & 1 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 & -1 & 0 \\[5pt] -15 & 6 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_1 → \text{R}_1 + \text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 6 & 15 \\[5pt] 0 & 1 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 & -1 & 1 \\[5pt] -15 & 6 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_2 → \text{R}_2 - 5\text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 1 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 & -1 & 1 \\[5pt] -15 & 6 & -5 \\[5pt] 0 & 0 & 1 \end{array}\right]}A
Applying {\text{R}_3 → \text{R}_3 - \text{R}_2}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 & -1 & 1 \\[5pt] -15 & 6 & -5 \\[5pt] 15 & -6 & 6 \end{array}\right]}A
Applying {\text{R}_3 → \dfrac13\text{R}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 & -1 & 1 \\[5pt] -15 & 6 & -5 \\[5pt] 5 & -2 & 2 \end{array}\right]}A
Thus {\text{A}^{-1} = \left[\begin{array}{cc} 3 & -1 & 1 \\[5pt] -15 & 6 & -5 \\[5pt] 5 & -2 & 2 \end{array}\right]}
Method 2: To find the inverse of the matrix {\left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] 5 & 1 & 0 \\[5pt] 0 & 1 & 3 \end{array}\right]} using the elementary column transformations:
Let A = {\left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] 5 & 1 & 0 \\[5pt] 0 & 1 & 3 \end{array}\right]}
Using the elementary row operations, we may write
A = AI
{⇒ \left[\begin{array}{ccc} 2 & 0 & -1 \\[5pt] 5 & 1 & 0 \\[5pt] 0 & 1 & 3 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
Applying {\text{C}_1 → \text{C}_1 + \text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & -1 \\[5pt] 5 & 1 & 0 \\[5pt] 3 & 1 & 3 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 0 & 1 \end{array}\right]}
Applying {\text{C}_3 → \text{C}_3 + \text{C}_1}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 5 & 1 & 5 \\[5pt] 3 & 1 & 6 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 1 & 0 & 1 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 0 & 2 \end{array}\right]}
Applying {\text{C}_1 → \text{C}_1 - \text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 5 \\[5pt] -3 & 1 & 6 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 0 & 1 & 0 \\[5pt] -1 & 0 & 2 \end{array}\right]}
Applying {\text{C}_3 → \text{C}_3 - 5\text{C}_2}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] -3 & 1 & 1 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 0 & 0 & 1 \\[5pt] 0 & 1 & -5 \\[5pt] -1 & 0 & 2 \end{array}\right]}
Applying {\text{C}_1 → \text{C}_1 + 3\text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 1 & 1 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 3 & 0 & 1 \\[5pt] -15 & 1 & -5 \\[5pt] 5 & 0 & 2 \end{array}\right]}
Applying {\text{C}_2 → \text{C}_2 - \text{C}_3}
{\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \text{A} \left[\begin{array}{ccc} 3 & -1 & 1 \\[5pt] -15 & 6 & -5 \\[5pt] 5 & -2 & 2 \end{array}\right]}
Thus {\text{A}^{-1} = \left[\begin{array}{cc} 3 & -1 & 1 \\[5pt] -15 & 6 & -5 \\[5pt] 5 & -2 & 2 \end{array}\right]}
18. Matrices A and B will be inverse of each other only if
(A) AB = BA
(B) AB = BA = O
(C) AB = O, BA = I
(D) AB = BA = I ✔
The following is the definition of Invertible Matrices.
If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A and it is denoted by {\text{A}^{-1}}
So, option D is the correct answer