SOLVED: If the matrix A below is invertible, find A-1. If it’s not invertible; explain why not: -1 A = 9 2 2

Invertible Matrix
Invertible Matrix

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If the matrix A below is invertible, find A-1. If it’s not invertible; explain why not:
-1 A = 9 2 2

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01:05

52 3A = 5 6 21 0 2 3 8

Without multiplying the matrices, explain why A is not invertible.

Observe how rows and columns of A.

Hint

01:03

use determinants to decide whether the given matrix is invertible.$$A=\left[\begin{array}{rrr}1 & 0 & -1 \\9 & -1 & 4 \\8 & 9 & -1\end{array}\right]$$

01:47

$$\begin{array}{l}\text { Why does the square matrix } A=\left[\begin{array}{rrr}1 & 2 & -1 \\2 & 4 & -2 \\0 & 1 & 3\end{array}\right] \text { not have } \\\text { an inverse? }\end{array}$$

15:04

decide whether the matrix is invertible, and if so, use the adjoint method to find its inverse.$$A=\left[\begin{array}{llll}1 & 3 & 1 & 1 \\2 & 5 & 2 & 2 \\1 & 3 & 8 & 9 \\1 & 3 & 2 & 2\end{array}\right]$$

Transcript

In this question we have to check whether the matrix A is invertible or not. If it is invertible we have to find its inverse. So we have the matrix A which is equals to 1 minus 1 2 minus 1 1 0 and then we have 2 2 1. We have to find its determinant. So determinant of it comes out to be 1 times 1 minus 0 plus 1 times minus 1 minus 0 and 2 times minus 2 minus 2. So this comes out to be 1 minus 1 and this will be minus 8. So obviously this is equals to minus 8 not 0. So I would say that since determinant is non -zero this implies that A inverse exists. Now we have to find out. So first of all I got that this matrix A is invertible. Next we have to check we have to find out what is that inverse of A. So I will now consider this augmented matrix. Find inverse. Consider augmented matrix that is A augmented with the identity matrix A is given to me 1 minus 1 2 minus 1 1 0 2 2 1 and this is augmented with 1 0 0 0 1 0 0 0 1. Correct. We’ll try to make this matrix to be either of a triangular or lower triangular and at that time what will be this matrix that will be the inverse. So my first operation is in R2 I will apply R1 plus R2 and in R3 I will apply R3 minus 2 times of R1. So I’m going to get the following 1 minus 1 2 0 0 2 and the same operation will be performed on this invertible in this identity matrix as well. So this becomes 1 1 0 then I have 0 4 minus 3 and this is…

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