The thing is, I can't seem to figure out what to do to achieve that identity matrix. [■8(1&0&17/11@0&1&19/11@0×(−11)/90&0×(−11)/90&(−90)/11×(−11)/90)] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@−4×(−11)/90&(−8)/11×(−11)/90&19/11×(−11)/90)] Matrix rank is calculated by reducing matrix to a row echelon form using elementary row operations. For a 4×4 Matrix we have to calculate 16 3×3 determinants. Make sure to perform the same operations on RHS so that you get I=BA. [■8(1&0&17/11@0&1&19/11@0&−8+8&−22+152/11)] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@−4+40/11&(−8)/11&9−80/11)] You can also find the inverse using an advanced graphing calculator. Validate the sum by performing the necessary row operations on LHS to get I in LHS. And as we'll see in the next video, calculating by the inverse of a 3x3 matrix … [■8(1&0&17/11@0&1&19/11@0&0&1)] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@2/45&4/45&(−19)/90)] [■8(1&0&0@0−19/11(0)&1−19/11(0)&19/11−19/11(1)@0&0&1)] = A [■8(1/45&2/11&13/90@5/11−19/11 (2/45)&(−1)/11−19/11 (4/45)&(−10)/11−19/11 ((−19)/11)@2/45&4/45&(−19)/90)] [■8(1&2&5@5−5&−1−10&6−25@4&0&−2)] = A [■8(1&0&−2@−5&1&10@0&0&1)] We will find inverse of a 2 × 2 & a 3 × 3 matrix Note:- While doing elementary operations, we use Only rows OR Only columns Not both Let's take some examples Next: Ex 3.4, 18→ Chapter 3 Class 12 Matrices; Concept wise; Inverse of matrix using elementary transformation. applying an elementary row operation has the same eﬀect as multiplying by the elementary matrix of the operation. The calculator will find the row echelon form (simple or reduced - RREF) of the given (augmented) matrix (with variables if needed), with steps shown. [■8(1&0&17/11@0&1&19/11@0+8(0)&−8+8(1)&−22+8(19/11) )] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@−4+8(5/11)&0+8((−1)/11)&9+8((−10)/11) )] You can copy and paste the entire matrix right here. Let’s learn how to find inverse of a matrix using it. But what if the reduced row echelon form of A is I? No headers. TI‐83 Plus/84 Plus: I will be using the TI‐83 Plus graphing calculator for these directions. Teachoo is free. Number of rows: m = . [■8(1&0&0@0&1&19/11@0&0&1)] = A [■8(1/11×11/45&2/11×11/45&1/11×143/90@5/11&(−1)/11&(−10)/11@2/45&4/45&(−19)/90)] [■8(1−17/11(0)&0−17/11(0)&17/11−17/11(1)@0&1&19/11@0&0&1)] = A [■8(1/11 (1−34/45) &2/11 (1−34/35)&1/11 (−2+323/90)@5/11&(−1)/11&(−10)/11@2/45&4/45&(−19)/90)] [■8(3−2 (1) &2−2(4)@1&4)] = A [■8(1−2 (0) &0−2(1)@0&1)] To calculate inverse matrix you need to do the following steps. Teachoo provides the best content available! _2 →_2− 5_1 From Thinkwell's College Algebra Chapter 8 Matrices and Determinants, Subchapter 8.4 Inverses of Matrices Why does this specific procedure of elementary row operations fail to calculate the determinant? Let's get a deeper understanding of what they actually are and how are they useful. [■8(1+6(0)&−6+6(1)@0&1)] = A [■8(1+6((−1)/10) &−2+6(3/10)@(−1)/10&3/10)] The only concept a student fears in this chapter, Matrices. [■8(9−2(4)&2−2(0)&1−2(−2)@5&−1&6@4&0&−2)] = A [■8(1−2(0)&0−2(0)&0−2(1)@0&1&0@0&0&1)] Since elementary row operations correspond to elementary matrices, the reverse of an operation (which is also an elementary row operation) should correspond to an elementary matrix… This is an inverse operation. We now turn our attention to a special type of matrix called an elementary matrix.An elementary matrix is always a square matrix. [■8(1−2(0)&2−2(1)&5−2(19/11)@0&1&19/11@0&−8&−22)] = A [■8(1−2(5/11)&0−2((−1)/11)&−2−2((−10)/11)@5/11&(−1)/11&(−10)/11@−4&0&9)] _1 →_1+ 6_2 [■8(1&−6@1&4)] = A [■8(1&−2@0&1)] _1 →_1− 2_2 [■8(1&−6@0&10)] = A [■8(1&−2@−1&3)] Learn All Concepts of Chapter 3 Class 12 Matrices - FREE. _2 →_2/(−11) [■8(1&2&5@0&−11&−19@4−4&−8&−2−20)] = A [■8(1&0&−2@−5&1&10@−4&0&1+8)] [■8(1&−6+6@0&1)] = A [■8(1−6/10&−2+18/10@(−1)/10&3/10)] one single elementary row operation on an identity matrix. [■8(1&2&5@0&−11&−19@4&0&−2)] = A [■8(1&0&−2@−5&1&10@0&0&1)] Making 2 as 0 [■8(1&0&5−38/11@0&1&19/11@0&−8&−22)] = A [■8(1−10/11&2/11&−2+20/11@5/11&(−1)/11&(−10)/11@−4&0&9)] Number of rows (equal to number of columns): n = . This becomes A−1 But not Each row must begin with a new line. Elementary Operations! Making −8 as 0 [■8(1&2&5@0&−11&−19@4−4(1)&0−4(2)&−2−4(5))] = A [■8(1&0&−2@−5&1&10@0−4(1)&0−4(0)&1−4(−2))]