By Prabhat Choudhary

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**Extra resources for A Practical Approach to Linear Algebra**

**Example text**

1 = 8 + 12 - 6 -1 = 13. Consider next q2I. 2 + 5. 3 = 3 + 2 + 0 + 6 = 11 55 Matries Consider next q22. To calculate this, we need the 2-nd row of A and the 2-nd column of B, so let us cover up all unnecessary information, so that x 4 x x x x 5 3 3 x 2 x -2 = x x x x x x X q22 X X j . 3 + 5. 1 = 12 + 3 -10 + 2 = 7. Consider next q31. 3 =-1 +0+0+ 18= 17. Consider finally q32. 1 = - 4 + 0 + (- 14) + 6 = - 12. We therefore conclude that AB=[ ~ 41 53 2-Ij o -1 7 6 1 4 2 3 0 -2 3 +~ 17 Example. Consider again the matrices A~P -1 4 3 1 5 0 7 -Ij ~ and B= 1 4 2 3 0 -2 3 137 1.

Bn} in a vector space V. Consider the identity transformation I = I v and its matrix [1]BA in these bases. By the definition [v]B = [1]BA [v] A , \Iv E V. , for any vector v E Vthe matrix [1]BA transforms its coordinates in the basis a into coordinates in the basis B. The matrix [1]BA is often called the change ofcoordinates (from the basis A to the basis B) matrix. The matrix [1]BA is easy to compute: according to the general rule of finding the matrix of a linear transformation, its kth column is the coordinate representation [aklB of kth element of the basis A Note that [1]AB = ([1] BAtl, (follows immediately from the mUltiplication of matrices rule), so any change of coordinate matrix is always invertible.

Here we just replaced the last vector by its sum with the second one. So, this formula is different from the solution we got from the row reduction, but it is nevertheless correct. The simplest way to check that give us correct solutions, is to check that the first vector (3, 1, 0, 2, ol satisfies the equation Ax = b, and that the other two (the ones with 40 Systems of Linear Equations the parameters x3 and Xs or sand t in front of them) should satisfy the associated homogeneous equation Ax = O.