Matrices and Systems of Linear Equations
Order 5908001
Note: In this chapter, uppercase letters (A, B, C, …) denote matrices, and
lowercase letters (a, b, c, x,y, z, …) denote scalars (numbers). Unless stated
otherwise, all scalars are real numbers.
Basic Concepts of Matrices
313. What is the size of matrix ?
314. In the matrix , identify the specified
element.
(A) a13
(B) a31
(C) a35
(D) a14
(E) a33
315. Exhibit the indicated matrix.
(A) The 3 × 1 column vector whose elements are 2, 3, and 8, in this order
(B) The 1 × 4 row vector whose elements are 0, –2, 1, 5, in this order
(C) The 3 × 3 matrix A = [aij]3×3 whose main diagonal elements are 3, –4, 5
and whose off-diagonal elements are 1’s
(D) I3x3
(E) The 2 × 3 zero matrix
For questions 316 to 319, compute as indicated.
316.
317.
318.
319.
320.
321.
322. .
(A) Find AB.
(B) Find BA.
(C) Does AB = BA?
Matrices and Systems of Linear Equations
Determinants and Inverses
For questions 323 to 327, find the determinant of the indicated matrix.
323.
324.
325.
326.
327.
For questions 328 to 331, let and .
328. Find (a) A–1 and (b) B–1
. Refer to the following guidelines, as needed.
The inverse of a square matrix A is a square matrix of the same size, A–1
, such
that AA–1 = A–1A = I. A simple method for finding the inverse of a 2 × 2 matrix is
given by the following procedure:
In general, if , then , provided ad
– bc ≠ 0. In words, switch the elements on A’s diagonal, negate the other two
elements (but don’t switch them), and then multiply the resulting matrix by .
Caution: This method works for 2 × 2 matrices only. (The inverses for larger
matrices are efficiently found using technological devices such as graphing
calculators). Note that this procedure for finding the inverse of a 2 × 2 matrix A
will not work if det A = 0. In fact, if the determinant of any square matrix is 0,
the matrix will not have an inverse.
329. Find (AB)
–1
.
330. Compute A–1B–1
.
331. Compute B–1 A–1
.
332. Indicate whether the statement is true or false.
(A) (AB)
–1 equals B–1 A–1
.
(B) (AB)
–1 equals A–1B–1
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