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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 6 - Matrices and DeterminantsProblem 40
Chapter 7, Problem 40

In Exercises 39–42, find A^(-1) Check that AA^-1 = I and A^(-1)A = I

Verified step by step guidance
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Step 1: Recall the definition of the inverse of a matrix. A matrix A has an inverse A^(-1) if and only if A is a square matrix (same number of rows and columns) and its determinant is non-zero. The inverse satisfies the property that AA^(-1) = I and A^(-1)A = I, where I is the identity matrix.
Step 2: Compute the determinant of the matrix A. For a 2x2 matrix A = [[a, b], [c, d]], the determinant is given by det(A) = ad - bc. If the determinant is zero, the matrix does not have an inverse.
Step 3: If the determinant is non-zero, calculate the inverse of the matrix. For a 2x2 matrix A = [[a, b], [c, d]], the inverse is given by A^(-1) = (1/det(A)) * [[d, -b], [-c, a]]. Replace the elements of the matrix and the determinant into this formula to find A^(-1).
Step 4: Verify that AA^(-1) = I by performing matrix multiplication. Multiply the original matrix A by its inverse A^(-1) and confirm that the result is the identity matrix I.
Step 5: Similarly, verify that A^(-1)A = I by performing matrix multiplication in the reverse order. Multiply the inverse A^(-1) by the original matrix A and confirm that the result is the identity matrix I.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse of a Matrix

The inverse of a matrix A, denoted A^(-1), is a matrix that, when multiplied by A, yields the identity matrix I. This means that A * A^(-1) = I and A^(-1) * A = I. Not all matrices have inverses; a matrix must be square and have a non-zero determinant to possess an inverse.
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Identity Matrix

The identity matrix, denoted I, is a special square matrix that has ones on the diagonal and zeros elsewhere. It acts as the multiplicative identity in matrix multiplication, meaning that for any matrix A, multiplying by I leaves A unchanged (A * I = A). The identity matrix is crucial for verifying the correctness of matrix inverses.
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Matrix Multiplication

Matrix multiplication involves combining two matrices to produce a third matrix. The number of columns in the first matrix must equal the number of rows in the second. The resulting matrix's elements are calculated by taking the dot product of the rows of the first matrix with the columns of the second, which is essential for checking the properties of inverses.
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Related Practice
Textbook Question

a. Write each linear system as a matrix equation in the form AX = B. b. Solve the system using the inverse that is given for the coefficient matrix.

{x−y+z=82y−z=−72x+3y=1The inverse of [1−1102−1230] is [33−1−2−21−4−52].\(\begin{cases}\)x - y + z = 8 \\2y - z = -7 \\2x + 3y = 1\(\end{cases}\]\text{The inverse of }\[\begin{bmatrix}\)1 & -1 & 1 \\0 & 2 & -1 \\2 & 3 & 0\(\end{bmatrix}\]\text{ is }\[\begin{bmatrix}\)3 & 3 & -1 \\-2 & -2 & 1 \\-4 & -5 & 2\(\end{bmatrix}\]\text{.}\)

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Textbook Question

In Exercises 37–44, use Cramer's Rule to solve each system. {4x−5y−6z=−1x−2y−5z=−122x−y=7\(\begin{cases}\)4x - 5y - 6z = -1 \(\x\) - 2y - 5z = -12 \\2x - y = 7\(\end{cases}\)

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Textbook Question

Find the cubic function f(x) = ax³ + bx² + cx + d for which ƒ( − 1) = 0, ƒ(1) = 2, ƒ(2) = 3, and ƒ(3) = 12.

Textbook Question

Find the quadratic function f(x) = ax² + bx + c for which ƒ( − 2) = −4, ƒ(1) = 2, and f(2) = 0.

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Textbook Question

In Exercises 37–44, use Cramer's Rule to solve each system. {x+y+z=4x−2y+z=7x+3y+2z=4\(\begin{cases}\)x + y + z = 4 \(\x\) - 2y + z = 7 \(\x\) + 3y + 2z = 4\(\end{cases}\)

Textbook Question

a. Write each linear system as a matrix equation in the form AX = B. b. Solve the system using the inverse that is given for the coefficient matrix.

{w−x+2y=−3x−y+z=4−w+x−y+2z=2−x+y−2z=−4The inverse of [1−12001−11−11−120−11−2] is [00−1−1141312120−10−1]\(\begin{cases}\)w - x + 2y \(\quad\]\quad\) = -3 \(\quad\[\quad\) x - y + z = 4 \\-w + x - y + 2z = 2 \(\quad\]\quad\) -x + y - 2z = -4\(\end{cases}\[\text{The inverse of }\]\begin{bmatrix}\)1 & -1 & 2 & 0 \\0 & 1 & -1 & 1 \\-1 & 1 & -1 & 2 \\0 & -1 & 1 & -2\(\end{bmatrix}\[\text{ is }\]\begin{bmatrix}\)0 & 0 & -1 & -1 \\1 & 4 & 1 & 3 \\1 & 2 & 1 & 2 \\0 & -1 & 0 & -1\(\end{bmatrix}\)