Use the fact that if A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, then A−1=1ad−bc[d−b−ca]$$A^{-1}$$ = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} to find the inverse of each matrix, if possible. Check that AA−1=$$I2AA^{-1}$$ = $$I_2$$ and A−1A=$$I2A^{-1} A$$ = $$I_2.A$$=[23−12]A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}

Ch. 6 - Matrices and Determinants

Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook

All textbooks

Blitzer 8th Edition

Ch. 6 - Matrices and Determinants

Problem 13

Chapter 7, Problem 13

Use the fact that if $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ , then $A^{- 1} = \frac{1}{a d - b c} [\begin{matrix} d & - b \\ - c & a \end{matrix}]$ to find the inverse of each matrix, if possible. Check that $A A^{- 1} = I_{2}$ and $A^{- 1} A = I_{2}$ .
$A = [\begin{matrix} 2 & 3 \\ - 1 & 2 \end{matrix}]$

Verified step by step guidance

Identify the elements of matrix $A$ as $a = 2$, $b = 3$, $c = -1$, and $d = 2$ from the given matrix $A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}$.

Calculate the determinant of matrix $A$ using the formula $ad - bc$. Substitute the values to get $\det(A) = (2)(2) - (3)(-1)$.

Check if the determinant is non-zero. If $\det(A) \neq 0$, the inverse matrix $A^{-1}$ exists. If it equals zero, the inverse does not exist.

Use the formula for the inverse of a 2x2 matrix: $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. Substitute the values of $a$, $b$, $c$, and $d$ into this formula.

Verify your result by multiplying $A$ and $A^{-1}$ in both orders: $AA^{-1}$ and $A^{-1}A$. Both products should equal the identity matrix $I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

10m

Was this helpful?

Key Concepts

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

Matrix Inverse

The inverse of a square matrix A, denoted A⁻¹, is a matrix that when multiplied by A yields the identity matrix I. For a 2x2 matrix, the inverse exists only if the determinant is non-zero, and it is calculated using a specific formula involving the elements of A.

Determinant of a 2x2 Matrix

The determinant of a 2x2 matrix A = [[a, b], [c, d]] is given by ad - bc. This scalar value determines whether the matrix is invertible; if the determinant is zero, the matrix has no inverse.

Identity Matrix and Verification

The identity matrix I₂ is a 2x2 matrix with ones on the diagonal and zeros elsewhere. To verify a matrix inverse, multiply A by A⁻¹ and A⁻¹ by A; both products should equal I₂, confirming the correctness of the inverse.

Recommended video:

Guided course

4:35

Introduction to Matrices

Related Practice

Textbook Question

In Exercises 9 - 16, find the following matrices: b. A - B

Textbook Question

In Exercises 9 - 16, find the following matrices: c. - 4A

views

Textbook Question

For Exercises 11–22, use Cramer's Rule to solve each system. $\begin{cases}\)12x + 3y = 15 \\2x - 3y = 13\(\end{cases}$

Textbook Question

Find the following matrices: - 3A + 2B

$A = $\begin{bmatrix}$3 & 1 & 1 \\-1 & 2 & 5$\end{bmatrix}$, B = $\begin{bmatrix}$2 & -3 & 6 \\-3 & 1 & -4$\end{bmatrix}$$

views

Textbook Question

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. $\begin{cases}\)w - 3x + y - 4z = 4 \\-2w + x + 2y = -2 \\3w - 2x + y - 6z = 2 \\-w + 3x + 2y - z = -6\(\end{cases}$

Textbook Question

Perform each matrix row operation and write the new matrix.

$$\begin{bmatrix}$2 & -6 & 4 & $\vert$ & 10 \\1 & 5 & -5 & $\vert$ & 0 \\3 & 0 & 4 & $\vert$ & 7$\end{bmatrix}\]\quad$ $\frac{1}{2}$R_1$

views