Use the fact that if A = [ a b c d ] A=\begin{bmatrix}a & b\\ c & d\end{bmatrix} , then A − 1 = 1 a d − b c [ d − b − c a ] 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 A A − 1 = I 2 AA^{-1} = I_2 and A − 1 A = I 2 A^{-1}A = I_2 . A = [ 10 − 2 − 5 1 ] A = \begin{bmatrix}10 & -2 \\-5 & 1\end{bmatrix} A = [ 10 − 5 − 2 1 ]

by the multiplication inverse of the matrix. Shown by using the formula for a inverse. Also verified that anytime's a universe equals I. And a inverse times A equals I where I too. As a multiple index of major supporter too. So let's look at all right. It's We have 4 -8 -6. 12. Now in the matrix we're gonna denote this as A B C. D. Like. So so A equals four V. Equals negative eight C equals negative six T equals 12. Good use of formula for our inverse. Now our first part we want is one over a D minus B. C. But he said it was four. So four Times D. Which is 12 minus R. B. -8 Time C. -6. This is multiplying our matrix. R. D. Is first now, so 12 negative B. That turns down to positive eight negative. See it turns out to be positive six. And A. Which turns this into four. I can simplify 1/ -98 times 96 is 48 times 12. Now we get 1/0 times 12. 864. We notice here we have a 1/0. We're not gonna be able to multiply that one of zero to the matrix because it's undefined. So, we're actually gonna get undefined answer for everything here, which means our inverse is undefined. Making this answer answer. D. The inverse does not exist. Okay, I'll help you solve the problem thank you for watching. Goodbye

7. Systems of Equations & Matrices

Determinants and Cramer's Rule

College Algebra

7. Systems of Equations & Matrices

Determinants and Cramer's Rule

2:20 minutes

Problem 17

Textbook Question

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{bmatrix}$10 & -2 \\-5 & 1$\end{bmatrix}$$

Verified step by step guidance

Identify the elements of the matrix $A$ as $a = 10$, $b = -2$, $c = -5$, and $d = 1$.

Calculate the determinant of $A$ using the formula $ad - bc$, which means compute $10 \times 1 - (-2) \times (-5)$.

Check if the determinant is non-zero. If it is zero, the inverse does not exist; if non-zero, proceed to find the inverse.

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

Verify the result by multiplying $A$ and $A^{-1}$ in both orders ($AA^{-1}$ and $A^{-1}A$) to check if the product is the identity matrix $I_2$.

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Key Concepts

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

Matrix Inverse Formula for 2x2 Matrices

The inverse of a 2x2 matrix A = [[a, b], [c, d]] exists if and only if the determinant (ad - bc) is nonzero. The inverse is given by A^(-1) = (1/(ad - bc)) * [[d, -b], [-c, a]]. This formula allows us to find the inverse by swapping and negating elements and scaling by the reciprocal of the determinant.

Determinant and Its Role in Invertibility

The determinant of a 2x2 matrix, calculated as ad - bc, determines whether the matrix is invertible. If the determinant is zero, the matrix has no inverse. A nonzero determinant ensures the matrix is invertible and the inverse can be computed using the formula.

Verification of Matrix Inverse Using Identity Matrix

To confirm that a matrix B is the inverse of A, multiply A by B and B by A. Both products should yield the 2x2 identity matrix I_2 = [[1, 0], [0, 1]]. This verification step ensures the correctness of the computed inverse matrix.

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