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If \(A=\left[\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right]\), then \(A^{-1}+A=\) ____. (1) I (2) \(2 \mathrm{I}\) (3) \(3 \mathrm{I}\) (4) \(4 \mathrm{I}\)

Short Answer

Expert verified
Based on the step by step solution provided, if \(A = \left[\begin{array}{ll}1 & 2 \\\ 1 & 3\end{array}\right]\), then \(A^{-1} + A = 4\mathrm{I}\). Thus, the correct answer is Option 4.

Step by step solution

01

Find the inverse of matrix A

To find the inverse of a 2x2 matrix, we can use the following formula: \(A^{-1} = \frac{1}{\mathrm{det}(A)} \left[\begin{array}{cc} d & -b \\\ -c & a \end{array}\right]\) Where \(A = \left[\begin{array}{cc} a & b \\\ c & d \end{array}\right]\) and \(\mathrm{det}(A) = ad - bc\). Here, \(A = \left[\begin{array}{ll}1 & 2 \\\ 1 & 3\end{array}\right]\), so \(a = 1\), \(b = 2\), \(c = 1\), and \(d = 3\). Compute the determinant and inverse of matrix A: \(\mathrm{det}(A) = (1)(3) - (2)(1) = 1\) \(A^{-1} = \frac{1}{\mathrm{1}} \left[\begin{array}{cc} 3 & -2 \\\ -1 & 1 \end{array}\right] = \left[\begin{array}{cc} 3 & -2 \\\ -1 & 1 \end{array}\right]\)
02

Add A and A^{-1}

Now, we will add the inverse matrix and the original matrix A: \(A^{-1}+A = \left[\begin{array}{cc} 3 & -2 \\\ -1 & 1 \end{array}\right] + \left[\begin{array}{ll}1 & 2 \\\ 1 & 3\end{array}\right]\) Perform the addition: \(A^{-1}+A = \left[\begin{array}{cc} 3+1 & -2+2 \\\ -1+1 & 1+3 \end{array}\right] = \left[\begin{array}{cc} 4 & 0 \\\ 0 & 4 \end{array}\right]\) This can be written as \(4\mathrm{I}\) where \(\mathrm{I}\) is the identity matrix: \(A^{-1}+A = 4\left[\begin{array}{cc} 1 & 0 \\\ 0 & 1 \end{array}\right]\)
03

Compare with the given options

Now, we will compare the result with the given options: Option 1: I Option 2: 2I Option 3: 3I Option 4: 4I The result \(A^{-1}+A = 4\mathrm{I}\) matches with Option 4. Therefore, the correct answer is: \((4) \ 4 \mathrm{I}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse of a Matrix
The inverse of a matrix is like finding the reciprocal of a number, except in the world of matrices, it's a bit more complex. For a matrix to have an inverse, which we denote as \(A^{-1}\), the matrix must be square (same number of rows and columns) and it must be non-singular, meaning it has a non-zero determinant.

An important property of an inverse matrix is that when it is multiplied with the original matrix, it returns the identity matrix \(I\). The process to find the inverse of a 2x2 matrix, such as in the given exercise, involves swapping the positions of the elements in the main diagonal, changing the signs of the off-diagonal elements, and then dividing by the determinant. Remember, the inverse of a matrix is not simply dividing 1 by the matrix but following the specified method for finding the inverse. When these steps are properly executed, the resulting matrix will effectively 'undo' the effects of the original matrix when multiplied together.
Matrix Addition
Adding matrices is like a walk in the park — if you stick to the path of matching elements! Just like combining apples with apples, you can only add matrices of the same dimensions. Matrix addition is done element-wise, which means you add the corresponding elements of the two matrices.

In our case, to add \(A^{-1}\) with \(A\), you simply look at each position in the matrices and add those numbers together. It's crucial to ensure that both matrices have the same dimensions; otherwise, they can't be friends. Matrix addition is commutative, meaning that \(A+B\) will always equal \(B+A\), and it's also associative, allowing you to group added matrices in any order without changing the result.
Identity Matrix
The identity matrix, denoted as \(I\), is the Mr. Neutral of matrices. It's a square matrix with ones on the main diagonal and zeroes everywhere else. No matter what compatible matrix you multiply it with, it keeps its identity, leaving the other matrix unchanged, just like multiplying a number by one.

The identity matrix serves as the multiplicative identity in the matrix world, meaning that any matrix multiplied by \(I\), or vice versa, will result in the original matrix. This property illustrates why finding the inverse of a matrix is so important; the product of a matrix and its inverse gives us this neutrally wonderful identity matrix.
Determinant of a Matrix
The determinant of a matrix, often shown as \(det(A)\) or \(|A|\), is akin to a magical number that holds tremendous power. In a 2x2 matrix, the determinant is calculated by crossing multiplying the diagonal elements and then subtracting one product from the other.

The determinant can reveal whether a matrix is invertible; a non-zero determinant indicates that the matrix has an inverse. It also has a role in determining the volume scaling factor for the linear transformation that the matrix represents. When working with inverses, the determinant appears in the denominator, further emphasizing the necessity of it not being zero to avoid diving into the realms of the undefined.

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Most popular questions from this chapter

If \(\left[\begin{array}{ll}4^{\mathrm{b}} & 288 \\ 72 & 18^{\mathrm{a}}\end{array}\right]\) is a singular matrix and \(2 \mathrm{~b}=\mathrm{a}+\frac{1}{\mathrm{c}}\), then \(\mathrm{c}\) is _______. (1) 4 (2) \(\frac{1}{4}\) (3) \(\frac{1}{6}\) (4) 6

\(M=\left(\begin{array}{ll}p & q \\ r & s\end{array}\right)\) is a singular matrix. Its determinant is equal to its trace, then \(p=\) ____. \((1)-\mathrm{q}\) (2) \(\mathrm{r}\) (3) 0 (4) -s

If \(A=\left[\begin{array}{cc}3 & 4 \\ -1 & 2\end{array}\right]\) and \(B=\left[\begin{array}{cc}2 & -3 \\ 4 & -5\end{array}\right]\), then find the determinant value of \(A B\). (1) 10 (2) 20 (3) 12 (4) 15

The following are the steps in finding the matrix \(\mathrm{B}\), if \(\mathrm{B}+\left(\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right)=\left(\begin{array}{ll}5 & 4 \\ 3 & 2\end{array}\right)\). Arrange them in sequential order. (a) \(\therefore\left(\begin{array}{ll}\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{s}\end{array}\right)+\left(\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right)=\left(\begin{array}{ll}5 & 4 \\ 3 & 2\end{array}\right)\) (b) Let \(\mathrm{B}=\left(\begin{array}{ll}\mathrm{p} & \mathrm{q} \\\ \mathrm{r} & \mathrm{s}\end{array}\right)\) (c) \(\left(\begin{array}{cc}\mathrm{p}+2 & \mathrm{q}+3 \\ \mathrm{r}+4 & \mathrm{~s}+5\end{array}\right)=\left(\begin{array}{cc}5 & 4 \\ 3 & 2\end{array}\right)\) (d) \(\mathrm{p}+2=5, \mathrm{q}+3=4, \mathrm{r}+4=3, \mathrm{~s}+5=2\) \(\quad \Rightarrow \mathrm{p}=3, \mathrm{q}=1, \mathrm{r}=-1, \mathrm{~s}=-3\) (e) \(\therefore \mathrm{B}=\left(\begin{array}{cc}3 & 1 \\ -1 & -3\end{array}\right)\) (1) bacde (2) badce (3) bdcae (4) badec

If \(A=a\left[\begin{array}{cc}-1 & 0 \\ -1 & 2\end{array}\right]\) nd \(B=\left[\begin{array}{cc}3 & 2 \\ -1 & 0\end{array}\right]\), then find \(B^{-1} \cdot A^{-1}\) (1) \(\left[\begin{array}{cc}\frac{-1}{2} & \frac{3}{4} \\ \frac{-1}{2} & \frac{-5}{4}\end{array}\right]\) (2) \(\left[\begin{array}{cc}\frac{-1}{2} & \frac{1}{2} \\ \frac{-5}{4} & \frac{3}{4}\end{array}\right]\) (3) \(\left[\begin{array}{ll}\frac{1}{2} & \frac{-1}{2} \\ \frac{-5}{4} & \frac{3}{4}\end{array}\right]\) (4) \(\left[\begin{array}{ll}\frac{1}{2} & \frac{-3}{4} \\ \frac{-1}{2} & \frac{-5}{4}\end{array}\right]\)

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