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Either compute the inverse of the given matrix, or else show that it is singular. \(\left(\begin{array}{rrr}{1} & {2} & {1} \\ {-2} & {1} & {8} \\ {1} & {-2} & {-7}\end{array}\right)\)

Short Answer

Expert verified
Question: Determine if the given matrix is singular or not. If not, find its inverse: Matrix: \begin{equation} \left(\begin{array}{ccc}{1} & {2} & {1} \\\ {-2} & {1} & {8} \\\ {1} & {-2} & {-7}\end{array}\right) \end{equation} Answer: The given matrix is not singular since its determinant is 19. Its inverse is: Inverse: \begin{equation} \left(\begin{array}{ccc}{-\frac{23}{19}} & {\frac{30}{19}} & {\frac{2}{19}} \\\ {-\frac{22}{19}} & {\frac{8}{19}} & {\frac{22}{19}} \\\ {0} & {-\frac{4}{19}} & {-\frac{3}{19}}\end{array}\right) \end{equation}

Step by step solution

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01

Determine if the matrix is singular or not

Compute the determinant of the given matrix: \begin{equation} \text{det}\left(\begin{array}{rrr}{1} & {2} & {1} \\\ {-2} & {1} & {8} \\\ {1} & {-2} & {-7}\end{array}\right)=1\cdot(1\cdot(-7)-2\cdot8)+2\cdot(-2\cdot(-7)-1\cdot8)+1\cdot(-2\cdot1-1\cdot(-2)) \end{equation} Calculate the result: \begin{equation} =-7-16+28+16-4+2=19 \end{equation} The determinant is not zero, so the matrix is not singular and we can proceed to find its inverse.
02

Find minors and cofactors of the given matrix

Find the minor matrix, by computing the determinant of the 2x2 submatrices: \begin{equation} \text{Minors}\left(\begin{array}{ccc}{1} & {2} & {1} \\\ {-2} & {1} & {8} \\\ {1} & {-2} & {-7}\end{array}\right)=\left(\begin{array}{rrr}{(-7-16)} & {(8+14)} & {(2-2)} \\\ {(-14-16)} & {(-7-1)} & {(-2-2)} \\\ {(-2+4)} & {(14+8)} & {(1-4)}\end{array}\right) \end{equation} The cofactor matrix is the same matrix with elements at the odd positions multiplied by -1: \begin{equation} \text{Cofactors}\left(\begin{array}{ccc}{1} & {2} & {1} \\\ {-2} & {1} & {8} \\\ {1} & {-2} & {-7}\end{array}\right)=\left(\begin{array}{rrr}{-23} & -22 & 0 \\\ {30} & 8 & -4 \\\ 2 & -22 & -3\end{array}\right) \end{equation}
03

Construct the adjoint matrix

Find the transpose of the cofactor matrix to obtain the adjoint matrix: \begin{equation} \text{adjoint}\left(\begin{array}{rrr}{1} & {2} & {1} \\\ {-2} & {1} & {8} \\\ {1}& {-2} & {-7}\end{array}\right)=\left(\begin{array}{ccc}{-23} & {30} & {2} \\\ {-22} & {8} & {22} \\\ {0} & {-4} & {-3}\end{array}\right) \end{equation}
04

Calculate the inverse matrix

To find the inverse of the given matrix, divide the elements of the adjoint matrix by the determinant: \begin{equation} \left(\begin{array}{ccc}{1} & {2} & {1} \\\ {-2} & {1} & {8} \\\ {1} &{-2} &{-7}\end{array}\right)^{-1}=\frac{1}{19}\left(\begin{array}{ccc}{-23} & {30} & {2} \\\ {-22} & {8} & {22} \\\ {0} & {-4} & {-3}\end{array}\right) \end{equation} So the inverse of the given matrix is: \begin{equation} \left(\begin{array}{ccc}{-\frac{23}{19}} & {\frac{30}{19}} & {\frac{2}{19}} \\\ {-\frac{22}{19}} & {\frac{8}{19}} & {\frac{22}{19}} \\\ {0} & {-\frac{4}{19}} & {-\frac{3}{19}}\end{array}\right) \end{equation}

Key Concepts

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

Determinant of a Matrix
The determinant of a matrix is a scalar value that is a fundamental property used in matrix theory. It tells us whether a square matrix is invertible or not, amongst other things. A singular matrix, one that does not have an inverse, will have a determinant of zero. To calculate the determinant of a 3x3 matrix, we use a method that involves cross-multiplication of the rows and columns, excluding the row and column of the current element. This method is essentially a sum of products of diagonals (positive and negative).

For example, for a matrix A, the determinant is calculated using the formula:
\[ \text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]
Where \(a_{ij}\) denotes the element in the i-th row and j-th column of matrix A. If this calculation yields any non-zero value, the matrix has an inverse; if it's zero, the matrix is singular and inverses cannot be found.
Cofactor Matrix
A cofactor matrix is an intermediate step in computing the inverse of a matrix. To find the cofactors, you must first calculate the minors of the matrix. Each minor is the determinant of a smaller matrix that you get by removing the i-th row and j-th column from the original matrix. The cofactor is obtained by applying a checkerboard pattern of plus and minus signs to these minors, known as the sign factor, given by \((-1)^{i+j}\) where \(i\) and \(j\) are row and column indices respectively.

The matrix of cofactors is simply the original matrix with each element replaced by its cofactor. This process reflects the position's influence on the determinant and entails coupling each minor with the appropriate sign.
Adjoint Matrix
Once we have obtained the cofactor matrix, we can find the adjoint of the original matrix. The adjoint, or adjugate, of a matrix is the transpose of its cofactor matrix. Transposing a matrix involves flipping it over its diagonal, which means writing the rows of the original matrix as columns.

The adjoint plays a key role in finding the inverse because the inverse is calculated by dividing the adjoint by the original matrix's determinant. The essence of using the adjoint in inversion is that it represents the unique combination of row operations that transform the original matrix into the identity matrix when multiplied by \(1/\text{det}(A)\).
Singular Matrix
A singular matrix is a square matrix that does not have an inverse. In terms of computation, it is identified by its determinant. If the determinant of a matrix is zero, then the matrix is singular. This indicates that there is no unique solution to a system of linear equations represented by this matrix.

Singular matrices cannot be used in division and therefore pose problems in various matrix operations, including solving systems of equations or finding eigenvalues. Recognizing a singular matrix before attempting inversion is crucial to avoid computational errors and to understand the limitations of a given set of linear equations.

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Most popular questions from this chapter

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-\frac{5}{4}} & {\frac{3}{4}} \\\ {\frac{3}{4}} & {-\frac{5}{4}}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{2 t} \\ {e^{t}}\end{array}\right) $$

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{1} & {1} \\ {4} & {-2}\end{array}\right) \mathbf{x} $$

Express the general solution of the given system of equations in terms of real-valued functions. In each of Problems 1 through 6 also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {-\frac{3}{2}} \\\ {\frac{9}{5}} & {-1}\end{array}\right) \mathbf{x} $$

The fundamental matrix \(\Phi(t)\) for the system ( 3) was found in Example \(2 .\) Show that \(\mathbf{\Phi}(t) \Phi(s)=\Phi(t+s)\) by multiplying \(\Phi(t)\) and \(\Phi(s) .\)

The clectric circuit shown in Figure 7.9 .1 is described by the system of differential equations $$ \frac{d \mathbf{x}}{d t}=\left(\begin{array}{cc}{-\frac{1}{2}} & {-\frac{1}{8}} \\ {2} & {-\frac{1}{2}}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{\frac{1}{2}} \\ {0}\end{array}\right) I(t) $$ where \(x_{1}\) is the current through the inductor, \(x_{2}\) is the voltage drop across the capacitor, and \(I(t)\) is the current supplied by the external source. (a) Determine a fundamental matrix \(\Psi(t)\) for the homogeneous system corresponding to Eq. (i). Refer to Problem 25 of Section \(7.6 .\) (b) If \(I(t)=e^{-t / 2}\), determine the solution of the system (i) that also satisfies the initial conditions \(\mathbf{x}(0)=0\).

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