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Magazine Chapter 5: Problem 36
Given the matrix $$ \mathbf{A}=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 1 & 0 & 1 \\ 2 & 2 & 2 \end{array}\right] $$ determine \(|\boldsymbol{A}|,\left|\boldsymbol{A} \boldsymbol{A}^{\mathrm{T}}\right|,\left|\boldsymbol{A}^{2}\right|\) and \(|\boldsymbol{A}+\boldsymbol{A}|\)
Short Answer
Expert verified
\(|\boldsymbol{A}| = 0\), \(|\boldsymbol{A}\boldsymbol{A}^{\mathrm{T}}| = 16\), \(|\boldsymbol{A}^{2}| = 20\), \(|\boldsymbol{A} + \boldsymbol{A}| = 0\).
Step by step solution
01
Calculate the Determinant of A
First, we need to calculate \(|\boldsymbol{A}|\). The matrix \(\boldsymbol{A}\) is a \(3 \times 3\) matrix. We use the rule of Sarrus to find the determinant for a \(3 \times 3\) matrix:\[ \left| \mathbf{A} \right| = \begin{vmatrix} 1 & 0 & -1 \ 1 & 0 & 1 \ 2 & 2 & 2 \end{vmatrix} \]Expand along the first row:\(1(0 \times 2 - 1 \times 2) - 0(1 \times 2 - 2 \times 1) - 1(1 \times 2 - 2 \times 0)\)which simplifies to \(1(0 - 2) - 0 + 1(2 - 0)\).Finally, this equals \(-2 + 2 = 0\).Thus, \(|\boldsymbol{A}| = 0\).
02
Calculate A Transpose
Next, we need to find \(\boldsymbol{A}^{\mathrm{T}}\), the transpose of \(\boldsymbol{A}\). Transposing a matrix involves flipping the matrix over its diagonal:\[ \boldsymbol{A}^{\mathrm{T}} = \begin{pmatrix} 1 & 1 & 2 \ 0 & 0 & 2 \ -1 & 1 & 2 \end{pmatrix} \]
03
Calculate AAT and its Determinant
Compute \(\boldsymbol{A}\boldsymbol{A}^{\mathrm{T}}\):\[ \begin{pmatrix} 1 & 0 & -1 \ 1 & 0 & 1 \ 2 & 2 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1 & 2 \ 0 & 0 & 2 \ -1 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \ 0 & 2 & 4 \ 0 & 4 & 12 \end{pmatrix} \]Now, compute the determinant of the resulting matrix:\[ \left| \begin{pmatrix} 2 & 0 & 0 \ 0 & 2 & 4 \ 0 & 4 & 12 \end{pmatrix} \right| = \left(2(2 \times 12 - 4 \times 4)\right) = 2(24 - 16) = 2 \times 8 = 16 \].Hence, \(|\boldsymbol{A}\boldsymbol{A}^{\mathrm{T}}| = 16\).
04
Calculate A Squared
Compute \(\boldsymbol{A}^2\) by multiplying \(\boldsymbol{A}\) by itself:\[ \begin{pmatrix} 1 & 0 & -1 \ 1 & 0 & 1 \ 2 & 2 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 & -1 \ 1 & 0 & 1 \ 2 & 2 & 2 \end{pmatrix} = \begin{pmatrix} -1 & -2 & -2 \ 3 & 2 & 0 \ 6 & 4 & 2 \end{pmatrix} \]The determinant of \(\boldsymbol{A}^2\):Expand using the first row: \( -1(2 \times 2 - 4 \times 0) - (-2)(3 \times 2 - 0 \times 6) + (-2)(3 \times 4 - 2 \times 6) \).This simplifies to \(-1(4) + 2(6) - 2(-6) = -4 + 12 + 12 = 20\).Thus, \(|\boldsymbol{A}^2| = 20\).
05
Calculate 2A and its Determinant
Compute \(\boldsymbol{A} + \boldsymbol{A} = 2 \boldsymbol{A}\):\[ 2 \times \begin{pmatrix} 1 & 0 & -1 \ 1 & 0 & 1 \ 2 & 2 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 0 & -2 \ 2 & 0 & 2 \ 4 & 4 & 4 \end{pmatrix} \]Now calculate the determinant of this scaled matrix (which is twice every element in \(\boldsymbol{A}\)):Notice that this matrix can be factored further simplifying to a situation similar to step 1:Thus, \(|2\boldsymbol{A}| = 2^3 \cdot |\boldsymbol{A}| = 8 \cdot 0 = 0\).Hence, \(|\boldsymbol{A} + \boldsymbol{A}| = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Multiplication
Matrix multiplication involves a systematic method to find the product of two matrices. This operation is not as straightforward as simple element-wise multiplication. To multiply two matrices, take each row of the first matrix and perform a dot product with each column of the second matrix. Crystallized in the formula:
- If you have matrices \( A \) with dimensions \( m \times n \) and \( B \) with dimensions \( n \times p \), the resulting matrix \( AB \) will have dimensions \( m \times p \).
- The element at the \((i, j)\) position of matrix \( AB \) is computed by: \[(AB)_{ij} = \sum_{k=1}^{n} A_{ik} B_{kj}\]This requires the columns of the first matrix to match the rows of the second.
Transpose of a Matrix
The transpose of a matrix is obtained by flipping the matrix over its main diagonal. This turns the rows of the original matrix into the columns of a new matrix. If you have a matrix \( \mathbf{A} \), its transpose, denoted by \( \mathbf{A}^{ ext{T}} \), is defined as follows:
- The element at row \( i \), column \( j \) in \( \mathbf{A} \) swaps to row \( j \), column \( i \) in \( \mathbf{A}^{ ext{T}} \).
- For a \( 3 \times 3 \) matrix:\[\mathbf{A} = \begin{pmatrix}a & b & c \d & e & f \g & h & i\end{pmatrix}\Rightarrow \mathbf{A}^{ ext{T}} = \begin{pmatrix}a & d & g \b & e & h \c & f & i\end{pmatrix}\]
Rule of Sarrus
The Rule of Sarrus is a handy shortcut to find the determinant of a \( 3 \times 3 \) matrix. This rule simplifies the process and is much faster than the cofactor expansion method. Here's how it works for a matrix \( \mathbf{A} \):
- Write down the first two columns of \( \mathbf{A} \) again beside the matrix.
- Multiply diagonally downwards and sum:\[(aei + bfg + cdh)\]
- Subtract the sum of diagonals going upwards:\[(ceg + afh + bdi)\]
- The determinant \( \left| \mathbf{A} \right| \) is simply:\[(aei + bfg + cdh - ceg - afh - bdi)\]
Inverse of a Matrix
The inverse of a matrix \( \mathbf{A} \) is labeled as \( \mathbf{A}^{-1} \) and is defined as the matrix that, when multiplied by \( \mathbf{A} \), results in the identity matrix. This concept is similar to finding the reciprocal of a number. Some key points about matrix inverses are:
- Only square matrices (matrices with the same number of rows and columns) can have inverses.
- The determinant \( |A| \) must be non-zero; otherwise, the matrix does not have an inverse.
- For a \( 2 \times 2 \) matrix \( \mathbf{A} = \begin{pmatrix}a & b \c & d\end{pmatrix}\), the inverse, if it exists, is given by:\[(\mathbf{A}^{-1} = \frac{1}{ad - bc} \begin{pmatrix}d & -b \-c & a\end{pmatrix})\]
