If \(A = \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right],\) then the expression A 3- 2A 2is
a null matrix
Solving Matrix Expression \(A^3 - 2A^2\)
We are given a matrix \(A = \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right]\) and asked to evaluate the expression \(A^3 - 2A^2\).
Step 1: Calculate \(A^2\)
To find \(A^2\), we multiply matrix \(A\) by itself:
\(A^2 = A \times A = \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right] \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right]\)
Performing the matrix multiplication:
| Element | Calculation | Result |
|---|---|---|
| (1,1) | \((1)(1) + (-1)(-1)\) | \(1 + 1 = 2\) |
| (1,2) | \((1)(-1) + (-1)(1)\) | \(-1 - 1 = -2\) |
| (2,1) | \((-1)(1) + (1)(-1)\) | \(-1 - 1 = -2\) |
| (2,2) | \((-1)(-1) + (1)(1)\) | \(1 + 1 = 2\) |
So, \(A^2 = \left[ {\begin{array}{c} 2&{ - 2}\\ { - 2}&2 \end{array}} \right]\). Notice that \(A^2 = 2 \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right] = 2A\).
Step 2: Calculate \(A^3\)
To find \(A^3\), we multiply \(A^2\) by \(A\):
\(A^3 = A^2 \times A = \left[ {\begin{array}{c} 2&{ - 2}\\ { - 2}&2 \end{array}} \right] \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right]\)
Performing the matrix multiplication:
| Element | Calculation | Result |
|---|---|---|
| (1,1) | \((2)(1) + (-2)(-1)\) | \(2 + 2 = 4\) |
| (1,2) | \((2)(-1) + (-2)(1)\) | \(-2 - 2 = -4\) |
| (2,1) | \((-2)(1) + (2)(-1)\) | \(-2 - 2 = -4\) |
| (2,2) | \((-2)(-1) + (2)(1)\) | \(2 + 2 = 4\) |
So, \(A^3 = \left[ {\begin{array}{c} 4&{ - 4}\\ { - 4}&4 \end{array}} \right]\). Alternatively, since \(A^2 = 2A\), we have \(A^3 = A^2 \times A = (2A) \times A = 2(A \times A) = 2A^2\). Substituting \(A^2 = 2A\) again, we get \(A^3 = 2(2A) = 4A\). \(4A = 4 \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right] = \left[ {\begin{array}{c} 4&{ - 4}\\ { - 4}&4 \end{array}} \right]\), which matches our direct calculation.
Step 3: Calculate \(2A^2\)
We found \(A^2 = \left[ {\begin{array}{c} 2&{ - 2}\\ { - 2}&2 \end{array}} \right]\). Now we multiply \(A^2\) by the scalar 2:
\(2A^2 = 2 \left[ {\begin{array}{c} 2&{ - 2}\\ { - 2}&2 \end{array}} \right] = \left[ {\begin{array}{c} 2 \times 2&{ 2 \times (-2)}\\ {2 \times (-2)}&{2 \times 2} \end{array}} \right] = \left[ {\begin{array}{c} 4&{ - 4}\\ { - 4}&4 \end{array}} \right]\).
Step 4: Evaluate \(A^3 - 2A^2\)
Now we subtract \(2A^2\) from \(A^3\):
\(A^3 - 2A^2 = \left[ {\begin{array}{c} 4&{ - 4}\\ { - 4}&4 \end{array}} \right] - \left[ {\begin{array}{c} 4&{ - 4}\\ { - 4}&4 \end{array}} \right]\)
Subtracting the corresponding elements:
\(A^3 - 2A^2 = \left[ {\begin{array}{c} 4-4&{ - 4-(-4)}\\ { - 4-(-4)}&{4-4} \end{array}} \right] = \left[ {\begin{array}{c} 0&{ 0}\\ { 0}&{0} \end{array}} \right]\).
Conclusion
The result of the expression \(A^3 - 2A^2\) is the matrix \(\left[ {\begin{array}{c} 0&{ 0}\\ { 0}&{0} \end{array}} \right]\). This matrix is known as the null matrix (or zero matrix).
Comparing our result to the given options:
- a null matrix - Matches our result.
- an identify matrix - The identity matrix for 2x2 is \(\left[ {\begin{array}{c} 1&0\\ 0&1 \end{array}} \right]\), which is not our result.
- equal to A - Our result \(\left[ {\begin{array}{c} 0&{ 0}\\ { 0}&{0} \end{array}} \right]\) is not equal to \(A = \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right]\).
- equal to -A - \(-A = -\left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right] = \left[ {\begin{array}{c} -1&{ 1}\\ { 1}&{-1} \end{array}} \right]\), which is not our result.
The expression \(A^3 - 2A^2\) is a null matrix.
Matrix Calculation Revision Table
| Concept | Description | Example |
|---|---|---|
| Matrix Multiplication | Multiplying two matrices \(M_1 (m \times n)\) and \(M_2 (n \times p)\) results in \(M_3 (m \times p)\). The element at position \((i, j)\) in \(M_3\) is the sum of the products of corresponding elements from row \(i\) of \(M_1\) and column \(j\) of \(M_2\). | \(\left[ {\begin{array}{c} a&b\\ c&d \end{array}} \right]\left[ {\begin{array}{c} e&f\\ g&h \end{array}} \right] = \left[ {\begin{array}{c} ae+bg&af+bh\\ ce+dg&cf+dh \end{array}} \right]\) |
| Matrix Power | For a square matrix \(A\), \(A^n\) is the result of multiplying \(A\) by itself \(n\) times (\(A \times A \times ... \times A\), \(n\) times). \(A^1 = A\). | \(A^2 = A \times A\) |
| Scalar Multiplication | Multiplying a matrix by a scalar \(k\) involves multiplying every element of the matrix by \(k\). | \(k \left[ {\begin{array}{c} a&b\\ c&d \end{array}} \right] = \left[ {\begin{array}{c} ka&kb\\ kc&kd \end{array}} \right]\) |
| Matrix Subtraction | Subtracting matrix \(B\) from matrix \(A\) (if they have the same dimensions) involves subtracting corresponding elements. | \(\left[ {\begin{array}{c} a&b\\ c&d \end{array}} \right] - \left[ {\begin{array}{c} e&f\\ g&h \end{array}} \right] = \left[ {\begin{array}{c} a-e&b-f\\ c-g&d-h \end{array}} \right]\) |
| Null Matrix | A matrix where all elements are zero. Denoted by \(\mathbf{0}\). The size of the null matrix is usually implied by the context. | \(\left[ {\begin{array}{c} 0&0\\ 0&0 \end{array}} \right]\) |
| Identity Matrix | A square matrix with ones on the main diagonal and zeros elsewhere. Denoted by \(I\) or \(I_n\) for size \(n \times n\). \(AI = IA = A\). | \(I_2 = \left[ {\begin{array}{c} 1&0\\ 0&1 \end{array}} \right]\) |
Additional Matrix Information
In this problem, we observed that \(A^2 = 2A\). This type of relationship is interesting for a matrix. It means the matrix \(A\) satisfies the polynomial equation \(x^2 - 2x = 0\). By the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. For matrix \(A = \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right]\), the characteristic equation is \(\det(A - \lambda I) = 0\).
\(\det\left(\left[ {\begin{array}{c} 1-\lambda&{ - 1}\\ { - 1}&1-\lambda \end{array}} \right]\right) = (1-\lambda)(1-\lambda) - (-1)(-1) = (1-\lambda)^2 - 1 = (1 - 2\lambda + \lambda^2) - 1 = \lambda^2 - 2\lambda\).
Setting the determinant to zero gives \(\lambda^2 - 2\lambda = 0\), which is the characteristic equation. By Cayley-Hamilton, substituting \(A\) for \(\lambda\) gives \(A^2 - 2A = \mathbf{0}\). Rearranging, we get \(A^2 = 2A\), which matches our earlier observation.
Using this property, the expression \(A^3 - 2A^2\) can be evaluated more simply:
\(A^3 - 2A^2 = A \times A^2 - 2A^2\)
Since \(A^2 = 2A\), substitute this into the expression:
\(A^3 - 2A^2 = A \times (2A) - 2(2A)\)
\(A^3 - 2A^2 = 2(A \times A) - 4A\)
\(A^3 - 2A^2 = 2A^2 - 4A\)
Substitute \(A^2 = 2A\) again:
\(A^3 - 2A^2 = 2(2A) - 4A\)
\(A^3 - 2A^2 = 4A - 4A\)
\(A^3 - 2A^2 = \mathbf{0}\) (the null matrix).
This confirms our result and shows how matrix properties derived from the characteristic equation can simplify calculations.
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