Calculating Matrix Expression with Identity Matrix

The problem asks us to evaluate the expression \(23A^3 - 19A^2 - 4A\) given the matrix \(A\). First, let's look at the matrix \(A\) provided:

\(A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)

This matrix is a square matrix of order 3. We recognize this specific matrix as the identity matrix of order 3, which is commonly denoted as \(I_3\).

The identity matrix \(I\) is a special matrix such that when multiplied by any matrix \(B\) of compatible dimensions, the result is \(B\) itself (\(IB = B\) and \(BI = B\)). A key property of the identity matrix is that any positive integer power of the identity matrix is the identity matrix itself. That is, \(I^n = I\) for any positive integer \(n\).

Powers of the Identity Matrix

Given that \(A = I_3\), we can find the powers of \(A\):

• \(A^1 = A = I_3\)
• \(A^2 = A \times A = I_3 \times I_3 = I_3\) (Using the property \(I \times I = I\))
• \(A^3 = A^2 \times A = I_3 \times I_3 = I_3\) (Again, using the property \(I \times I = I\))

In general, for any positive integer \(n\), \(A^n = (I_3)^n = I_3\).

Evaluating the Matrix Expression

Now, let's substitute these powers of \(A\) back into the given expression \(23A^3 - 19A^2 - 4A\):

\(23A^3 - 19A^2 - 4A = 23(I_3) - 19(I_3) - 4(I_3)\)

We can treat the identity matrix \(I_3\) as a common factor here, similar to how we handle variables in algebraic expressions:

\((23 - 19 - 4)I_3\)

Now, perform the scalar arithmetic inside the parentheses:

\(23 - 19 = 4\)

\(4 - 4 = 0\)

So the expression simplifies to:

\(0 \times I_3\)

Multiplying any matrix by the scalar zero results in the null matrix (or zero matrix) of the same order. The null matrix of order 3 is a 3x3 matrix where all elements are zero.

\(0 \times I_3 = 0 \times \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{lll}0 \times 1 & 0 \times 0 & 0 \times 0 \\ 0 \times 0 & 0 \times 1 & 0 \times 0 \\ 0 \times 0 & 0 \times 0 & 0 \times 1\end{array}\right] = \left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\)

This is the null matrix of order 3.

Comparing with Options

Let's compare our result with the given options:

• Option 1: Null matrix of order 3. This matches our calculated result.
• Option 2: Identity matrix of order 3. This is \(I_3\), which is not our result.
• Option 3: \(\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]\). This matrix is \(2I_3\), not our result.
• Option 4: \(\left[\begin{array}{lll}7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7\end{array}\right]\). This matrix is \(7I_3\), not our result.

Therefore, the expression \(23A^3 - 19A^2 - 4A\) is equal to the null matrix of order 3.

Revision Table: Key Matrix Concepts

Additional Information on Identity Matrices

The identity matrix plays a crucial role in matrix algebra, similar to the number 1 in regular arithmetic. Some important properties of the identity matrix \(I\) (of appropriate size) include:

• For any matrix \(B\), \(BI = B\) and \(IB = B\).
• The identity matrix is always square.
• The diagonal elements are 1, and off-diagonal elements are 0.
• The determinant of an identity matrix is always 1.
• The inverse of an identity matrix is the identity matrix itself: \(I^{-1} = I\).
• The identity matrix is a diagonal matrix, a symmetric matrix, and an idempotent matrix (since \(I^2 = I\)).

Understanding these properties simplifies many matrix calculations, especially those involving powers of the identity matrix, as seen in this problem.