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Q15:
Find $A^2 – 5A + 6I$, if $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$

Solution :

Step 1: Computing $A^2$

Given the matrix $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$, we define $A^2$ as the product $A \times A$:

$A^2 = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$

[By Matrix Multiplication Definition: $(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$]

  • $R_1 C_1: (2)(2) + (0)(2) + (1)(1) = 4 + 0 + 1 = 5$
  • $R_1 C_2: (2)(0) + (0)(1) + (1)(-1) = 0 + 0 - 1 = -1$
  • $R_1 C_3: (2)(1) + (0)(3) + (1)(0) = 2 + 0 + 0 = 2$
  • $R_2 C_1: (2)(2) + (1)(2) + (3)(1) = 4 + 2 + 3 = 9$
  • $R_2 C_2: (2)(0) + (1)(1) + (3)(-1) = 0 + 1 - 3 = -2$
  • $R_2 C_3: (2)(1) + (1)(3) + (3)(0) = 2 + 3 + 0 = 5$
  • $R_3 C_1: (1)(2) + (-1)(2) + (0)(1) = 2 - 2 + 0 = 0$
  • $R_3 C_2: (1)(0) + (-1)(1) + (0)(-1) = 0 - 1 + 0 = -1$
  • $R_3 C_3: (1)(1) + (-1)(3) + (0)(0) = 1 - 3 + 0 = -2$

$A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$

Step 2: Calculating Scalar Multiples $5A$ and $6I$

Per the property of Scalar Multiplication $k A = [k a_{ij}]$:

$5A = 5 \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}$

The identity matrix $I$ of order 3 is $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Thus:

$6I = 6 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

Step 3: Final Matrix Evaluation

Constructing the final expression $A^2 - 5A + 6I$:

$= \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

Performing element-wise addition/subtraction:

$5 - 10 + 6$$-1 - 0 + 0$$2 - 5 + 0$
$9 - 10 + 0$$-2 - 5 + 6$$5 - 15 + 0$
$0 - 5 + 0$$-1 - (-5) + 0$$-2 - 0 + 6$

Simplifying the elements:

= $\begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$

Final Solution: $\begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$


More Questions from Class 12 Mathematics Matrices EXERCISE 3.2


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