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Q5:
If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 – 5A + 7I = 0$.
Solution :
Step 1: Calculating $A^2$
Given the matrix $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, the square of the matrix $A^2$ is computed by the product $A \cdot A$.
$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$
By the definition of matrix multiplication [Sum of row-column products]:
$A^2 = \begin{bmatrix} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \\ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{bmatrix}$
$A^2 = \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$
Step 2: Calculating Scalar Multiples
We evaluate the term $-5A$ using the scalar multiplication property $k \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} ka & kb \\ kc & kd \end{bmatrix}$:
$-5A = -5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -15 & -5 \\ 5 & -10 \end{bmatrix}$
We define the identity matrix $I$ for a $2 \times 2$ order as $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. Thus, the term $7I$ is:
$7I = 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$
Step 3: Summing the Components
We now evaluate the expression $LHS = A^2 - 5A + 7I$ by adding the matrices computed in the previous steps:
$LHS = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} + \begin{bmatrix} -15 & -5 \\ 5 & -10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$
Summing the corresponding elements [By Matrix Addition Property]:
| Element Position | Summation | Result |
|---|---|---|
| (1, 1) | $8 - 15 + 7$ | $0$ |
| (1, 2) | $5 - 5 + 0$ | $0$ |
| (2, 1) | $-5 + 5 + 0$ | $0$ |
| (2, 2) | $3 - 10 + 7$ | $0$ |
Step 4: Conclusion of Proof
The resulting matrix is:
$LHS = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$
Since the result is the Zero Matrix $O$, the identity $A^2 - 5A + 7I = 0$ is satisfied.
Final Solution: $A^2 - 5A + 7I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$. Hence, the statement is proved.
More Questions from Class 12 Mathematics Matrices Miscellaneous Exercise on Chapter 3
- Q1: If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
- Q2: Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
- Q3: Find the values of $x, y, z$ if the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfy the equation $A′A = I$.
- Q4: For what values of $x$ : $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O$?
- Q6: Find $x$, if $\begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = O$
- Q7(a): A manufacturer produces three products $x, y, z$ which he sells in two markets. Annual sales are indicated below: (a) If unit sale prices of $x, y$ and $z$ are ` 2.50, ` 1.50 and ` 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
- Q7(b): A manufacturer produces three products $x, y, z$ which he sells in two markets. Annual sales are indicated below: (b) If the unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively. Find the gross profit.
- Q8: Find the matrix $X$ so that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$
- Q9: If $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that $A^2 = I$, then
- Q10: If the matrix A is both symmetric and skew symmetric, then
- Q11: If A is square matrix such that $A^2 = A$, then $(I + A)^3 – 7 A$ is equal to
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