#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 8

Hence proved $(A-2I)(A-3I)=0$

Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given: $A =\begin{bmatrix} 4 &2 \\ -1& 1 \end{bmatrix}$

Prove: $(A-2I)(A-3I)=0$

Consider,

$A-2I$

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

\begin{aligned} &A-2 I=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]-\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right] \\ &=\left[\begin{array}{cc} 4-2 & 2-0 \\ -1-0 & 1-2 \end{array}\right] \end{aligned}

$=\begin{bmatrix} 2 &2 \\ -1 &-1 \end{bmatrix}$                                               ...(i)

Now consider,

\begin{aligned} &A-3 I=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]-\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] \\ &=\left[\begin{array}{cc} 4-3 & 2-0 \\ -1-0 & 1-3 \end{array}\right] \end{aligned}

$=\begin{bmatrix} 1 &2 \\ -1 &-2 \end{bmatrix}$                                              ...(ii)

Now multiply equation (i) & (ii)

\begin{aligned} &{\left[\begin{array}{cc} 2 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 2 \times 1+2 \times(-1) & 2 \times 2+2 \times(-2) \\ -1 \times 1+(-1) \times(-1) & -1 \times 2+(-1) \times(-2) \end{array}\right]} \\ &=\left[\begin{array}{cc} 2-2 & 4-4 \\ -1+1 & -2+2 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}

Hence proved$(A-2I)(A-3I)=0$

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