Download App

Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
If $A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right]$, verify that $A^3 - 6A^2 + 9A - 4I = 0$ and hence find $A^{-1}$
✓

Answer

Given: $A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&-1&2 \end{array}} \right]$$\therefore {A^2} = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&-1&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right]$
$\Rightarrow {A^2}=\left[ {\begin{array}{*{20}{c}} {4 + 1 + 1}&{ - 2 - 2 - 1}&{2 + 1 + 2} \\ { - 2 - 2 - 1}&{1 + 4 + 1}&{ - 1 - 2 - 2} \\ {2 + 1 + 2}&{ - 1 - 2 - 2}&{1 + 1 + 4} \end{array}} \right]$ $=\left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right]$
Now $A^3 = A^2.A = \left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {12 + 5 + 5}&{ - 6 - 10 - 5}&{6 + 5 + 10} \\ { - 10 - 6 - 5}&{5 + 12 + 5}&{ - 5 - 6 - 10} \\ {10 + 5 + 6}&{ - 5 - 10 - 6}&{5 + 5 + 12} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {22}&{ - 21}&{21} \\ { - 21}&{22}&{ - 21} \\ {21}&{ - 21}&{22} \end{array}} \right]$
L.H.S. $ = {A^3} - 6{A^2} + 9A - 4I$
$= \left[ {\begin{array}{*{20}{c}} {22}&{ - 21}&{21} \\ { - 21}&{22}&{ - 21} \\ {21}&{ - 21}&{22} \end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right] $ $+ 9\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right] - 4\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {22}&{ - 21}&{21} \\ { - 21}&{22}&{ - 21} \\ {21}&{ - 21}&{22} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {36}&{ - 30}&{30} \\ { - 30}&{36}&{ - 30} \\ {30}&{ - 30}&{36} \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} {18}&{ - 9}&9 \\ { - 9}&{18}&{ - 9} \\ 9&{ - 9}&{18} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&0&0 \\ 0&4&0 \\ 0&0&4 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {22 - 36}&{ - 21 + 30}&{21 - 30} \\ { - 21 + 30}&{22 - 36}&{ - 21 + 30} \\ {21 - 30}&{ - 21 + 30}&{22 - 36} \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} {18 - 4}&{ - 9 - 0}&{9 - 0} \\ { - 9 - 0}&{18 - 4}&{ - 9 - 0} \\ {9 - 0}&{ - 9 - 0}&{18 - 4} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} { - 14}&9&{ - 9} \\ 9&{ - 14}&9 \\ { - 9}&9&{14} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - 14}&9&{ - 9} \\ 9&{ - 14}&9 \\ { - 9}&9&{14} \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right] = 0$= R.H.S.
Now, to find ${A^{ - 1}}$, multiplying ${A^3} - 6{A^2} + 9A - 4{I^{ - 1}} = 0.{A^{ - 1}}$by ${A^{ - 1}}$
$\Rightarrow {A^3}{A^{ - 1}} - 6{A^2}{A^{ - 1}} + 9A{A^{ - 1}} - 4I.{A^{ - 1}} = 0{A^{ - 1}}$
$ \Rightarrow {A^2} - 6A + 9I - 4{A^{ - 1}} = 0$
$\Rightarrow 4{A}^{-1} = {A^2} - 6A + 9I$
$\Rightarrow 4{A}^{-1} = \left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right]$ $ - 6\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right] + 9\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$
$\Rightarrow 4{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right] $ $- \left[ {\begin{array}{*{20}{c}} {12}&{ - 6}&6 \\ { - 6}&{12}&{ - 6} \\ 6&{ - 6}&{12} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 9&0&0 \\ 0&9&0 \\ 0&0&9 \end{array}} \right]$
$\Rightarrow 4{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {6 - 12 + 9}&{ - 5 + 6 + 0}&{5 - 6 + 0} \\ { - 5 + 6 + 0}&{6 - 12 + 9}&{ - 5 + 6 + 0} \\ {5 - 6 + 0}&{ - 5 + 6 + 0}&{6 - 12 + 9} \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} 3&1&{ - 1} \\ 1&3&1 \\ { - 1}&1&3 \end{array}} \right]$
$ \Rightarrow {A^{ - 1}} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 3&1&{ - 1} \\ 1&3&1 \\ { - 1}&1&3 \end{array}} \right]$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Determinants→Determinants — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Differentiate the following functions with respect to x:
$10^{\log\sin\text{x}}$
→
Find the equation of the normals to the curve $y = x^3 + 2x + 6$ which are parallel to the line $x + 14y + 4 = 0.$
→
Show that the function $\text{f(x)}=\begin{cases}|2\text{x}-3||\text{x}|, & \text{x}\geq1\\\sin\Big(\frac{\pi\text{x}}{2}\Big),& \text{x}>1\end{cases}$ is continuous but not differentiable at x = 1.
→
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos\text{x}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
→
Find the real values of $\lambda$ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions:
$2\lambda\text{x}-2\text{y}+3\text{z}=0,$
$\text{x}+\lambda\text{y}+2\text{z}=0,$
$2\text{x}+0\text{y}+\lambda\text{z}=0$
→
A die is thrown three times, find the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on first two tosses.
→
If $f(x) = x^2 - 2x$, find f(A), where $\text{A}=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
→
Find the area of the region bounded by the curves $y = x^2 + 2, y = x, x = 0$ and $x = 3.$
→
$\text{If y} = \sqrt{\text{x + 1}} - \sqrt{\text{x - 1}}, \ \text{Prove that}\ (\text{x}^{2} - 1) \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + \text{x} \frac{\text{dy}}{\text{dx}} - \frac{1}{4} \text{y}= 0.$
→
A bag contains $4$ balls. Two balls are drawn at random, and are found to be white. What is the probability that all balls are white?
→